题意：

给定一个有向图，每条边都有一个权值，每次你可以选择一个结点和一个整数的，把所有以v为终点的边的权值减去d，

把所有以v为起点的边的权值加上d

最后要让所有边的权的最小值非负且尽量大

代码

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<vector>
#include<queue>
using namespace std;
const int N = 1e3;
const int inf  = 0x3f3f3f3f;
int dist[N],inq[N],cnt[N];
struct node
{int v,w;node(int v,int w):v(v),w(w) {};
};
vector<node> G[N];
void init(int n)
{for(int i = 0; i<=n; i++) G[i].clear();
}
bool spfa(int n)
{queue<int> q;q.push(0);memset(inq,0,sizeof(inq));memset(cnt,0,sizeof(cnt));memset(dist,inf,sizeof(dist));dist[0] = 0;inq[0] = 1;cnt[0] = 1;while(!q.empty()){int u = q.front();q.pop();inq[u] = 0;for(int i = 0; i<G[u].size(); i++){int v = G[u][i].v, w= G[u][i].w;if(dist[v]>dist[u] + w){dist[v] = dist[u] + w;if(!inq[v]) q.push(v),inq[v] = 1;if(++cnt[v]>n) return true;}}}return false;
}
bool test(int mid,int n)
{for(int i = 1; i<=n; i++)for(int j = 0; j<G[i].size(); j++)G[i][j].w -=mid;int ret = spfa(n);for(int i = 1; i<=n; i++)for(int j = 0; j<G[i].size(); j++)G[i][j].w +=mid;return ret;
}
int main()
{int V,E;while(~scanf("%d%d",&V,&E)){init(V);int ub = 0;int u ,v ,w;for(int i = 0; i<E; i++){scanf("%d%d%d",&u,&v,&w);G[u].push_back(node(v,w));ub = max(w,ub);}for(int i = 1; i<=V; i++)G[0].push_back(node(i,0));int L = 1, R = ub;if(test(L,V)) puts("No Solution");else if(!test(ub+1,V)) puts("Infinite");else{while(L<R){int mid = L + (R -L)/2;if(test(mid,V)) R = mid ;else  L = mid + 1;}printf("%d\n",L-1);}}return 0;
}