uvalive 6800 - The Mountain of Gold? 判负环

题意：给出一个图，判断能不能从0出发，再回到0，经过边的权值为负。

判断图中有没有负环，并且负环上的点能回到0即可。

#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 1000 + 10;
int dist[maxn];
struct Edge{
int u, v, cost;
Edge(){}
Edge(int u, int v, int cost) : u(u), v(v), cost(cost) {}
};
vector<Edge> e;
vector<int> ee[1005];
int vis[1005];
int dfs(int u) {
if(u == 0)
return 1;
vis[u] = 1;
for(int i = 0; i < ee[u].size(); i++) {
if(!vis[ee[u][i]])
if(dfs(ee[u][i]))
return 1;
}
return 0;
}
bool bellman_ford(int s, int n){
for(int i = 0; i < n; ++i) dist[i] = INF;
dist[s] = 0;
for(int i = 1; i < n; ++i){
bool flag = false;
for(int j = 0; j < e.size(); ++j){
int u = e[j].u; int v = e[j].v;
int cost = e[j].cost;
if(dist[v] > dist[u] + cost){
dist[v] = dist[u] + cost;
flag = true;
}
}
if(!flag) return true;
}
for(int j = 0; j < e.size(); ++j) {
memset(vis, 0, sizeof(vis));
if(dfs(e[j].v) && dist[e[j].v] > dist[e[j].u] + e[j].cost)
return false;
}

return true;
}

int main()
{
int kase = 0;
int T; scanf("%d", &T);
while(T --){
e.clear();

int n, m;
scanf("%d%d", &n, &m);
for(int i = 0; i < n; i++)
ee[i].clear();
while(m --){
int u, v, cost;
scanf("%d%d%d",&u, &v, &cost);
ee[u].push_back(v);
e.push_back(Edge(u, v, cost));
}
printf("Case #%d: %s\n",++kase, !bellman_ford(0, n) ? "possible" : "not possible");
}
return 0;
}