Poj 3259 Wormholes【spfa 负环判断】
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 40566 | Accepted: 14864 |
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 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
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 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
题意：
给出n 顶点，以及某些点之间移动需要消耗的时间，另有一些单向的边，权值为负，表示可以时间可以倒流多少，给出整个图的信息，问这个人能否在任意起点出发，最终回到起点时的时间在出发的时间之前
题解：“
不难发现，如果图中出现了边的总权值为负的环，那么肯定可以满足题意（无限在环上跑，时间不是问题），也就是需要进行是否有负环的判断，
用spfa 进行最短路操作的时候，可以顺带发现负环...
/*
http://blog.csdn.net/liuke19950717
*/
#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
using namespace std;
const int inf=0x3f3f3f3f;
const int maxn=2505;
int edgenum,head[maxn];
struct node
{
int from,to,val;
int next;
}edge[maxn*maxn];
void init()
{
edgenum=0;
memset(head,-1,sizeof(head));
}
void add(int a,int b,int c)
{
node tp={a,b,c,head[a]};
edge[edgenum]=tp;
head[a]=edgenum++;
}
void slove(int n,int st)
{
int dist[maxn],vis[maxn]={0},num[maxn]={0};
memset(dist,inf,sizeof(dist));
vis[st]=1;num[st]=1;dist[st]=0;
queue<int> q;
q.push(st);
while(!q.empty())
{
int u=q.front();q.pop();
vis[u]=0;
for(int i=head[u];i!=-1;i=edge[i].next)
{
int v=edge[i].to;
if(dist[v]>dist[u]+edge[i].val)
{
dist[v]=dist[u]+edge[i].val;
if(!vis[v])
{
vis[v]=1;++num[v];
q.push(v);
if(num[v]>n)//负环
{
printf("YES\n");
return;
}
}
}
}
}
printf("NO\n");
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int n,m,w;
scanf("%d%d%d",&n,&m,&w);
init();
for(int i=0;i<m+w;++i)
{
int a,b,c;
scanf("%d%d%d",&a,&b,&c);
i<m?add(a,b,c),add(b,a,c):add(a,b,-c);
}
slove(n,1);
}
return 0;
}
ps：
本来该是学最短路的时候做这样的题目，到现在才做，真是够了.....图论也忘记的差不多了....