Znzryb's Blog

Codeforces Round 933 (Div. 3)——G. Rudolf and Subway

Posted on 2025-09-03 Edited on 2026-02-18

思路讲解

https://acm.hdu.edu.cn/contest/problem?cid=1172&pid=1005

这个题目的思路还是比较简单的，建虚拟点，

传送门

while (!pq.empty()) {
    auto [node,dis]=pq.front();
    pq.pop_front();
    if(dis>Dis[node]) continue;
    Dis[node]=dis;
    for(auto&v:g[node]){
        if(node<=N){
            if(dis+1>=Dis[v]) continue;
            Dis[v]=dis+1;
            pq.PB({v,dis+1});
        }else{
            if(dis>=Dis[v]) continue;
            Dis[v]=dis;
            pq.PF({v,dis});
        }
    }
}

AC代码

https://codeforces.com/contest/1941/submission/336750153

源代码

// by Gospel_rock
// https://codeforces.com/problemset/problem/1941/G
#include <algorithm>
#include <bits/stdc++.h>
#include <queue>
#include <tuple>
#include <utility>
#include <vector>
#define all(vec) vec.begin(),vec.end()
#define PB push_back
#define EB emplace_back
#define EM emplace
#define FI first
#define SE second
#define pct __builtin_popcountll
#define ctz __builtin_ctzll
#define SZ(a) ((long long) a.size())
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (int i = (a); i >= (b); --i)
#define debug(var) cerr << #var <<":"<<var<<"\n";
#define cend cerr<<"\n-----------\n"
#define fsp(x) fixed<<setprecision(x)
#define minn(vec) *min_element(vec.begin(),vec.end())
#define maxx(vec) *max_element(vec.begin(),vec.end())
using namespace std;
#ifdef LOCAL
template<typename T> void _print(T x) { cerr << x; }
// 可以打印多层数组，但是格式不是很好
template<typename T> void _print(vector<T> v) { cerr << "[ "; for (T i : v) _print(i), cerr << " "; cerr << "]"; }
template<typename T> void _print(set<T> s) { cerr << "{ "; for (T i : s) _print(i), cerr << " "; cerr << "}"; }
template<typename T, typename U> void _print(pair<T, U> p) { cerr << "{"; _print(p.first); cerr << ", "; _print(p.second); cerr << "}"; }
template<typename T, typename U> void _print(map<T, U> m) { cerr << "{ "; for (auto &p : m) _print(p), cerr << " "; cerr << "}"; }
// 打印两层数组，格式好一些
template<typename T> void _print(vector<vector<T>> v){cerr<<"{\n";for (auto i : v) _print(i), cerr << "\n";cerr<<" }";}
#endif

using  ll = long long;using ull = unsigned long long;
using ld=double;using LD=long double;
using i128 = __int128;
using pdd = pair<ld,ld>;using plb = pair<ll,bool>;
using pll = pair<ll,ll>;using vll=vector<ll>;using vb=vector<bool>;
using tll = tuple<ll,ll,ll>;using vii=vector<int>;
using vpll = vector<pll>;using vtll = vector<tll>;
using vdb = vector<ld>;using vc=vector<char>;
using arr3 = array<ll,3> ;using arr2 = array<ll,2>;
using CD=complex<double>;
static constexpr ll MAXN=(ll)1e6+10,INF=(ll)1e17+3; // 1e17+3是素数 1e18+9是素数
static constexpr ll mod=(ll)1e9+7; // 998244353;
static constexpr double eps=1e-8;const double pi=acos(-1.0);
ll N,M,Q,X,K,T,lT; 
inline void __();
signed main()
{
    ios::sync_with_stdio(false); 
    cin.tie(0);cout.tie(0);
    
