poj 3259 Wormholes //SPFA
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 10267 | Accepted: 3640 |
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
题目描述的太麻烦了
就是测试有没有负权回路。
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
const int V=510;
const int E=10000;
const int INF=0x7fffffff;
struct EDGE
{
int link,val,next;
}edge[E];
int map[V][V];
int head[V],cnt[V],dist[V];
bool vis[V];
int e;
void addedge(int x,int y,int c)
{
edge[e].link=y;
edge[e].val=c;
edge[e].next=head[x];
head[x]=e++;
}
bool relax(int u,int v,int val)
{
if(dist[u]+val<dist[v])
{
dist[v]=dist[u]+val;
return true;
}
return false;
}
bool SPFA(int src,int n)
{
memset(vis,false,sizeof(vis));
memset(cnt,0,sizeof(cnt));
for(int i=1;i<=n;i++) dist[i]=INF;
dist[src]=0;
vis[src]=true;
queue<int> q;
q.push(src);
++cnt[src];
while(!q.empty())
{
int u,v;
u=q.front();
q.pop();
vis[u]=false;
for(int i=head[u];i!=-1;i=edge[i].next)
{
v=edge[i].link;
if(relax(u,v,edge[i].val)&&!vis[v])
{
q.push(v);
vis[v]=true;
++cnt[v];
if(cnt[v]>=n) return true;;
}
}
}
return false;
}
int main()
{
int T,n,m,w;
scanf("%d",&T);
while(T--)
{
e=0;
memset(map,-1,sizeof(map));
memset(head,-1,sizeof(head));
scanf("%d%d%d",&n,&m,&w);
for(int i=1;i<=m;i++)
{
int x,y,c;
scanf("%d%d%d",&x,&y,&c);
if(map[x][y]==-1) map[x][y]=map[y][x]=c;
else map[x][y]=map[y][x]=min(map[x][y],c);
}
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
if(map[i][j]!=-1) addedge(i,j,map[i][j]);
for(int i=1;i<=w;i++)
{
int x,y,c;
scanf("%d%d%d",&x,&y,&c);
addedge(x,y,-c);
}
if(SPFA(1,n)) printf("YES/n");
else printf("NO/n");
}
return 0;
}