POJ-3259-Wormholes
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 35294 | Accepted: 12881 |
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.
题目大意是说John从出发点通过一些路径和昆虫洞，最后能返回出发点，并且时间早于出发时间，要求你判断所给的路径和昆虫洞是否能够满足这一要求
其实想明白了以后就是判断是否存在负权回路，我用的Spfa做的；
这里注意一下：path是双向的，wormhole是单向的。
Source
#include <stdio.h>
#include <stdlib.h>
#include <queue>
#include <string.h>
using namespace std;
const int nmax=2505;
const int inf=0xfffffff;
queue<int>q;
struct edge{
int to;
int val;
int next;
}edge[nmax*nmax];
int dis[nmax];
int cout[nmax];
int hand[nmax];
int top;
int n,m,w;
void addedge(int f,int t,int v) //静态邻接表
{
edge[top].to=t;
edge[top].val=v;
edge[top].next=hand[f];
hand[f]=top++;
}
int spfa(int start)
{
int i;
int inqueue[nmax];
memset(inqueue,0,sizeof(inqueue));
for(i=0;i<=n;i++)
dis[i]=inf;
dis[start]=0;
memset(cout,0,sizeof(cout));
int now=1;
q.push(start);
cout[start]++;
inqueue[start]=1;
while(!q.empty())
{
now=q.front();
q.pop();
inqueue[now]=0;
for(i=hand[now];i!=-1;i=edge[i].next)
{
int t=edge[i].to;
int val=edge[i].val;
if(dis[now]+val<dis[t])
{
dis[t]=dis[now]+val;
if(!inqueue[t])
{
cout[t]++;
q.push(t);
inqueue[t]=1;
if(cout[t]>n)
return 1;
}
}
}
}
return 0;
}
int main()
{
int t,i,j;
scanf("%d",&t);
while(t--)
{
scanf("%d%d%d",&n,&m,&w);
int a,b,c;
memset(hand,-1,sizeof(hand));
top=0;
while(m--)
{
scanf("%d%d%d",&a,&b,&c);//双向的；
addedge(a,b,c);
addedge(b,a,c);
}
while(w--)
{
scanf("%d%d%d",&a,&b,&c);//单向的，把它负的就可以了；
addedge(a,b,-c);
}
if(spfa(1))
printf("YES\n");//存在负环说明能提前返回。
else
printf("NO\n");
}
return 0;
}