Znzryb's Blog
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Posted on Edited on

AC代码与思路

相比于最后一次TLE背包提交，加了以下这些

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for(int j=1;j<=x-sumYen;j++) {  // 背包容量为钱，背包价值为生产力
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if(sumYen>x)
     return false;

主要思路是二分答案加背包dp

背包容量为钱，背包价值为能加工多少产品

通过背包dp让加工产品最大化

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// https://atcoder.jp/contests/abc374/tasks/abc374_e
#include <iostream>
#include <cmath>
typedef long long ll;
using namespace std;
const ll MAXN=105,MAXANS=static_cast<ll>(1e9)+7,MAXA=1e7+7;
int n,x,mach[MAXN][6];
int dp[MAXA];
inline bool check(ll needCap) {
    ll sumYen=0;
    for(int i=1;i<=n;i++) {
        dp[0]=0;
        int cap1=mach[i][1],p1=mach[i][2],cap2=mach[i][3],p2=mach[i][4];
        bool isBreak=false;
        if(sumYen>x)
            return false;
        for(int j=1;j<=x-sumYen;j++) {  // 背包容量为钱，背包价值为能加工多少产品
            int up1=j-p1,up2=j-p2;
            if(up1>=0)
                dp[j]=dp[up1]+cap1;
            if(up2>=0)
                dp[j]=max(dp[j],dp[up2]+cap2);
            if(dp[j]>=needCap) {
                sumYen+=j;isBreak=true;break;
            }
        }
        if(!isBreak)
            return false;
    }
    if(sumYen<=x)
        return true;
    return false;
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    cin>>n>>x;
    for(int i=1;i<=n;i++) {
            // cap1      price1 per U
        cin>>mach[i][1]>>mach[i][2]>>mach[i][3]>>mach[i][4];
    }
    ll l=0,r=MAXANS;
    while(l<r) {
        ll mid=l+r+1>>1;
        if(check(mid)) l=mid;
        else r=mid-1;
    }
    cout<<l<<endl;
    return 0;
}
// AC 64 个点 https://atcoder.jp/contests/abc374/submissions/58537183 已经比较接近了
// 就TLE了3个点 https://atcoder.jp/contests/abc374/submissions/58545771
// AC https://atcoder.jp/contests/abc374/submissions/58545889
// AC https://www.luogu.com.cn/record/180827335

以下是错误提交记录与心路历程

这个程序实际上没啥太大问题，但是即便开long long minYen还是超范围溢出了。

这个题目再次证明了，二分的题目r不能设太大，一个是复杂度，还有一个是溢出。

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// https://atcoder.jp/contests/abc374/tasks/abc374_e
#include <iostream>
typedef long long ll;
using namespace std;
const ll MAXN=105,MAXANS=static_cast<ll>(1e13)+7;
ll n,x,mach[MAXN][6];
inline bool check(ll needCap) {
    ll sumYen=0;
    for(int i=1;i<=n;i++) {
        if(sumYen>x)
            return false;
        ll cap1=mach[i][1],p1=mach[i][2],cap2=mach[i][3],p2=mach[i][4];
        ll minYen=0;ll needMach1=(needCap-1)/cap1+1,needMach2=(needCap-1)/cap2+1; // (a-1)/b +1
        minYen=needMach1*p1;          // 向上取整的trick
        minYen=min(minYen,needMach2*p2); 
        sumYen+=minYen;
    }
    if(sumYen<=x)
        return true;
    return false;
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    cin>>n>>x;
    for(int i=1;i<=n;i++) {
            // cap1      price1 per U
        cin>>mach[i][1]>>mach[i][2]>>mach[i][3]>>mach[i][4];
    }
    ll l=0,r=MAXANS;
    while(l<r) {
        ll mid=l+r+1>>1;
        if(check(mid)) l=mid;
        else r=mid-1;
    }
    cout<<l<<endl;
    return 0;
}
// AC 34 WA 64 https://atcoder.jp/contests/abc374/submissions/58535735

原来两台机子都能用！那check()函数写起来是比较麻烦

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Both machines S_i and T_i can be used for process i.
If you purchase both machines S_i and T_i, the number of items that can be processed in process i will be the sum of the items processed by both machines.

