POJ 3259 Wormholes bellman-ford判负环
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: A single integer, F. F farm descriptions follow.
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
Lines 1..F: For each farm, output “YES” if FJ can achieve his goal, otherwise output “NO” (do not include the quotes).
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 1, FJ cannot travel back in time.
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.
题意：给出一个图，问是否有负环。N个点，M个正边，W个负边，后面M行M行正边，双向，W条负边，单向。有负环，输出YES，没有输出NO。跑一个bellman_ford就行了。
bellman_ford原理：n个点，如果由起点到每一个点有最短路，那么由邻近的边开始类似dij的更新（他们都叫松弛，我不懂），那么剩余n-1个点，最多n-1次更新就可以找出到每个点的最短路。如果存在负环，那么这个负环可以一直跑，最短路为-inf，这样就不存在到任何点的最短路。因为已经跑了n-1次了，能最短路的都最短怒了，所以，只需要循环找一遍是否存在可以继续更新的边，如果存在，则代表存在有负环。
中间使用了一个flag，如果有点能更新就一直继续，如果不能则直接退出循环，能提高运算效率。使用结构体的储存边之间关系在这道题中效率比邻接矩阵效率高，邻接表还不会，不会讨论复杂度。而且感觉也挺好写易懂。
/// poj3259
/// 2015/11/4
#include"stdio.h"
#include"iostream"
#include"algorithm"
#include"string.h"
#define inf 77777777;
using namespace std;
struct node
{
int u,v,w; ///u -> v 有一条边 u到v边的权值为w
}edge[6010];
int n,m,w; ///n个点，m条正边，w条负边
int dis[510]; ///用来更新是否存在最短路
bool bellman_ford()
{
bool flag;
for(int i = 1;i <= 505;i++) ///初始化
dis[i] = inf;
dis[1] = 0;
for(int i = 1;i <= n-1 && !flag;i++) ///n-1次更新
{
flag = true;
for(int j = 1;j <= m;j++)
{
if(dis[edge[j].v] > dis[edge[j].u]+edge[j].w) ///是否能更新
{
dis[edge[j].v] = dis[edge[j].u]+edge[j].w;
flag = false; ///能更新赋值为false
}
}
}
for(int i = 1;i <= m;i++) ///找是否存在能继续更新的点，有就存在负环 返回true
if(dis[edge[i].v] > dis[edge[i].u]+edge[i].w)
return true;
return false;
}
int main(void)
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d%d%d",&n,&m,&w);
for(int i = 1;i <= m;i++) ///m条正边
{
scanf("%d%d%d",&edge[i].u,&edge[i].v,&edge[i].w);
edge[i+m].u = edge[i].v; ///双向边
edge[i+m].v = edge[i].u;
edge[i+m].w = edge[i].w;
}
for(int i = m*2+1;i <= m*2+w;i++) ///w条负边
{
scanf("%d%d%d",&edge[i].u,&edge[i].v,&edge[i].w);
edge[i].w = -edge[i].w;
}
m = m*2+w; ///把m更新为总的边一共有m*2+w条
if(bellman_ford()) ///有负边输出YES,没有输出NO
printf("YES\n");
else
printf("NO\n");
}
}