    cin>>lT;
    while(lT--) 
        __();
    return 0;
}
/*

*/
struct odis{
    ll o,dis;
    // odis(ll o,ll dis):o(o),dis(dis) {}
    // odis(){}
    bool operator<(const odis& other)const{
        return dis>other.dis;
    }
};
inline void __(){
    cin>>N>>M;
    vector<pair<ll,ll>> li;
    vector<tll> edge;
    FOR(i,1,M){
        ll u,v,c;
        cin>>u>>v>>c;
        li.EB(u,c);
        li.EB(v,c);
        edge.EB(u,v,c);
    }
    priority_queue<odis> pq;
    ll st=1,ed=N;
    cin>>st>>ed;
    // li.EB(st,0);
    sort(all(li));
    li.resize(unique(all(li))-li.begin());
    auto get=[&](pll a){
        return lower_bound(all(li),a)-li.begin();
    };
    vector<vector<int> > g(SZ(li)+N+5);
    for(auto&[u,v,c]:edge){
        int idx=get({u,c})+N+1,idx2=get({v,c})+N+1;
        g[idx].EB(idx2);
        g[idx2].EB(idx);
        // g[v].EB(idx);
        g[u].EB(idx);
        g[idx].EB(v);
        // g[u].EB(idx2);
        g[v].EB(idx2);
        g[idx2].EB(u);
    }
    vll Dis(SZ(li)+N+5,INF);
    // vb vis(SZ(li)+N+5);
    pq.push({st,0});
    while (!pq.empty()) {
        auto [node,dis]=pq.top();
        pq.pop();
        if(Dis[node]!=INF) continue;
        Dis[node]=dis;
        for(auto&v:g[node]){
            if(node<=N){
                pq.push({v,dis+1});
            }else{
                pq.push({v,dis});
            }
            // if(v>=N+1){
            //     ll col=li[v-N-1].SE;
            //     if(col==c){
            //         pq.push({v,dis});
            //     }else{
                    
            //     }
            // }else{
            //     pq.push({v,dis});
            // }
        }
    } 
	cout<<Dis[ed]<<"\n";
}
/*
AC
https://codeforces.com/contest/1941/submission/336750153
AC  不保证联通
https://vjudge.net/solution/63392276
1
4 3
1 2 1
1 3 1
4 1 1
2 3

5
6 6
1 2 1
2 3 1
5 2 2
2 4 2
4 6 2
3 6 3
1 3
6 6
1 2 1
2 3 1
5 2 2
2 4 2
4 6 2
3 6 3
1 6
6 6
1 2 1
2 3 1
5 2 2
2 4 2
4 6 2
3 6 3
6 6
4 3
1 2 1
1 3 1
4 1 1
2 3
6 7
1 2 43
1 3 34
4 6 43
6 3 43
2 3 43
5 3 43
4 5 43
1 6

1
2
0
1
1

*/

心路历程（WA，TLE，MLE……）

Untitled

Posted on 2025-09-02 Edited on 2026-02-18

思路讲解

题解：

https://codeforces.com/gym/105922/attachments/download/31723/slide.pdf

AC代码

源代码

心路历程（WA，TLE，MLE……）

P11361 [NOIP2024] 编辑字符串

Posted on 2025-09-02 Edited on 2026-02-18

思路讲解

要多写，如果实在没有思路就数学化。

参考思路如下。

https://www.luogu.com.cn/article/7xrsn748

不难发现，我们要最小化下面👇。

$\sum_{i=1}^{n}(s[i]-s'[i])^2 \Rightarrow \sum_{i=1}^{n}s[i]^2+s'[i]^2-2s[i]s'[i]$

Codeforces Round 1046 (Div. 2)——D. For the Champion

常数分离以后，我们其实就是要最大化👇。

相当于我们要最大化 $\sum_{i=1}^{n}s[i]s'[i]$ 。不过我们发现好像没什么用？非常有用，我们就知道了，1 1 非常重要，要尽量多的出现1 1，其他不重要。

