Wormholes 最短路判断有无负权值

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.

 #include<stdio.h>
 #include<string.h>
 #include<stdlib.h>

 const int EM = ;
 const int VM = ;
 const int INF = ;
 struct node
 {
     int u,v,w;
 }map[EM];

 int cnt,dis[VM];
 int n,m,k;

 void addedge(int au,int av,int aw)
 {
     map[cnt].u = au;
     map[cnt].v = av;
     map[cnt].w = aw;
     cnt++;
 }

 int Bellman_ford()
 {
     int flag ,i;
     //初始化
     for( i = ; i <= n; i++)
     {
         dis[i] = INF;
     }
     dis[] =;

     for( i = ; i <= n; i++)
     {
         flag = ;
         for(int j = ; j < cnt; j++)
         {
             if(dis[map[j].v] > dis[map[j].u]+map[j].w)
             {
                 dis[map[j].v] = dis[map[j].u]+map[j].w;
                 flag = ;
             }
         }
         if(flag== ) break;
     }
     if(i == n+) return ;//若第n次还可以松弛说明存在负环
     else return ;
 }

 int main()
 {
     int t,u,v,w,ans;
     scanf("%d",&t);
     while(t--)
     {
         cnt = ;
         scanf("%d %d %d",&n,&m,&k);
         while(m--)
         {
             scanf("%d %d %d",&u,&v,&w);
             //添加双向边
             addedge(u,v,w);
             addedge(v,u,w);
         }
         while(k--)
         {
             scanf("%d %d %d",&u,&v,&w);
             //添加单向边
             addedge(u,v,-w);
         }
         ans = Bellman_ford();
         if(ans == )
             printf("YES\n");
         else printf("NO\n");
     }
     return ;
 }

 //spfa判断有无负环
 #include<stdio.h>
 #include<string.h>
 #include<stdlib.h>
 #include<iostream>
 #include<queue>
 using namespace std;

 const int MAX = ;
 const int INF = ;
 int n,m,w;
 int map[MAX][MAX];
 queue<int>que;
 int inque[MAX];
 int vexcnt[MAX];
 int dis[MAX];

 bool spfa()
 {
     memset(inque,,sizeof(inque));
     memset(vexcnt,,sizeof(vexcnt));
     for(int i = ; i <= n; i++)
         dis[i] = INF;
     dis[] = ;
     que.push();
     inque[] = ;
     vexcnt[]++;
     while(!que.empty())
     {
         int tmp = que.front();
         que.pop();
         inque[tmp] = ;
         for(int i = ; i <= n; i++)
         {
             if(dis[tmp] < INF && dis[i] > dis[tmp] + map[tmp][i])
             {
                 dis[i] = dis[tmp] + map[tmp][i];
                 if(inque[i] == )
                 {
                     inque[i] = ;
                     vexcnt[i]++;
                     que.push(i);
                     if(vexcnt[i] >= n)
                     {
                         return false;
                     }
                 }
             }
         }
     }
     return true;
 }
 int main()
 {
     int t;
     int x,y,z;
     scanf("%d",&t);
     while(t--)
     {
         while(!que.empty())que.pop();
         scanf("%d %d %d",&n,&m,&w);
         for(int i = ; i <= n; i++)
             for(int j = ; j <= n; j++)
             {
                 if(i == j) map[i][j] = ;
                 else map[i][j] = INF;
             }
         for(int i = ; i <= m; i++)
         {
             scanf("%d %d %d",&x,&y,&z);
             if(map[x][y] > z)
             {
                 map[x][y] = z;
                 map[y][x] = z;
             }
         }
         for(int i = ; i <= w; i++)
         {
             scanf("%d %d %d",&x,&y,&z);
             if(map[x][y] > -z)
                 map[x][y] = -z;
         }
         if(spfa())
             printf("NO\n");
         else printf("YES\n");
     }
     return ;
 }

<Bellman-Ford算法>

Dijkstra算法无法判断负权回路，而负权回路的含义是，回路的权值和为负。若不为负，即便有负权的边，也可正确求出最短路径，如果遇到负权，则可以采用Bellman-Ford算法。

Bellman-Ford算法能在更普遍的情况下（存在负权边）解决单源点最短路径问题。对于给定的带权（有向或无向）图 G=（V,E），其源点为s，加权函数 w是 边集 E 的映射。对图G运行Bellman-Ford算法的结果是一个布尔值，表明图中是否存在着一个从源点s可达的负权回路。若不存在这样的回路，算法将给出从源点s到 图G的任意顶点v的最短路径d[v]。

适用条件&范围

1.单源最短路径(从源点s到其它所有顶点v);

2.有向图&无向图(无向图可以看作(u,v),(v,u)同属于边集E的有向图);

3.边权可正可负(如有负权回路输出错误提示);

4.差分约束系统;

Bellman-Ford算法描述：

1,.初始化：将除源点外的所有顶点的最短距离估计值 d[v] ←+∞, d[s] ←0;

2.迭代求解：反复对边集E中的每条边进行松弛操作，使得顶点集V中的每个顶点v的最短距离估计值逐步逼近其最短距离；（运行|v|-1次）

3.检验负权回路：判断边集E中的每一条边的两个端点是否收敛。如果存在未收敛的顶点，则算法返回false，表明问题无解；否则算法返回true，并且从源点可达的顶点v的最短距离保存在 d[v]中。

描述性证明：

首先指出，图的任意一条最短路径既不能包含负权回路，也不会包含正权回路，因此它最多包含|v|-1条边。

其次，从源点s可达的所有顶点如果 存在最短路径，则这些最短路径构成一个以s为根的最短路径树。Bellman-Ford算法的迭代松弛操作，实际上就是按顶点距离s的层次，逐层生成这棵最短路径树的过程。

在对每条边进行1遍松弛的时候，生成了从s出发，层次至多为1的那些树枝。也就是说，找到了与s至多有1条边相联的那些顶点的最短路径；对每条边进行第2遍松弛的时候，生成了第2层次的树枝，就是说找到了经过2条边相连的那些顶点的最短路径……。因为最短路径最多只包含|v|-1 条边，所以，只需要循环|v|-1 次。

每实施一次松弛操作，最短路径树上就会有一层顶点达到其最短距离，此后这层顶点的最短距离值就会一直保持不变，不再受后续松弛操作的影响。（但是，每次还要判断松弛，这里浪费了大量的时间，怎么优化？单纯的优化是否可行？）

注意：上述只对正权图有效。如果存在负权不一定第i次就能确定最短路，且与边的顺序有关。

如果没有负权回路，由于最短路径树的高度最多只能是|v|-1，所以最多经过|v|-1遍松弛操作后，所有从s可达的顶点必将求出最短距离。如果 d[v]仍保持 +∞，则表明从s到v不可达。

如果有负权回路，那么第 |v| 遍松弛操作仍然会成功，这时，负权回路上的顶点不会收敛。