poj 3259 Wormholes 【SPFA&&判断负环】
2015-08-21 09:43
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Wormholes
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
分析:
英语不好,就不在翻译了。意思就是求最短路,如果最短路中有负环,输出yes,没有输出no。
代码:
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 36852 | Accepted: 13502 |
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
分析:
英语不好,就不在翻译了。意思就是求最短路,如果最短路中有负环,输出yes,没有输出no。
代码:
#include<cstdio> #include<cstring> #include<queue> #include<algorithm> #include<cstdlib> #define INF 0x3f3f3f3f #define maxn 10010 using namespace std; int N,M,W; int pnum; int dis[maxn]; int vis[maxn]; int head[maxn]; int used[maxn]; bool cnt; struct node{ int from; int to; int val; int next; }; node pp[2*maxn]; void addpp(int u,int v,int w) { node E ={u,v,w,head[u]}; pp[pnum]=E; head[u]=pnum++; } void init() { int a,b,c; while(M--) { scanf("%d%d%d",&a,&b,&c); addpp(a,b,c); addpp(b,a,c); } } void gmap() { int a,b,c; while(W--) { scanf("%d%d%d",&a,&b,&c); addpp(a,b,-c); } } void SPFA(int s) { queue<int>q; memset(dis,INF,sizeof(dis)); memset(vis,0,sizeof(vis)); memset(used,0,sizeof(used)); dis[s]=0; vis[s]=1; used[s]++; q.push(s); while(!q.empty()) { int u=q.front(); q.pop(); vis[u]=0; for(int i=head[u];i!=-1;i=pp[i].next) { int v=pp[i].to; if(dis[v]>dis[u]+pp[i].val) { dis[v]=dis[u]+pp[i].val; if(!vis[v]) { vis[v]=1; q.push(v); used[v]++; if(used[v]>N) { cnt=true; return; } } } } } } int main(){ int T; scanf("%d",&T); while(T--) { pnum=0; cnt=false; memset(head,-1,sizeof(head)); scanf("%d%d%d",&N,&M,&W); init(); gmap(); SPFA(1); if(cnt) printf("YES\n"); else printf("NO\n"); } return 0; }
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