当然，如果光有这个想法是不行的。我们还需要用到分块思想。

首先那，分块，我们自然是同时只能按照一个字符串分块的，因此我们按照 $s$ 可不可以交换分块。

但是，我们按照 $s$ 分块以后，我们可以再对 $s'$ 进行分块。这个听起来很抽象是吧？其实也简单，就是说 $s’$ 其实如果不是落在整段 $s$ 可交换块上，可以部分落在 s 的可交换块上，然后我们的策略其实就是尽量多的出现1 1。

AC代码

源代码

心路历程（WA，TLE，MLE……）

至少有K个重复字符的最长子串

Posted on 2025-09-01 Edited on 2026-02-18

思路讲解

https://leetcode.cn/problems/longest-substring-with-at-least-k-repeating-characters/description/

其实就是双指针。

void cal(ll uni){
    vll cnt(28);
    ll l=0,r=0;
    ll cur=0,bigk=0;
    while (r<SZ(s)) {
        if(cur<=uni){
            int idx=s[r]-'a';
            if(cnt[idx]==0) ++cur;
            cnt[idx]++;
            if(cnt[idx]==K) ++bigk;
            ++r;
        }else{
            int idx=s[l]-'a';
            if(cnt[idx]==K) --bigk;
            --cnt[idx];
            if(cnt[idx]==0) --cur;
            ++l;
        }
        if(cur==uni && bigk==cur){
            ans=max(ans,r-l);
        }
    }
}

AC代码

https://leetcode.cn/problems/longest-substring-with-at-least-k-repeating-characters/submissions/658772401/

源代码

// by Gospel_rock
#include <bits/stdc++.h>
#define all(vec) vec.begin(),vec.end()
#define PB push_back
#define EB emplace_back
#define EM emplace
#define FI first
#define SE second
#define pct __builtin_popcountll
#define ctz __builtin_ctzll
#define SZ(a) ((long long) a.size())
#define FOR(i, a, b) for (long long i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (long long i = (a); i >= (b); --i)
#define debug(var) cerr << #var <<":"<<var<<"\n";
#define cend cerr<<"\n-----------\n"
#define lson(var) (var<<1)
#define rson(var) (var<<1|1)
#define fsp(x) fixed<<setprecision(x)
#define minn(vec) *min_element(vec.begin(),vec.end())
#define maxx(vec) *max_element(vec.begin(),vec.end())
using namespace std;
#ifdef LOCAL
template<typename T> void _print(T x) { cerr << x; }
// 可以打印多层数组，但是格式不是很好
template<typename T> void _print(vector<T> v) { cerr << "[ "; for (T i : v) _print(i), cerr << " "; cerr << "]"; }
template<typename T> void _print(set<T> s) { cerr << "{ "; for (T i : s) _print(i), cerr << " "; cerr << "}"; }
template<typename T, typename U> void _print(pair<T, U> p) { cerr << "{"; _print(p.first); cerr << ", "; _print(p.second); cerr << "}"; }
template<typename T, typename U> void _print(map<T, U> m) { cerr << "{ "; for (auto &p : m) _print(p), cerr << " "; cerr << "}"; }
// 打印两层数组，格式好一些
template<typename T> void _print(vector<vector<T>> v){cerr<<"{\n";for (auto i : v) _print(i), cerr << "\n";cerr<<" }";}
#endif

using  ll = long long;using ull = unsigned long long;
using DB=double;using LD=long double;
// using i128 = __int128;
using pdd = pair<DB,DB>;using plb = pair<ll,bool>;
using pll = pair<ll,ll>;using vll=vector<ll>;using vb=vector<bool>;
using tll = tuple<ll,ll,ll>;using vii=vector<int>;
using vpll = vector<pll>;using vtll = vector<tll>;
using vdb = vector<DB>;using vc=vector<char>;
using arr3 = array<ll,3> ;using arr2 = array<ll,2>;
using CD=complex<double>;
static constexpr ll MAXN=(ll)1e6+10,INF=(ll)1e17+3; // 1e17+3是素数 1e18+9是素数
static constexpr ll mod=(ll)1e9+7; // 998244353;
static constexpr double eps=1e-8;const double pi=acos(-1.0);
ll N,M,Q,X,K,T,lT;

inline void __();
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    
    // cin>>lT;
    // while(lT--)
        __();
    return 0;
}
/*

