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POJ 3259 Wormholes（SPFA判负权回路）

2016-09-08 19:47 363 查看

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Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 44849		Accepted: 16528

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

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

题意：
John的农场里N块地，M条路连接两块地，W个虫洞；路是双向的，虫洞是一条单向路，会在你离开之前把你传送到目的地，

就是当你过去的时候时间会倒退Ts。我们的任务是知道会不会在从某块地出发后又回来，看到了离开之前的自己。

简言之，就是看图中有没有负权环。

思路：
判断每个点的入队次数，如果大于N（图中总的点数），就是有负环，SPFA随便搞一下，是个基础结论题

代码：

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
using namespace std;

//SPFA   实际上是BFS
const  int  inf=1e9;
const  int  maxn=555;
vector<pair<int,int> >E[maxn];//邻接表
int n,m;
int cnt[maxn];
int d[maxn],inq[maxn];

void  init(){
for(int i=0; i<maxn; i++)E[i].clear();
for(int i=0; i<maxn; i++)inq[i]=0;
for(int i=0; i<maxn; i++)d[i]=inf;
for(int i=0; i<maxn; i++)cnt[i]=0;
}

bool spfa(int n){
queue<int>q;
q.push(1),d[1]=0,inq[1]=1;
while(!q.empty()){
int now=q.front();
q.pop();inq[now]=0;
for(int i=0; i<E[now].size(); i++){
int  v=E[now][i].first;
if(d[v]>d[now]+E[now][i].second){
d[v]=d[now]+E[now][i].second;
if(inq[v]==1)continue;
inq[v]=1;
q.push(v);
if(++cnt[v]>n)return true;
}
}
}
return false;
}

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; i++){
int x,y,z;scanf("%d %d %d",&x,&y,&z);
E[x].push_back(make_pair(y,z));  //边x,y,边权z
E[y].push_back(make_pair(x,z));
}
for(int i=0; i<W; i++){
int x,y,z;scanf("%d %d %d",&x,&y,&z);
E[x].push_back(make_pair(y,-z));
}
if(spfa(n))printf("YES\n");
else printf("NO\n");
}
return  0;
}

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