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.
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.
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。
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。