*/
string s;
ll ans=0;
void cal(ll uni){
    vll cnt(28);
    ll l=0,r=0;
    ll cur=0,bigk=0;
    while (r<SZ(s)) {
        if(cur<=uni){
            int idx=s[r]-'a';
            if(cnt[idx]==0) ++cur;
            cnt[idx]++;
            if(cnt[idx]==K) ++bigk;
            ++r;
        }else{
            int idx=s[l]-'a';
            if(cnt[idx]==K) --bigk;
            --cnt[idx];
            if(cnt[idx]==0) --cur;
            ++l;
        }
        if(cur==uni && bigk==cur){
            ans=max(ans,r-l);
        }
    }
}
inline void __(){
    cin>>s;
    cin>>K;
    FOR(i,1,26){
        cal(i);
    }
    cout<<ans<<"\n";
}
/*

*/

心路历程（WA，TLE，MLE……）

ABC-421-D - RLE Moving

Posted on 2025-08-31 Edited on 2026-02-18

思路讲解

1
2

// 整体思路就是我们如果发现如果两者起始位置不同，那么是不会有相交的，或者相交一次
// 简单而言，就是必须先有一次相交，接着运动方向一样，才能有相交，否则就直接穿过去了

// 在同一点并且操作序列相同
if(a==b && opa==opb){
    ans+=mn;
}else{
    auto f=[&](ll dis){
        Point c=oper(a, opa,dis),d=oper(b, opb, dis);
        return mandis(c, d);
    };
    ll l=1,r=mn;
    while (l<r) {
        ll mid=l+r>>1;
        // if(f(mid)<=f(mid+1)){
        // 从mid点（或其前）开始，函数开始单调递增
        if(f(mid+1)-f(mid)>=0){
            r=mid;
        }else{
            l=mid+1;
        }
    }
    if(f(l)==0){
        ++ans;
    }
}

AC代码

https://atcoder.jp/contests/abc421/submissions/68978595

源代码

// by Gospel_rock
#include <bits/stdc++.h>
#include <deque>
#include <istream>
#include <vector>
#define all(vec) vec.begin(),vec.end()
#define PB push_back
#define EB emplace_back
#define EM emplace
#define FI first
#define SE second
#define pct __builtin_popcountll
#define ctz __builtin_ctzll
#define SZ(a) ((long long) a.size())
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (int i = (a); i >= (b); --i)
#define debug(var) cerr << #var <<":"<<var<<"\n";
#define cend cerr<<"\n-----------\n"
#define fsp(x) fixed<<setprecision(x)
#define minn(vec) *min_element(vec.begin(),vec.end())
#define maxx(vec) *max_element(vec.begin(),vec.end())
using namespace std;
#ifdef LOCAL
template<typename T> void _print(T x) { cerr << x; }
// 可以打印多层数组，但是格式不是很好
template<typename T> void _print(vector<T> v) { cerr << "[ "; for (T i : v) _print(i), cerr << " "; cerr << "]"; }
template<typename T> void _print(set<T> s) { cerr << "{ "; for (T i : s) _print(i), cerr << " "; cerr << "}"; }
template<typename T, typename U> void _print(pair<T, U> p) { cerr << "{"; _print(p.first); cerr << ", "; _print(p.second); cerr << "}"; }
template<typename T, typename U> void _print(map<T, U> m) { cerr << "{ "; for (auto &p : m) _print(p), cerr << " "; cerr << "}"; }
// 打印两层数组，格式好一些
template<typename T> void _print(vector<vector<T>> v){cerr<<"{\n";for (auto i : v) _print(i), cerr << "\n";cerr<<" }";}
#endif

