poj3259 Wormholes
2014-08-14 09:44
162 查看
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
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.
题意:N块地,M条路,W个虫洞。判断有没有可以是时间倒流的路径。注意把W次输入的w[i]改成-w[i]。
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.
题意:N块地,M条路,W个虫洞。判断有没有可以是时间倒流的路径。注意把W次输入的w[i]改成-w[i]。
SPFA:判断有没有入队次数超过N的点。
特别提醒:
不要相信给出的数据范围,这是个大坑!!测试数据绝对有超过很多的,我开了20000的数组。
#include<cstdio> #include<cstring> #include<string> #include<queue> #include<algorithm> #include<queue> #include<map> #include<stack> #include<iostream> #include<list> #include<set> #include<cmath> #define INF 0x3f // 一个大数 #define eps 1e-6 using namespace std; #define maxn 20000 //数组开大些! #define Maxn 20000 int n,m,W; int first[maxn]; int next[maxn]; int u[Maxn]; int v[Maxn]; int w[Maxn]; int dis[maxn]; int go[maxn]; int vist[maxn]; int conut[maxn]; int ttt; queue<int> Q; int spfa(int x) { vist[x]=1; memset(dis,INF,sizeof(dis)); dis[x]=0; Q.push(x); conut[x]++; while(!Q.empty()) { int t=Q.front(); Q.pop(); vist[t]=0; conut[t]++; for(int i=first[t];i!=-1;i=next[i]) { if(dis[t]+w[i]<dis[v[i]]) { dis[v[i]]=dis[t]+w[i]; if(!vist[v[i]]) { if(conut[t]>=n) //如果某点入队超过n次,则可判定存在负环。。。。 return 1; Q.push(v[i]); ++conut[v[i]]; } } } } return 0; } int main() { int T; cin>>T; while(T--) { memset(vist,0,sizeof(vist)); memset(conut,0,sizeof(conut)); while(!Q.empty()) Q.pop(); // cout<<"二货@"<<endl; scanf("%d%d%d",&n,&m,&W); for(int i=1;i<=n;i++) first[i]=-1; for(int i=1;i<=m;i++) { scanf("%d%d%d",&u[i],&v[i],&w[i]); u[i+m]=v[i]; v[i+m]=u[i]; w[i+m]=w[i]; } for(int i=m*2+1;i<=2*m+W;i++) { scanf("%d%d%d",&u[i],&v[i],&w[i]); w[i]=-w[i]; } for(int i=1;i<=2*m+W;i++) { next[i]=first[u[i]]; first[u[i]]=i; } int qq=spfa(1); if(qq) puts("YES"); else puts("NO"); } return 0; }
相关文章推荐
- poj3259 Wormholes
- POJ3259-Wormholes(SPFA 判负环)
- poj3259 Wormholes
- POJ3259 Wormholes(虫洞)(Bellman-floyd法解决)
- POJ3259----Wormholes(最短路)
- poj3259 Wormholes (Bellman-Ford算法)
- POJ3259 - Wormholes - 最短路判负环
- POJ3259 Wormholes 洛谷P3385 【模板】负环
- POJ3259 Wormholes(最短路,有无负环,spfa,模板)
- POJ3259-Wormholes
- POJ3259——Wormholes
- POJ3259 Wormholes 找负环
- POJ3259---Wormholes(最短路:验证存在负环)
- poj3259——Wormholes
- POJ3259 Wormholes
- POJ3259 Wormholes 【Bellmanford判断是否存在负回路】
- POJ3259 Wormholes (Bellman-Ford最短路径算法)
- 【最短路】poj3259-Wormholes(Bellman-Ford 最短路)
- POJ3259-Wormholes
- 【poj3259】Wormholes 【USACO 2006 December Gold】