Wormholes (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

23 3 11 2 21 3 42 3 13 1 33 2 11 2 32 3 43 1 8

Sample Output

NOYES

#include<iostream>#include<algorithm>#include<vector>#include<queue>#include<cstring>using namespace std;//BY 郭炜老师；//Bellman-Ford算法，超时；int F, N, M, W;const int INF = 1 << 30;struct Edge{    int s, e, w;//起点，终点。权；    Edge (int ss, int ee, int ww): s(ss), e(ee), w(ww){};};vector<Edge> edges;//储存边；int dist[510];int Bellman_Ford(int v) //v是起点；{    for (int i = 1; i <= N; i++)    {        dist[i] = INF;    }    dist[v] = 0;    for (int k = 1; k < N; k++) //经过不超过k条边；    {// dist(k)[u] = min( dist(k-1)[u], min(dist(k-1)[j] + Edge[j][u])), j = 0, 1, ……, n-1; j不能是终点；     // 类似滚动数组的处理方法；        bool changed = false; //改进；        for (int i = 0; i < edges.size(); i++)        {            int s = edges[i].s;            int e = edges[i].e;            if (dist[s] + edges[i].w < dist[e])            {                dist[e] = dist[s] + edges[i].w; // 从起点v->e的最短路径长度被更新为v->s加上s->e的路径长度之和；                changed = true;            }        }        if (changed == false)            return false;    }    // k == N    for (int i = 0; i < edges.size(); i++)    {        int s = edges[i].s;        int e = edges[i].e;        /*        如果成立，说明找到了经过n条边的v->e的路径，且比任何少于n条边的从v->e的路径都短；        一共n个顶点，路径却经过了n条边，则必有一个顶点m被至少经过两次，则m是一个回路的起点和终点；        走这个回路比不走时路径更短，说明这个回路是负权回路；        */        if (dist[s] + edges[i].w < dist[e])            return true;    }    return false;}int main(){    cin >> F;    while (F--)    {        edges.clear();        cin >> N >> M >> W;        for (int i = 0; i < M; i++)        {            int s, e, t;            cin >> s >> e >> t;            edges.push_back(Edge(s, e, t));            edges.push_back(Edge(e, s, t));        }        for (int i = 0; i < W; i++)        {            int s, e, t;            cin >> s >> e >> t;            edges.push_back(Edge(s, e, -t));        }        if (Bellman_Ford(1)) //因为正权边都是双向，所以从1可达所有点；            cout << "YES" << endl;        else            cout << "NO" << endl;    }}

#include<iostream>#include<algorithm>#include<vector>#include<queue>#include<cstring>using namespace std;//SPFA算法是对Bellman-Ford算法的改进。更容易理解；int F, N, M, W;const int INF = 1 << 30;struct Edge{    int e, w; //e表示终点，w表示权值；    Edge(int ee, int ww): e(ee), w(ww){};};vector<Edge> G[1000]; //整个有向图；int updateTimes[1000]; //最短路的改进次数；int dist[1000]; //dist[i]是源点到i的目前最短路长度；int spfa(int v) //起点为v；{    for (int i = 1; i <= N; i++)        dist[i] = INF;    dist[v] = 0;    queue<int> que;    que.push(v);    memset(updateTimes, 0, sizeof(updateTimes));    while (!que.empty())    {        //每次迭代，取出队头的s，依次枚举，看是否存在改进空间；        int s = que.front();        que.pop();        for (int i = 0; i < G[s].size(); i++)        {            int e = G[s][i].e;            if (dist[e] > dist[s] + G[s][i].w) //更新抵达e的最短路径长度；            {                dist[e] = dist[s] + G[s][i].w;                que.push(e); //e有可能改进其它的点，于是将其放入队尾；                updateTimes[e]++;                /*                若一个点最短路被改进的次数达到n，则存在负权环；                其原因与Bellma-Ford算法是类似的；                n次改进说明找到了一条经过n条边的从源点s到特定点e的路径；                一共只有n个点，说明e是一个环的起点和终点；                走这个回路比不走这个回路路径更短，说明它是负权回路；                */                if (updateTimes[e] >= N)                    return true;            }        }    }    return false;}int main(){    cin >> F;    while (F--)    {        cin >> N >> M >> W;        for (int i = 1; i < 1000; i++)            G[i].clear();        int s, e, t;        for (int i = 0; i < M; i++)        {            cin >> s >> e >> t;            G[s].push_back(Edge(e, t));            G[e].push_back(Edge(s, t));        }        for (int i = 0; i < W; i++)        {            cin >> s >> e >> t;            G[s].push_back(Edge(e, -t));        }        if (spfa(1))            cout << "YES" << endl;        else            cout << "NO" << endl;    }}