Wormholes
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 24249 | Accepted: 8652 |
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.
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.
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
这题就是判断存不存在负环回路。
前M条是双向边，后面的W是单向的负边。
为了防止出现不连通，
增加一个结点作为起点。
起点到所有点的长度为0
bellman_ford算法：
/* * POJ 3259 * 判断图中是否存在负环回路。 * 为了防止图不连通的情况，增加一个点作为起点，这个点和其余的点都相连。 */ #include <iostream> #include <stdio.h> #include <string.h> #include <algorithm> #include <vector> #include <queue> using namespace std; /* * 单源最短路bellman_ford算法，复杂度O(VE) * 可以处理负边权图。 * 可以判断是否存在负环回路。返回true,当且仅当图中不包含从源点可达的负权回路 * vector<Edge>E;先E.clear()初始化，然后加入所有边 * 点的编号从1开始(从0开始简单修改就可以了) */ const int INF=0x3f3f3f3f; const int MAXN=550; int dist[MAXN]; struct Edge { int u,v; int cost; Edge(int _u=0,int _v=0,int _cost=0):u(_u),v(_v),cost(_cost){} }; vector<Edge>E; bool bellman_ford(int start,int n)//点的编号从1开始 { for(int i=1;i<=n;i++)dist[i]=INF; dist[start]=0; for(int i=1;i<n;i++)//最多做n-1次 { bool flag=false; for(int j=0;j<E.size();j++) { int u=E[j].u; int v=E[j].v; int cost=E[j].cost; if(dist[v]>dist[u]+cost) { dist[v]=dist[u]+cost; flag=true; } } if(!flag)return true;//没有负环回路 } for(int j=0;j<E.size();j++) if(dist[E[j].v]>dist[E[j].u]+E[j].cost) return false;//有负环回路 return true;//没有负环回路 } int main() { // freopen("in.txt","r",stdin); // freopen("out.txt","w",stdout); int T; int N,M,W; int a,b,c; scanf("%d",&T); while(T--) { scanf("%d%d%d",&N,&M,&W); E.clear(); while(M--) { scanf("%d%d%d",&a,&b,&c); E.push_back(Edge(a,b,c)); E.push_back(Edge(b,a,c)); } while(W--) { scanf("%d%d%d",&a,&b,&c); E.push_back(Edge(a,b,-c)); } for(int i=1;i<=N;i++) E.push_back(Edge(N+1,i,0)); if(!bellman_ford(N+1,N+1))printf("YES "); else printf("NO "); } return 0; }
SPFA算法：
//============================================================================ // Name : POJ.cpp // Author : // Version : // Copyright : Your copyright notice // Description : Hello World in C++, Ansi-style //============================================================================ #include <iostream> #include <string.h> #include <stdio.h> #include <algorithm> #include <vector> #include <queue> using namespace std; /* * 单源最短路SPFA * 时间复杂度 0(kE) * 这个是队列实现，有时候改成栈实现会更加快，很容易修改 * 这个复杂度是不定的 */ const int MAXN=1010; const int INF=0x3f3f3f3f; struct Edge { int v; int cost; Edge(int _v=0,int _cost=0):v(_v),cost(_cost){} }; vector<Edge>E[MAXN]; void addedge(int u,int v,int w) { E[u].push_back(Edge(v,w)); } bool vis[MAXN]; int cnt[MAXN]; int dist[MAXN]; bool SPFA(int start,int n) { memset(vis,false,sizeof(vis)); for(int i=1;i<=n;i++)dist[i]=INF; dist[start]=0; vis[start]=true; queue<int>que; while(!que.empty())que.pop(); que.push(start); memset(cnt,0,sizeof(cnt)); cnt[start]=1; while(!que.empty()) { int u=que.front(); que.pop(); vis[u]=false; for(int i=0;i<E[u].size();i++) { int v=E[u][i].v; if(dist[v]>dist[u]+E[u][i].cost) { dist[v]=dist[u]+E[u][i].cost; if(!vis[v]) { vis[v]=true; que.push(v); if(++cnt[v]>n)return false; //有负环回路 } } } } return true; } int main() { // freopen("in.txt","r",stdin); // freopen("out.txt","w",stdout); int T; int N,M,W; int a,b,c; scanf("%d",&T); while(T--) { scanf("%d%d%d",&N,&M,&W); for(int i=1;i<=N+1;i++)E[i].clear(); while(M--) { scanf("%d%d%d",&a,&b,&c); addedge(a,b,c); addedge(b,a,c); } while(W--) { scanf("%d%d%d",&a,&b,&c); addedge(a,b,-c); } for(int i=1;i<=N;i++) addedge(N+1,i,0); if(!SPFA(N+1,N+1))printf("YES "); else printf("NO "); } return 0; }