AC 64 个点 https://atcoder.jp/contests/abc374/submissions/58537475

加上了对于组合的判定

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// https://atcoder.jp/contests/abc374/tasks/abc374_e
#include <iostream>
typedef long long ll;
using namespace std;
const ll MAXN=105,MAXANS=static_cast<ll>(1e10)+7;
ll n,x,mach[MAXN][6];
inline bool check(ll needCap) {
    ll sumYen=0;
    for(int i=1;i<=n;i++) {
        if(sumYen>x)
            return false;
        ll cap1=mach[i][1],p1=mach[i][2],cap2=mach[i][3],p2=mach[i][4];
        ll minYen;ll needMach1=(needCap-1)/cap1+1,needMach2=(needCap-1)/cap2+1; // (a-1)/b +1
        minYen=min(needMach1*p1,needMach2*p2);
        ll firstMach1=needCap/cap1,remainMach2=(needCap-firstMach1*cap1-1)/cap2+1;
        ll priceComb1=firstMach1*p1+remainMach2*p2;
        ll firstMach2=needCap/cap2,remainMach1=(needCap-firstMach2*cap2-1)/cap1+1;
        ll priceComb2=firstMach2*p2+remainMach1*p1;;
        minYen=min(minYen,min(priceComb1,priceComb2));
        sumYen+=minYen;
    }
    if(sumYen<=x)
        return true;
    return false;
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    cin>>n>>x;
    for(int i=1;i<=n;i++) {
            // cap1      price1 per U
        cin>>mach[i][1]>>mach[i][2]>>mach[i][3]>>mach[i][4];
    }
    ll l=0,r=MAXANS;
    while(l<r) {
        ll mid=l+r+1>>1;
        if(check(mid)) l=mid;
        else r=mid-1;
    }
    cout<<l<<endl;
    return 0;
}
// AC 64 个点 https://atcoder.jp/contests/abc374/submissions/58537183 已经比较接近了

我怀疑是我的向上取整模版的问题

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#include <iostream>
#include <cmath>
using namespace std;
long long a=0;
int main()
{
    cout<<-1/782<<endl;
    std::cout << (0-1)/782+1 << std::endl;   // 模版行
    cout<<ceil(a/1.0/782)<<endl;
    return 0;
}
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0

毫无疑问，0/782=0，你就算加ceil()也应该是0，但显然我们的模版出了问题。

哎，用系统ceil（）提交了一遍，发现又WA了，和前面一模一样，只能说就这样吧，累了。

https://atcoder.jp/contests/abc374/submissions/58539387

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// https://atcoder.jp/contests/abc374/tasks/abc374_e
#include <iostream>
#include <cmath>
typedef long long ll;
using namespace std;
const ll MAXN=105,MAXANS=static_cast<ll>(1e10)+7;
ll n,x,mach[MAXN][6];
inline bool check(ll needCap) {
    ll sumYen=0;
    for(int i=1;i<=n;i++) {
        if(sumYen>x)
            return false;
        ll cap1=mach[i][1],p1=mach[i][2],cap2=mach[i][3],p2=mach[i][4];
        ll minYen;ll needMach1=ceil(needCap/1.0/cap1),needMach2=ceil(needCap/1.0/cap2); // (a-1)/b +1
        minYen=min(needMach1*p1,needMach2*p2);
        ll firstMach1=needCap/cap1,remainMach2=ceil((needCap-firstMach1*cap1)/1.0/cap2);
        ll priceComb1=firstMach1*p1+remainMach2*p2;
        ll firstMach2=needCap/cap2,remainMach1=ceil((needCap-firstMach2*cap2)/1.0/cap1);
        ll priceComb2=firstMach2*p2+remainMach1*p1;;
        minYen=min(minYen,min(priceComb1,priceComb2));
        sumYen+=minYen;
    }
    if(sumYen<=x)
        return true;
    return false;
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    cin>>n>>x;
    for(int i=1;i<=n;i++) {
            // cap1      price1 per U
        cin>>mach[i][1]>>mach[i][2]>>mach[i][3]>>mach[i][4];
    }
    ll l=0,r=MAXANS;
    while(l<r) {
        ll mid=l+r+1>>1;
        if(check(mid)) l=mid;
        else r=mid-1;
    }
    cout<<l<<endl;
    return 0;
}
// AC 64 个点 https://atcoder.jp/contests/abc374/submissions/58537183 已经比较接近了

