• 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;
    }
    低调做人,高调做事。
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  • 原文地址:https://www.cnblogs.com/Asimple/p/6137586.html
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