poj 3259 Wormholes （负权最短路，SPAF）

[align=center]Wormholes[/align]

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 36641		Accepted: 13405

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.
Source

题目链接：http://poj.org/problem?id=3259

题目大意：时空旅行，前m条路是双向的，旅行时间为正值，w条路是虫洞，单向的，旅行时间是负值，也就是能回到过去。求从一点出发，判断能否在”过去“回到出发点，即会到出发点的时间是负的。

解题思路：裸的负权最短路问题，SPAF解决。枚举出发点即可。

代码如下：

#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
#include <algorithm>
#define inf 1e9
using namespace std;
int const maxn=505;
int n,p;
int dist[maxn],vis[maxn],num[maxn];
struct node
{
int v,w;
node(int vv,int ww)
{
v=vv;
w=ww;
}
};
vector <node> vt[maxn];
void SPAF(int v0)
{
memset(vis,0,sizeof(vis));
memset(num,0,sizeof(num));
for(int i=0;i<maxn;i++)
dist[i]=-inf;
queue <int>q;
q.push(v0);
dist[v0]=0;
while(!q.empty())
{
int u=q.front();
q.pop();
if(num[u]>n)
continue;
num[u]++;
if(num[u]>=n)
dist[u]=inf;
vis[u]=0;
int len=vt[u].size();
for(int i=0;i<len;i++)
{
int v=vt[u][i].v;
int w=vt[u][i].w;
if(dist[v]<dist[u]+w)
{
dist[v]=dist[u]+w;
if(v==v0&&dist[v]>0)
{
p=1;
return ;
}
if(!vis[v])
{
vis[v]=1;
q.push(v);
}
}
}
}
}
int main(void)
{
int m1,m2,u,v,w,t;
scanf("%d",&t);
while(t--)
{
scanf("%d%d%d",&n,&m1,&m2);
int ans=0;
p=0;
for(int i=0;i<maxn;i++)
vt[i].clear();
for(int i=0;i<m1;i++)
{
scanf("%d%d%d",&u,&v,&w);    //双向路径
vt[u].push_back(node(v,-w));
vt[v].push_back(node(u,-w));
}
for(int i=0;i<m2;i++)
{
scanf("%d%d%d",&u,&v,&w);
vt[u].push_back(node(v,w));    //单向路径
}
for(int i=1;i<=n;i++)
{
SPAF(i);
if(p)
break;
}
if(p)
printf("YES\n");
else
printf("NO\n");
}
}