Znzryb's Blog
0%

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

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
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

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
#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)

1
2
3
4
5
6
7
8
7 5
100
400
300
100
500
101
400
1
500

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。

1
2
3
4
5
6
7
8
9
10
11
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。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
// 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 
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
// 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

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
#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

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
#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

Posted on Edited on

题目大意

给定一个数组，通过若干次“除以 2 取整”的操作把元素压到一定范围内，要求最终能覆盖 0 到 x-1 的所有数（每个数至少出现一次），求最大的可行 x。通常用二分答案 + 贪心检查。

AC 代码 二分答案 相比于那个过了93.33%的还是修改了很多。

总体的check函数的贪心思路是 ≥x 的没用，之前就到过的（vis中有的）也没用。

没用，那怎么让他有用那？那当然是>>1才有用，不断的>>1，直到有用（注意条件，不要死循环了）。

https://ac.nowcoder.com/acm/contest/view-submission?submissionId=71786526

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
#include <iostream>
#include <vector>
#include <bitset>
#include <algorithm>
#include <set>
using namespace std;
const int maxn=1e5+7,maxA=static_cast<int>(1e6)+7;
int n,t,a[maxA],A;
set<int> vis;
inline bool check(int x) {
    for(int i=n;i>=1;i--) {
        int ai=a[i];
        while(ai>=x)
            ai>>=1;
        while(vis.count(ai)!=0 && ai!=0)  // 到过以前到过的值不算有用 ai！=0防止死循环的发生
            ai>>=1;
        vis.insert(ai);
    }
    for(int i=0;i<=x-1;i++) {
        if(vis.count(i)==0)
            return false;
    }
    return true;
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    cin>>t;
    for(int _=1;_<=t;_++) {
        cin>>n;
        for(int i=1;i<=n;i++) {
            cin>>a[i];
        }
        sort(&a[1],&a[n+1],less<int>());
        int l=0,r=n+1;  // 如果n个数连续，那么就是n
        while(l<r) {
            vis.clear();
            int mid=l+r+1>>1;
            if(check(mid)) l=mid;
            else r=mid-1;
        }
        cout<<l<<endl;
    }
    return 0;
}
// TLE https://ac.nowcoder.com/acm/contest/view-submission?submissionId=71785349
// 53.33% WA https://ac.nowcoder.com/acm/contest/view-submission?submissionId=71785676
// 93.33% WA https://ac.nowcoder.com/acm/contest/view-submission?submissionId=71785979
// AC https://ac.nowcoder.com/acm/contest/view-submission?submissionId=71786526

TLE pass 2/30

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
#include <iostream>
#include <vector>
#include <bitset>
#include <algorithm>
using namespace std;
const int maxn=1e5+7,maxA=static_cast<int>(1e6)+7;
int n,t,a[maxA],A;
bitset<maxA> ovis,vis;
inline bool check(int x) {
    for(int i=n;i>=1;i--) {
        int ai=a[i];
        if(ai<x)
            break;
        while(ai>>=1) {
            if(!vis[ai] && ai<x) {vis[ai]=true;break;}
        }
    }
    for(int i=1;i<=x-1;i++) {
        if(!vis[i])
            return false;
    }
    return true;
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    cin>>t;
    for(int _=1;_<=t;_++) {
        cin>>n;
        vis.reset();ovis.reset();   // 初始化
        for(int i=1;i<=n;i++) {
            cin>>a[i];A=max(A,a[i]);ovis[a[i]]=true;
        }
        sort(&a[1],&a[n+1],less<int>());
        int l=0,r=A;
        while(l<r) {
            for(int i=1;i<=A;i++)
                vis[i]=ovis[i];
            int mid=l+r+1>>1;
            if(check(mid)) l=mid;
            else r=mid-1;
        }
        cout<<l<<endl;
    }
    return 0;
}
// TLE https://ac.nowcoder.com/acm/contest/view-submission?submissionId=71785349

int l=0,r=n+1; // 如果n个数连续，那么就是n

for(int i=1;i<=n+1;i++)
vis[i]=ovis[i];

把上界换成n+1过了53.33% https://ac.nowcoder.com/acm/contest/view-submission?submissionId=71785676

但出现了WA

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
#include <iostream>
#include <vector>
#include <bitset>
#include <algorithm>
using namespace std;
const int maxn=1e5+7,maxA=static_cast<int>(1e6)+7;
int n,t,a[maxA],A;
bitset<maxA> ovis,vis;
inline bool check(int x) {
    for(int i=n;i>=1;i--) {
        int ai=a[i];
        if(ai<x)
            break;
        while(ai>>=1) {
            if(!vis[ai] && ai<x) {vis[ai]=true;break;}
        }
    }
    for(int i=1;i<=x-1;i++) {
        if(!vis[i])
            return false;
    }
    return true;
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    cin>>t;
    for(int _=1;_<=t;_++) {
        cin>>n;
        vis.reset();ovis.reset();   // 初始化
        for(int i=1;i<=n;i++) {
            cin>>a[i];ovis[a[i]]=true;
        }
        sort(&a[1],&a[n+1],less<int>());
        int l=0,r=n+1;  // 如果n个数连续，那么就是n
        while(l<r) {
            for(int i=1;i<=n+1;i++)
                vis[i]=ovis[i];
            int mid=l+r+1>>1;
            if(check(mid)) l=mid;
            else r=mid-1;
        }
        cout<<l<<endl;
    }
    return 0;
}
// TLE https://ac.nowcoder.com/acm/contest/view-submission?submissionId=71785349

WA pass93.33% https://ac.nowcoder.com/acm/contest/view-submission?submissionId=71785979

应该是快了

增加了

1
2
3
4
for(int i=2;i<=n;i++) {
        if(a[i]>=x) break;
        if(a[i]==a[i-1]) vis[0]=true;break;
    }
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
#include <iostream>
#include <vector>
#include <bitset>
#include <algorithm>
using namespace std;
const int maxn=1e5+7,maxA=static_cast<int>(1e6)+7;
int n,t,a[maxA],A;
bitset<maxA> ovis,vis;
inline bool check(int x) {
    for(int i=2;i<=n;i++) {
        if(a[i]>=x) break;
        if(a[i]==a[i-1]) vis[0]=true;break;
    }
    for(int i=n;i>=1;i--) {
        int ai=a[i];
        if(ai<x)
            break;
        while(true) {
            ai>>=1;
            if(!vis[ai] && ai<x) {vis[ai]=true;break;}
            if(ai==0)
                break;
        }
    }
    for(int i=0;i<=x-1;i++) {
        if(!vis[i])
            return false;
    }
    return true;
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    cin>>t;
    for(int _=1;_<=t;_++) {
        cin>>n;
        vis.reset();ovis.reset();   // 初始化
        for(int i=1;i<=n;i++) {
            cin>>a[i];ovis[a[i]]=true;
        }
        sort(&a[1],&a[n+1],less<int>());
        int l=0,r=n+1;  // 如果n个数连续，那么就是n
        while(l<r) {
            for(int i=0;i<=n+1;i++)
                vis[i]=ovis[i];
            int mid=l+r+1>>1;
            if(check(mid)) l=mid;
            else r=mid-1;
        }
        cout<<l<<endl;
    }
    return 0;
}
// TLE https://ac.nowcoder.com/acm/contest/view-submission?submissionId=71785349
// 53.33% WA https://ac.nowcoder.com/acm/contest/view-submission?submissionId=71785676