| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 41728 | Accepted: 15325 |
Description
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.
Input
Line 1 of each farm: Three space-separated integers respectively: N, M, and W
Lines 2..M+1 of each farm: Three space-separated numbers (S, E, T) that describe, respectively: a bidirectional path between S and E that requires T seconds to traverse. Two fields might be connected by more than one path.
Lines M+2..M+W+1 of each farm: Three space-separated numbers (S, E, T) that describe, respectively: A one way path from S to E that also moves the traveler back T seconds.
Output
Sample Input
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 8
Sample Output
NO
YES
Hint
For farm 2, FJ could travel back in time by the cycle
1->2->3->1, arriving back at his starting location 1 second
before he leaves. He could start from anywhere on the cycle to
accomplish this.
就是当你过去的时候时间会倒退Ts。我们的任务是知道会不会在从某块地出发后又回来，看到了离开之前的自己。
简化下，就是看图中有没有负权环。
#include <cstdio>
#define maxn 502
#define INF 0x3fffffff ///边
typedef struct Edge{
int u, v; ///起点， 终点
int weight; ///权值
}Edge; Edge edge[]; ///双向边，保存边的权值
int dist[maxn]; ///节点到原点的最小距离
int edgenum; ///边数
///插入边
void insert(int u, int v, int w)
{
edge[edgenum].u = u;
edge[edgenum].v = v;
edge[edgenum++].weight = w;
}
bool Bellman_Ford(int source, int nodenum) ///原点和结点个数
{
for(int i=; i < nodenum; ++i)
dist[i] = INF;
dist[source] = ;
for(int i=; i < nodenum; ++i){
for(int j=; j < edgenum; ++j)
{
if(dist[edge[j].v] > dist[edge[j].u] + edge[j].weight)///松弛计算
dist[edge[j].v] = dist[edge[j].u] + edge[j].weight;
}
}
bool flag = false;
/// 判断是否有负权环
for(int i=; i < edgenum; ++i){
if(dist[edge[i].v] > dist[edge[i].u] + edge[i].weight)
{
flag = true; ///有负权环
break;
}
}
return flag;
}
int main()
{
int T;
scanf("%d", &T);
while(T--)
{
int n, m, c;
scanf("%d%d%d", &n, &m, &c);
edgenum = ;
for(int i=; i<m; i++)
{
int u, v, w;
scanf("%d%d%d", &u, &v, &w);
insert(u, v, w);
insert(v, u, w);
}
for(int i=; i<c; i++)
{
int u, v, w;
scanf("%d%d%d", &u, &v, &w);
insert(u, v, -w);
}
if(Bellman_Ford(, n))
printf("YES\n");
else
printf("NO\n");
}
return ;
}