Bellman-Ford||SPFA-POJ-3259-Wormholes

Wormholes
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 37628 Accepted: 13850
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

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

题目真的很难读懂，大致意思是路径是双向的，而虫洞是单向的，要求是否存在负环，用Bellman-Ford算法就可以了。

////  main.cpp//  最短路练习-F-Wormholes////  Created by 袁子涵 on 15/10/11.//  Copyright (c) 2015年 袁子涵. All rights reserved.////  532ms   800KB#include <iostream>#include <string.h>#include <stdio.h>#include <stdlib.h>#define INF 999999999using namespace std;typedef struct edge{    int S,E;    long long int T;}Edge;int F,N,M,W;Edge eg[10000];long long int sum=0,dis[1000];bool bellman_ford(){    bool flag=0;    for (int i=1; i<N; i++) {        dis[i]=INF;    }    dis[0]=0;    for (int i=0; i<N; i++) {        flag=0;        for (int j=0; j<sum; j++) {            if (dis[eg[j].E]>dis[eg[j].S]+eg[j].T && dis[eg[j].S]<INF) {                dis[eg[j].E]=dis[eg[j].S]+eg[j].T;                flag=1;            }        }        if (flag==0) {            break;        }    }    for (int i=0; i<sum; i++) {        if (dis[eg[i].E]>dis[eg[i].S]+eg[i].T && dis[eg[i].S]<INF) {            dis[eg[i].E]=dis[eg[i].S]+eg[i].T;            return 1;        }    }    return 0;}int main(int argc, const char * argv[]) {    cin >> F;    int s,e,t;    while (F--) {        cin >> N >> M >> W;        sum=0;        for (int i=0; i<M; i++) {            cin >> s >> e >> t;            eg[sum].S=s-1;            eg[sum].E=e-1;            eg[sum++].T=t;            eg[sum].S=e-1;            eg[sum].E=s-1;            eg[sum++].T=t;        }        for (int i=0; i<W; i++) {            cin >> s >> e >> t;            eg[sum].S=s-1;            eg[sum].E=e-1;            eg[sum++].T=-t;        }        if (bellman_ford()) {            cout << "YES" << endl;        }        else            cout << "NO" << endl;    }    return 0;}

只是判断负环而已，所以还能用SPFA做。

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////  main.cpp//  POJ-3259-Wormholes////  Created by 袁子涵 on 15/11/28.//  Copyright © 2015年 袁子涵. All rights reserved.////  125ms   812KB#include <iostream>#include <stdio.h>#include <string.h>#include <algorithm>#include <math.h>#include <stdlib.h>#include <vector>#include <queue>#define MAXN 505#define INF 0x3f3f3f3fusing namespace std;int f,n,m,w;struct Edge{    int v,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){    memset(vis, 0, sizeof(vis));    for (int i=1; i<=n; i++) {        dist[i]=INF;    }    dist[start]=0;    vis[start]=1;    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]=1;                    que.push(v);                    if (++cnt[v]>n) {                        return 0;                    }                }            }        }    }    return 1;}int main(int argc, const char * argv[]) {    cin >> f;    int a,b,c;    while (f--) {        cin >> n >> m >> w;        for (int i=1; i<=n; i++) {            E[i].clear();        }        for (int i=1; i<=m; i++) {            scanf("%d%d%d",&a,&b,&c);            addedge(a, b, c);            addedge(b, a, c);        }        for (int i=1; i<=w; i++) {            scanf("%d%d%d",&a,&b,&c);            addedge(a, b, -c);        }        if(SPFA(1))            printf("NO\n");        else            printf("YES\n");    }    return 0;}