引入了性价比概念，但发现连样例都过不了😂，还是要用dp。

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// https://atcoder.jp/contests/abc374/tasks/abc374_e
#include <iostream>
#include <cmath>
typedef long long ll;
using namespace std;
const ll MAXN=105,MAXANS=static_cast<ll>(1e9)+7,MAXA=1e7+7;
ll n,x,mach[MAXN][6];
int dp[MAXA];
inline bool check(ll needCap) {
    ll sumYen=0;
    for(int i=1;i<=n;i++) {
        if(sumYen>x)
            return false;
        ll sumCap=0,capA=mach[i][1],pA=mach[i][2],capB=mach[i][3],pB=mach[i][4];
        while (true) {
            if(sumCap>=needCap)
                break;
            ll remainCap=needCap-sumCap;
            double effA=pA*1.0/min(remainCap,capA);   // 采用A机器的零件均摊价格(总消耗钱数/每台机器能做的有贡献零件数)
            double effB=pB*1.0/min(remainCap,capB);   // 采用B机器零件均摊价格
            if(effA<effB)
                sumCap+=capA,sumYen+=pA;
            else
                sumCap+=capB,sumYen+=pB;
        }
    }
    if(sumYen<=x)
        return true;
    return false;
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    cin>>n>>x;
    for(int i=1;i<=n;i++) {
            // cap1      price1 per U
        cin>>mach[i][1]>>mach[i][2]>>mach[i][3]>>mach[i][4];
    }
    // for(int i=1;i<=n;i++) {
    //     if(mach[i][2]/1.0/mach[i][1]>mach[i][4]/1.0/mach[i][3]) {
    //         swap(mach[i][1],mach[i][3]);
    //         swap(mach[i][2],mach[i][4]);
    //     }
    // }
    ll l=0,r=MAXANS;
    while(l<r) {
        ll mid=l+r+1>>1;
        if(check(mid)) l=mid;
        else r=mid-1;
    }
    cout<<l<<endl;
    return 0;
}

将中间的判定改为了背包dp，果然这种通用做法就是应用性广泛且经过证明
// 就TLE了3个点 https://atcoder.jp/contests/abc374/submissions/58545771

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// https://atcoder.jp/contests/abc374/tasks/abc374_e
#include <iostream>
#include <cmath>
typedef long long ll;
using namespace std;
const ll MAXN=105,MAXANS=static_cast<ll>(1e9)+7,MAXA=1e7+7;
int n,x,mach[MAXN][6];
int dp[MAXA];
inline bool check(ll needCap) {
    ll sumYen=0;
    for(int i=1;i<=n;i++) {
        dp[0]=0;
        int cap1=mach[i][1],p1=mach[i][2],cap2=mach[i][3],p2=mach[i][4];
        bool isBreak=false;
        if(sumYen>x)
            return false;
        for(int j=1;j<=x-sumYen;j++) {  // 背包容量为钱，背包价值为生产力
            dp[j]=0;
            int up1=j-p1,up2=j-p2;
            if(up1>=0)
                dp[j]=dp[up1]+cap1;
            if(up2>=0)
                dp[j]=max(dp[j],dp[up2]+cap2);
            if(dp[j]>=needCap) {
                sumYen+=j;isBreak=true;break;
            }
        }
        if(!isBreak)
            return false;
    }
    if(sumYen<=x)
        return true;
    return false;
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    cin>>n>>x;
    for(int i=1;i<=n;i++) {
            // cap1      price1 per U
        cin>>mach[i][1]>>mach[i][2]>>mach[i][3]>>mach[i][4];
    }
    // for(int i=1;i<=n;i++) {
    //     if(mach[i][2]/1.0/mach[i][1]>mach[i][4]/1.0/mach[i][3]) {
    //         swap(mach[i][1],mach[i][3]);
    //         swap(mach[i][2],mach[i][4]);
    //     }
    // }
    ll l=0,r=MAXANS;
    while(l<r) {
        ll mid=l+r+1>>1;
        if(check(mid)) l=mid;
        else r=mid-1;
    }
    cout<<l<<endl;
    return 0;
}
// AC 64 个点 https://atcoder.jp/contests/abc374/submissions/58537183 已经比较接近了
// 就TLE了3个点 https://atcoder.jp/contests/abc374/submissions/58545771

Posted on Edited on

当年用python写的代码

193.18pts TLE 3/22 WA 2/22

https://ac.nowcoder.com/acm/contest/view-submission?submissionId=70565258

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from collections import deque

# 输入
y_, x_ = map(int, input().split())
maze = [list(input()) for _ in range(y_)]

# 各个方向都能走  不能走上       不能走下     不能走左      不能走右
qList = [(0, 0, 0, 4, 1), (0, 0, 0, 3, 1), (0, 0, 0, 2, 1), (0, 0, 0, 1, 1), (0, 0, 0, 0, 1)]  # x, y, d, 健全, 穿墙次数
direction5 = [[(-1, 0), (0, 1), (0, -1)], [(1, 0), (0, 1), (0, -1)], [(1, 0), (-1, 0), (0, 1)], 
              [(1, 0), (-1, 0), (0, -1)], [(1, 0), (-1, 0), (0, 1), (0, -1)]]

def bfs():
    visited = [[[False] * 2 for _ in range(x_)] for _ in range(y_)]
    q = deque(qList)
    while q:
        x0, y0, d, dI, throughAb = q.popleft()
        if visited[y0][x0][throughAb]:
            continue
        visited[y0][x0][throughAb] = True

        for dx, dy in direction5[dI]:
            nx, ny = x0 + dx, y0 + dy
            if 0 <= nx < x_ and 0 <= ny < y_:
                # 终点判断
                if nx == x_ - 1 and ny == y_ - 1:
                    if maze[ny][nx] == '.' or (maze[ny][nx] == 'X' and throughAb == 1):
                        return d + 1
                
