Bellman-Ford算法模板题——POJ 3259
2014-12-02 14:25
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Wormholes
Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%I64d
& %I64u
Submit Status Practice POJ
3259
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
懒得自己打题意~
这道题是是求是否存在负权环。
题意 : 一个famer有一些农场,这些农场里面有一些田地,田地里面有一些虫洞,田地和田地之间有路,虫洞有这样的性质: 时间倒流。问你这个农民能不能看到他自己,也就是说,有没有这样一条路径,能利用虫洞的时间倒流的性质,让这个人能在这个点出发前回去,这样他就是能看到他自己
解题思路:使用Bellman-Ford算法,看图中有没有负权环。有的话就是可以,没有的话就是不可以了。
输入N,M,W,表示N个点,M表示两点间无向,W表示两点间有向
Wormholes
Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%I64d
& %I64u
Submit Status Practice POJ
3259
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
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
懒得自己打题意~
这道题是是求是否存在负权环。
题意 : 一个famer有一些农场,这些农场里面有一些田地,田地里面有一些虫洞,田地和田地之间有路,虫洞有这样的性质: 时间倒流。问你这个农民能不能看到他自己,也就是说,有没有这样一条路径,能利用虫洞的时间倒流的性质,让这个人能在这个点出发前回去,这样他就是能看到他自己
解题思路:使用Bellman-Ford算法,看图中有没有负权环。有的话就是可以,没有的话就是不可以了。
输入N,M,W,表示N个点,M表示两点间无向,W表示两点间有向
#include <iostream> #include <cstdio> #include <cstdlib> #include <cmath> #include <cstring> #include <queue> #include <algorithm> using namespace std; const int MAXN=10010; const int INF=1<<30; int dis[505]; struct edge { int u,v,t; }E[6000]; bool Bellman_Ford(int n, int m) { int i,j; for(i=1; i<=n; i++) dis[i]=INF; dis[1]=0; for(i=1; i<=n-1; i++){ for(j=0; j<m; j++){ if(dis[E[j].u]+E[j].t < dis[E[j].v]) dis[E[j].v] = dis[E[j].u]+E[j].t; //cout<<E[j].u<<" "<<E[j].v<<" "<<dis[E[j].u]<<" "<<dis[E[j].v]<<endl; } } for(j=0; j<m; j++){ if(dis[E[j].v] > dis[E[j].u]+E[j].t) return 1; } return 0; } int main() { //freopen("in.txt","r",stdin); int T; scanf("%d", &T); while(T--) { int n,m,w; scanf("%d%d%d", &n,&m,&w); int size=0; int i,j; for(i=1; i<=m; i++){ int u,v,t; scanf("%d%d%d", &u,&v,&t); E[size].u=u; E[size].v=v; E[size].t=t; size++; E[size].v=u; E[size].u=v; E[size].t=t; size++; } for(i=1; i<=w; i++){ int u,v,t; scanf("%d%d%d", &u,&v,&t); E[size].u=u; E[size].v=v; E[size].t=-t; size++; } int ok=Bellman_Ford(n,size); if(ok) printf("YES\n"); else printf("NO\n"); //cout<<m+w<<" "<<size<<endl; } return 0; }
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