标签：inpu 负环 require 示例 struct 就是 fan main move
While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is very peculiar because it is a one-way path that delivers you to its destination at a time that is BEFORE you entered the wormhole! Each of FJ‘s farms comprises N (1 ≤ N ≤ 500) fields conveniently numbered 1..N, M (1 ≤ M ≤ 2500) paths, and W (1 ≤ W ≤ 200) wormholes.
As FJ is an avid time-traveling fan, he wants to do the following: start at some field, travel through some paths and wormholes, and return to the starting field a time before his initial departure. Perhaps he will be able to meet himself :) .
To help FJ find out whether this is possible or not, he will supply you with complete maps to F (1 ≤ F ≤ 5) of his farms. No paths will take longer than 10,000 seconds to travel and no wormhole can bring FJ back in time by more than 10,000 seconds.
2 3 3 1 1 2 2 1 3 4 2 3 1 3 1 3 3 2 1 1 2 3 2 3 4 3 1 8Sample Output
NO YESHint
const int inf = 1 << 29;
int n,m, w;
struct ed
{
int to, cost;
ed(int _t = 0, int _c = 0):to(_t),cost(_c){}
};
vector<ed>edge[505];
bool vis[505];
int d[505];
int cnt[505];
bool spfa(){
queue<int>que;
memset(cnt, 0, sizeof(cnt));
memset(vis, false, sizeof(vis));
for(int i = 1; i <= 500; i++) d[i] = inf;
d[1] = 0;
cnt[1] = 1;
que.push(1);
while(!que.empty()){
int u = que.front();
que.pop();
vis[u] = false;
for(int i = 0; i < edge[u].size(); i++){
int to = edge[u][i].to;
int cost = edge[u][i].cost;
if (d[u]+cost < d[to]){
d[to] = d[u]+cost;
if (!vis[to]){
vis[to] = true;
que.push(to);
cnt[to]++;
if (cnt[to] > n) return true;
}
}
}
}
return false;
}
int main() {
int t;
int a, b, c;
cin >>t;
while(t--){
scanf("%d%d%d", &n, &m, &w);
for(int i = 1; i <= 500; i++) edge[i].clear();
for(int i = 1; i <= m; i++){
scanf("%d%d%d", &a, &b, &c);
edge[a].push_back(ed(b,c));
edge[b].push_back(ed(a,c));
}
for(int i = 1; i <= w; i++){
scanf("%d%d%d", &a, &b, &c);
edge[a].push_back(ed(b, -c));
}
if (spfa()) printf("YES\n");
else printf("NO\n");
}
return 0;
}
标签：inpu 负环 require 示例 struct 就是 fan main move
原文地址：http://www.cnblogs.com/ccut-ry/p/7786661.html