using  ll = long long;using ull = unsigned long long;
using ld=double;using LD=long double;
using i128 = __int128;
using pdd = pair<ld,ld>;using plb = pair<ll,bool>;
using pll = pair<ll,ll>;using vll=vector<ll>;using vb=vector<bool>;
using tll = tuple<ll,ll,ll>;using vii=vector<int>;
using vpll = vector<pll>;using vtll = vector<tll>;
using vdb = vector<ld>;using vc=vector<char>;
using arr3 = array<ll,3> ;using arr2 = array<ll,2>;
using CD=complex<double>;
static constexpr ll MAXN=(ll)1e6+10,INF=(ll)1e17+3; // 1e17+3是素数 1e18+9是素数
static constexpr ll mod=(ll)1e9+7; // 998244353;
static constexpr double eps=1e-8;const double pi=acos(-1.0);
ll N,M,Q,X,K,T,lT,L;

inline void __();
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    
    // cin>>lT;
    // while(lT--)
        __();
    return 0;
}
/*

*/
using Real = ll;
int sgn(Real x) {
    if (-eps < x && x < eps) return 0;
    return x < 0 ? -1 : 1;
}
struct Point{
    Real x,y;
    Point(){}
    Point(Real x,Real y):x(x),y(y){}
    //                      & 是指只对左值起效果，不能对Point(1,2)这样的临时值
    // friend Point& operator+=(Point &a,Point b){
    //     a.x+=b.x;
    //     a.y+=b.y;
    //     return a;
    // }
    friend Point operator +(Point a,Point b){
        return Point(a.x+b.x,a.y+b.y);
    }
    friend Point operator -(Point a,Point b){
        return Point(a.x-b.x,a.y-b.y);
    }
    friend bool operator <(Point a,Point b) {
        return sgn(a.x-b.x)?a.x<b.x:a.y<b.y; 
    }
    friend bool operator ==(Point a,Point b){
        if(sgn(a.x-b.x)) return false;
        if(sgn(a.y-b.y)) return false;
        return true;
    }
    friend Point operator *(Point a,Real k){
        return Point(a.x*k,a.y*k);
    }
    friend Point operator /(Point a,Real k){
        return Point(a.x/k,a.y/k);
    }
    friend bool operator !=(Point a,Point b){
        return !(a==b);
    }
    friend istream& operator >>(istream& is,Point &e){
        Real a,b;is>>a>>b;
        e=Point(b,a);
        return is;
    }
    friend ostream &operator<<(ostream &os, Point p) {
        return os << "(" << p.x << ", " << p.y << ")";
    }
};

Real cross(Point a,Point b){
    return a.x*b.y-a.y*b.x;
}
Real cross(Point p1, Point p2, Point p0) { // 叉乘 (p1 - p0) x (p2 - p0);
    return cross(p1 - p0, p2 - p0);
}
Real dot(Point a,Point b){
    return a.x*b.x+a.y*b.y;
}
Real mandis(Point a,Point b){
    Real res=abs(a.x-b.x)+abs(a.y-b.y);
    return res;
}
// 采用两点式
struct Line{
    Point p,q;
    Line(){}
    Line(Point a,Point b):p(a),q(b) {}
    friend bool operator<(Line a,Line b){
        if(a.p!=b.p) return a.p<b.p;
        return a.q<b.q; 
    }
    friend bool operator==(Line a,Line b){
        if(a.p!=b.p) return false;
        if(a.q!=b.q) return false;
        return true;
    } 
    friend istream& operator>>(istream& is, Line& e) {
        Real a,b,c,d;
        is >> a >> b >> c >> d;
        e = Line(Point(a,b),Point(c,d));
        return is;
    }
    friend ostream&operator<<(ostream &os, Line l) {
        return os << "<" << l.p << ", " << l.q << ">";
    }
};

