poj 3259 Wormholes （SPFA 判断有无负权回路）
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Wormholes
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
USACO 2006 December Gold
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第一次用SPFA做最短路，还是贴一下吧，基本上就是裸题，就不多说了。
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#define maxn 505
#define maxx 5500
using namespace std;
int n,m,cxx,ans,cnt,sx;
bool vis[maxn];
int dist[maxn],p[maxn],num[maxn];
struct Node
{
int r,cost;
int next;
}edge[maxx];
queue<int>q;
void init()
{
memset(vis,0,sizeof(vis));
memset(p,0,sizeof(p));
memset(num,0,sizeof(num));
memset(dist,0x3f,sizeof(dist));
}
void addage(int u,int v,int w)
{
cnt++;
edge[cnt].r=v;
edge[cnt].cost=w;
edge[cnt].next=p[u];
p[u]=cnt;
}
bool SPFA()
{
int i,j,nx;
sx=1;
while(!q.empty()) q.pop();
dist[sx]=0;
vis[sx]=1;
num[sx]++;
q.push(sx);
while(!q.empty())
{
nx=q.front();
vis[nx]=0;
q.pop();
for(i=p[nx];i;i=edge[i].next)
{
if(dist[edge[i].r]>dist[nx]+edge[i].cost)
{
dist[edge[i].r]=dist[nx]+edge[i].cost;
if(!vis[edge[i].r])
{
vis[edge[i].r]=1;
num[edge[i].r]++;
if(num[edge[i].r]>n) return true ;
q.push(edge[i].r);
}
}
}
}
return false ;
}
int main()
{
int i,j,t,u,v,w;
scanf("%d",&t);
while(t--)
{
scanf("%d%d%d",&n,&m,&cxx);
init();
cnt=0;
for(i=1;i<=m;i++)
{
scanf("%d%d%d",&u,&v,&w);
addage(u,v,w);
addage(v,u,w);
}
for(i=1;i<=cxx;i++)
{
scanf("%d%d%d",&u,&v,&w);
addage(u,v,-w);
}
if(SPFA()) printf("YES\n");
else printf("NO\n");
}
return 0;
}