poj 3259 Wormholes （spfa求最短路）

[align=center]Wormholes[/align]

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 27964		Accepted: 10058

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 toF (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 toE 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.

解题思路：首先，应知道图不一定是连通的，只有图中存在负环就输出“YES”。因此，用spfa算法时，如果一次从源点的松弛不能到达所有边，必须把剩下的再用一次spfa算法。

#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
#define N 505
const int inf=10005;
int map

;         //记录点之间关系
int dis
;       //记录当前点到源点的距离
int mark
;     //标记该点是否在队列内
int num
,n;       //记录一次搜索中某点入队的次数
int Min(int a,int b)
{
return a<b?a:b;
}
int spfa()
{
int i,s=1;
queue<int>q;
memset(num,0,sizeof(num));
while(1)
{
if(q.empty())
{
for(i=1;i<=n;i++)
if(!num[i])      //若源点没有和所有的点连通，
break;
if(i<=n)
{
q.push(i);    //则把后面的点入队，在进行一次spfa算法
memset(mark,0,sizeof(mark));
memset(num,0,sizeof(num));
for(;i>0;i--)
num[i]=1;      //先把前面的不可到点标记
}
else
return 0;
}
s=q.front();
q.pop();
mark[s]=0;
for(i=1;i<=n;i++)
{
if(dis[i]>dis[s]+map[s][i])
{
dis[i]=dis[s]+map[s][i];
if(!mark[i])
{
mark[i]=1;
num[i]++;
q.push(i);
if(num[i]>n)      //存在负环
return 1;
}
}
}
}
return 0;
}
int main()
{
int T,a,b,c,m,w,i,j;
scanf("%d",&T);
while(T--)
{
scanf("%d%d%d",&n,&m,&w);
for(i=1;i<=n;i++)
{
for(j=1;j<=n;j++)
map[i][j]=inf;
dis[i]=inf;
}
while(m--)
{
scanf("%d%d%d",&a,&b,&c);
map[a][b]=Min(map[a][b],c);    //两点存在多条路
map[b][a]=Min(map[b][a],c);     //bidirectional双向边
}
while(w--)
{
scanf("%d%d%d",&a,&b,&c);
map[a][b]=Min(map[a][b],0-c);
}
if(spfa())
printf("YES\n");
else
printf("NO\n");
}
return 0;
}