// p+tv=q+sw
// (p-q) x w+tv x w = 0
// t=-(p-q)x w/(vxw)
// 传入的是方向式  拿到的是两直线交点
Point getintersect(const Line &a,const Line &b){
    Point p=a.p, v=a.q-a.p,q=b.p,w=b.q-b.p;
    Point u=p-q;
    ld t=cross(w,u)/cross(v,w);
    return p+v*t;
}
bool seginterline(Line seg,Line e){
    if(sgn(cross(e.p-seg.p, e.q-seg.p))*sgn(cross(e.p-seg.q, e.q-seg.q))>0){
        return false;
    }
    return true;
}
bool onseg(Point p,Line s){
    return sgn(cross(p-s.p, p-s.q))==0 && sgn(dot(p-s.p,p-s.q))<=0;
}
// AC https://vjudge.net/problem/OpenJ_Bailian-1410
bool seginterseg(Line l1, Line l2) {
    Point s1 = l1.p, e1 = l1.q;
    Point s2 = l2.p, e2 = l2.q;
    // s2 相对于 s1->e1 其在哪一侧？
    int o1 = sgn(cross(e1, s2, s1));
    int o2 = sgn(cross(e1, e2, s1));
    int o3 = sgn(cross(e2, s1, s2));
    int o4 = sgn(cross(e2, e1, s2));
    // 防止把共线但不重叠的两线段判定为相交
    if (o1 == 0 && onseg(s2, l1)) return true;
    if (o2 == 0 && onseg(e2, l1)) return true;
    if (o3 == 0 && onseg(s1, l2)) return true;
    if (o4 == 0 && onseg(e1, l2)) return true;
    // 需要都在不同侧
    return (o1 * o2 < 0) && (o3 * o4 < 0);
}

Point oper(Point a,char op,ll dis){
    if(op=='U'){
        a.y-=dis;
    }else if(op=='D'){
        a.y+=dis;
    }else if(op=='R'){
        a.x+=dis;
    }else{
        a.x-=dis;
    }
    return a;
}
struct OP{
    char op;ll dis;
    friend istream& operator >>(istream &is,OP &e){
        is>>e.op>>e.dis;
        return is;
    }
};
inline void __(){
    Point a,b;
    cin>>a>>b;
    cin>>N>>M>>L;
    // 整体思路就是我们如果发现如果两者起始位置不同，那么是不会有相交的，或者相交一次
    // 简单而言，就是必须先有一次相交，接着运动方向一样，才能有相交，否则就直接穿过去了
    // vector<OP> A(N+5),B(N+5);
    queue<OP> A,B;
    FOR(i,1,M){
        OP a;
        cin>>a;
        A.push(a);
    }
    FOR(i,1,L){
        OP b;
        cin>>b;
        B.push(b);
    }
    ll ans=0;
    while (!A.empty()) {
        auto [opa,disa]=A.front();
        auto [opb,disb]=B.front();
        ll mn=min(disa,disb);
        A.front().dis-=mn;B.front().dis-=mn;
        if(A.front().dis==0){
            A.pop();
        }
        if(B.front().dis==0){
            B.pop();
        }
        if(a==b && opa==opb){
            ans+=mn;
        }else{
            auto f=[&](ll dis){
                Point c=oper(a, opa,dis),d=oper(b, opb, dis);
                return mandis(c, d);
            };
            ll l=1,r=mn;
            while (l<r) {
                ll mid=l+r>>1;
                // if(f(mid)<=f(mid+1)){
                // 从mid点（或其前）开始，函数开始单调递增
                if(f(mid+1)-f(mid)>=0){
                    r=mid;
                }else{
                    l=mid+1;
                }
            }
            if(f(l)==0){
                ++ans;
            }
        }
        a=oper(a, opa, mn);
        b=oper(b, opb, mn);
    }
    cout<<ans;
}
/*
AC
https://atcoder.jp/contests/abc421/submissions/68978595
*/