POJ

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.

题目大意就是：农夫约翰有F个农场，每个农场有N块地，其间有M条路（无向），W条时光隧道（有向且时间倒流即：权值为负）。问是否可能回到过去？

经典的bellman_Ford理解题，不知道的可以去百度！

//Asimple
#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cctype>
#include <cstdlib>
#include <stack>
#include <cmath>
#include <set>
#include <map>
#include <string>
#include <queue>
#include <limits.h>
#include <time.h>
#define INF 0x3f3f3f3f
using namespace std;
const int maxn = 6005;
typedef long long ll;
int n, m, num, T, k, x, y, len;
int Map[maxn][maxn];
int dis[maxn];
typedef struct node {
    int begin;
    int end;
    int weight;
    node(){}
    node(int begin, int end, int weight) {
        this->begin = begin;
        this->end = end;
        this->weight = weight;
    }
}eee;
eee edg[maxn];
//Bellman-Ford算法：求含负权图的单源最短路径算法 
//单源最短路径(从源点s到其它所有顶点v)
bool bellmanFord() {
    memset(dis, 0, sizeof(dis));
    for(int i=0; i<n; i++) {
        for(int j=0; j<len; j++) {
            eee e = node(edg[j].begin, edg[j].end, edg[j].weight);
            if( dis[e.end] > dis[e.begin] + e.weight) {
                dis[e.end] = dis[e.begin] + e.weight;
                if( i == n-1 ) return true;
            }
        }
    }
    return false;
}

void input() {
    cin >> T ;
    while( T -- )  {
        cin >> n >> m >> k;
        len = 0;
        for(int i=0; i<m; i++) {
            cin >> x >> y >> num;
            edg[len].begin = x;
            edg[len].end = y;
            edg[len].weight = num;
            len ++;
            edg[len].begin = y;
            edg[len].end = x;
            edg[len].weight = num;
            len ++;
        }
        for(int i=0; i<k; i++) {
            cin >> x >> y >> num ;
            edg[len].begin = x;
            edg[len].end = y;
            edg[len].weight = -num;
            len ++;
        }
        if( bellmanFord() ) cout << "YES" << endl;
        else cout << "NO" << endl; 
    }
}

int main(){
    input();
    return 0;
}

2017-5-26　　修改：

自己写了一个邻接矩阵的SPFA解法

坑点：可能会出现重复的路径，这个时候需要取小值。

#include <iostream>
#include <cstring>
#include <queue>
using namespace std;
const int maxn = 500+5;
const int INF = (1 << 20 );
int n, m, x, y, num, T, k;
int Map[maxn][maxn], dis[maxn], c[maxn];

void init(){
    for(int i=0; i<=n; i++) {
        dis[i] = INF;
        c[i] = 0;
        for(int j=0; j<=n; j++) {
            Map[i][j] = INF;
        }
    }
}

bool spfa(){
    bool vis[maxn];
    queue<int> q;
    memset(vis, false, sizeof(vis));
    q.push(1);
    vis[1] = true;
    c[1] = 1;
    dis[1] = 0;
    while( !q.empty() ) {
        x = q.front();q.pop();
        vis[x] = false;
        for(int i=1; i<=n; i++) {
            if( dis[i]>dis[x]+Map[x][i] ) {
                dis[i] = dis[x]+Map[x][i];
                if( !vis[i] ) {
                    vis[i] = true;
                    c[i] ++;
                    if( c[i]>=n ) return true;
                    q.push(i);
                }
            }
        }
    }
    return false;
}

int main(){
    cin >> T;
    while( T -- ) {
        cin >> n >> m >> k;
        init();
        while( m -- ) {
            cin >> x >> y >> num;
            Map[x][y] = min(Map[x][y], num);
            Map[y][x] = Map[x][y];
        }
        while( k -- ) {
            cin >> x >> y >> num;
            Map[x][y] = min(Map[x][y], -num);
        }
        if( spfa() ) cout << "YES" << endl;
        else cout << "NO" << endl;
    }
    return 0;
}