qq_44866969的博客

#include<bits/stdc++.h>#define ll long longusing namespace std;const ll inf=1e18;const int N=510;int n,m,t;ll d;ll f[N][N];int p[N];int main(){ scanf("%d%d%d",&n,&m,&t); scanf("%lld",&d); for(int i=1;i<=n;i++) for(in.

//恰好经过k条边的最短距离：允许存在负环。//d[k][i][j]表示i到j恰好经过k条边的最短距离//d[a+b,i,j]=min(d[a,i,k]+d[b,k,j]),k=1~n//满足结合律，因此可用快速幂优化到n^3logn#include<bits/stdc++.h>using namespace std;const int N=210;int g[N][N],res[N][N],n,m,k,s,t;void mul(int c[][N],int a[][N],in.

#include <cstring>#include <iostream>#include <algorithm>using namespace std;const int N = 110, INF = 0x3f3f3f3f;int n, m;int d[N][N], g[N][N];int pos[N][N];int path[N], cnt;void get_path(int i, int j){ if (pos[i][j] == 0) .

//传递闭包(O(n*m^3))#include <cstring>#include <iostream>#include <algorithm>using namespace std;const int N = 26;int n, m;bool g[N][N], d[N][N];bool st[N];void floyd(){ memcpy(d, g, sizeof d); for (int k = 0; k < n.