                # 走路与穿墙判断
                if maze[ny][nx] == '.' and not visited[ny][nx][throughAb]:
                    q.append((nx, ny, d + 1, dI, throughAb))
                elif maze[ny][nx] == 'X' and throughAb == 1 and not visited[ny][nx][0]:
                    q.append((nx, ny, d + 1, dI, 0))
                
    return -1

print(bfs())

Posted on Edited on

60pts TLE 可能是check函数复杂度太高了

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#include <iostream>
using namespace std;
typedef long long ll;
const ll maxn=1e5+10,maxans=1e13+7;
ll n,m,a[maxn];
inline bool check(ll x) {
    ll s=1,subCnt=0;
    while (s<=n) {
        ll res=0,cnt=0;
        for(ll i=s;i<=n;i++) {
            if(res+a[i]>x)
                break;
            res+=a[i];
            cnt+=1;
        }
        s+=cnt;
        subCnt+=1;
    }
    if(subCnt<=m) {
        return true;
    }
    return false;
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    cin>>n>>m;
    for(int i=1;i<=n;i++) {
        cin>>a[i];
    }
    ll l=1,r=maxans;
    while(l<r) {
        ll mid=l+r>>1;
        if(check(mid)) r=mid;
        else l=mid+1;
    }
    cout<<l<<endl;
    return 0;
}
// 60pts https://www.luogu.com.cn/record/180491211

可以二分里套二分，当然那个二分我们就不写了，直接上stl

P1182_2.in

P1182_2.out

不过我仔细分析了一下时间复杂度，以及这个样例(另一道题目https://www.luogu.com.cn/problem/P2884)

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会使上面这个程序死循环，让我确信不是check（）的问题

仔细分析过后，发现还是check（）的问题，check死循环了。

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while (s<=n) {
        ll res=0,cnt=0;
        for(ll i=s;i<=n;i++) {
            if(res+a[i]>x)    // 某些情况下，cnt走不到cnt+1，导致s停滞不前，导致死循环
                break;
            res+=a[i];
            cnt+=1;
        }
        s+=cnt;
        subCnt+=1;
    }

解决起来倒也简单

ll l=maxa,r=maxans;

让左端点比数组中最大的元素大，就可以了，这样可以保证cnt的前进。

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// https://www.luogu.com.cn/problem/P2884
// https://www.luogu.com.cn/problem/P1182  双倍经验
#include <iostream>
using namespace std;
typedef long long ll;
const ll maxn=1e5+10,maxans=1e13+7;
ll n,m,a[maxn];
inline bool check(ll x) {
    ll s=1,subCnt=0;
    while (s<=n) {
        ll res=0,cnt=0;
        for(ll i=s;i<=n;i++) {
            if(res+a[i]>x)
                break;
            res+=a[i];
            cnt+=1;
        }
        s+=cnt;
        subCnt+=1;
    }
    if(subCnt<=m) {
        return true;
    }
    return false;
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    cin>>n>>m;
    ll maxa=0;
    for(int i=1;i<=n;i++) {
        cin>>a[i],maxa=max(maxa,a[i]);
    }
    ll l=maxa,r=maxans;
    while(l<r) {
        ll mid=l+r>>1;
        if(check(mid)) r=mid;
        else l=mid+1;
    }
    cout<<l<<endl;
    return 0;
}
// 60pts TLE https://www.luogu.com.cn/record/180491211
// 40pts TLE https://www.luogu.com.cn/record/180498527
// AC https://www.luogu.com.cn/record/180509878
// AC https://www.luogu.com.cn/record/180510298

Posted on Edited on 
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// https://www.acwing.com/problem/content/791/
#include <iostream>
using namespace std;
const int maxn=100007;
int n,q,k,a[maxn];

int main()
{
    cin>>n>>q;
    for(int i=1;i<=n;i++) {
        cin>>a[i];
    }
    for(int _=1;_<=q;_++) {
        cin>>k;
        int l=1,r=n;
        while (l<r) {
            int mid=l+r>>1;
            if(a[mid]>=k) r=mid;
            else l=mid+1;
        }
        if(a[l]!=k) {
            cout<<"-1 -1"<<endl;
            continue;
        }
        int l2=1,r2=n;
        while (l2<r2) {
            int mid=l2+r2+1>>1;
            if(a[mid]<=k) l2=mid;
            else r2=mid-1;
        }
        cout<<l-1<<" "<<l2-1<<endl;
    }
    return 0;
}
// AC  https://www.acwing.com/problem/content/submission/code_detail/37312161/


Posted on Edited on

题目大意

给定 n 条线段，每条线段有两个端点坐标。移动时分为两种速度：移动到某条线段的起点/终点使用速度 s，沿着线段绘制使用速度 t。需要选择绘制线段的顺序及每条线段从哪一端开始，求完成全部绘制的最短总时间。

Pass 10/74 WA https://atcoder.jp/contests/abc374/submissions/58484468 赛时代码

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#include <iostream>
#include <bitset>
#include <vector>
#include <algorithm>
#include <cmath>
#include <set>
#include <deque>
using namespace std;
int n,s,t,pos[10][6];  // s->not emitting laser && t-> emitting ......
bool vis[10][6],vis2[10];
double ans=10000000,dis;
void dfs(int x,int y,int cnt){
    if(cnt>=n){
        ans=min(dis,ans);
        return;
    }
    for(int i=1;i<=n;i++){
        if(vis2[i]) continue;
        vis2[i]=true;
        double squareS1=(pos[i][3]-pos[i][1])*(pos[i][3]-pos[i][1]);
        double squareS2=(pos[i][4]-pos[i][2])*(pos[i][4]-pos[i][2]);
        double lenSeg=sqrt(squareS1+squareS2)/t;
        for(int j=1;j<=3;j+=2){
            double square1=(pos[i][j]-pos[x][y])*(pos[i][j]-pos[x][y]);
            double square2=(pos[i][j+1]-pos[x][y+1])*(pos[i][j+1]-pos[x][y+1]);
            double ldis=sqrt(square1+square2)/s;
            dis=dis+(ldis+lenSeg);
            dfs(i,j,cnt+1);
            dis=dis-(ldis+lenSeg);
        }
        vis2[i]=false;
    }
}
int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    cin>>n>>s>>t;
    for(int i=1;i<=n;i++){
        cin>>pos[i][1]>>pos[i][2]>>pos[i][3]>>pos[i][4];
    }
    dis=0;
    dfs(0,0,0);
    cout<<ans<<endl;
    //cin>>n;
}
// WA https://atcoder.jp/contests/abc374/submissions/58484468

只改了一个地方就AC了
dfs(i,4-j,cnt+1); // 进入点和终止点不是一个，这条线段的终止点就是下一条线段的起始点

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#include <iostream>
#include <bitset>
#include <vector>
#include <algorithm>
#include <cmath>
#include <set>
#include <deque>
using namespace std;
int n,s,t,pos[10][6];  // s->not emitting laser && t-> emitting ......
bool vis[10][6],vis2[10];
double ans=10000000,dis;
void dfs(int x,int y,int cnt){
    if(cnt>=n){
        ans=min(dis,ans);
        return;
    }
    for(int i=1;i<=n;i++){
        if(vis2[i]) continue;
        vis2[i]=true;
        double squareS1=(pos[i][3]-pos[i][1])*(pos[i][3]-pos[i][1]);
        double squareS2=(pos[i][4]-pos[i][2])*(pos[i][4]-pos[i][2]);
        double lenSeg=sqrt(squareS1+squareS2)/t;
        for(int j=1;j<=3;j+=2){
            double square1=(pos[i][j]-pos[x][y])*(pos[i][j]-pos[x][y]);
            double square2=(pos[i][j+1]-pos[x][y+1])*(pos[i][j+1]-pos[x][y+1]);
            double ldis=sqrt(square1+square2)/s;
            dis=dis+(ldis+lenSeg);
            dfs(i,4-j,cnt+1);   // 进入点和终止点不是一个，这条线段的终止点就是下一条线段的起始点
            dis=dis-(ldis+lenSeg);
        }
        vis2[i]=false;
    }
}
int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    cin>>n>>s>>t;
    for(int i=1;i<=n;i++){
        cin>>pos[i][1]>>pos[i][2]>>pos[i][3]>>pos[i][4];
    }
    dis=0;
    dfs(0,0,0);
    cout.precision(17);   // 打印double全精度
    cout<<ans<<endl;
    //cin>>n;
}
//AC https://atcoder.jp/contests/abc374/submissions/58506793
//AC https://www.luogu.com.cn/record/180435741