POJ3259----Wormholes(最短路)

Wormholes

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 33948 Accepted: 12376

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 toF (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 toE 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注：主要用spfa判断负圈的存在，测试数据有点水，因为各点并非是全连通的，此时可以通过由0点向每个点加一条权值为0的边。source code://双端队列实现的SLF优化的spfa
#include <cstdio>#include <cstring>#include <deque>using namespace std;const int inf = 30000000;const int maxn = 505;const int maxm = 50000;int head[maxn], next[maxm], vis[maxn], dis[maxn], cnt[maxn];int n, m, w, e;struct node{    int u, v, c;    node(){}    node(int u, int v, int c):u(u), v(v), c(c){}}p[maxm];void addnode(int u, int v, int c){    p[e] = node(u, v, c);    next[e] = head[u];    head[u] = e++;}bool relax(int u, int v, int c){    if (dis[v] > dis[u] + c) {        dis[v] = dis[u] + c;        return true;    }    return false;}void init(){    e = 0;    memset(head, -1, sizeof(head));    memset(next, -1, sizeof(next));}bool spfa(){    for (int i = 1; i <= n; i++)        dis[i] = inf;    memset(vis, 0, sizeof(vis));    memset(cnt, 0, sizeof(cnt));    dis[1] = 0;    vis[1] = 1;    deque<int> que;    while (!que.empty())        que.pop_front();    que.push_back(1);    while (!que.empty()) {        int u = que.front();        que.pop_front();        vis[u] = 0;        for (int i = head[u]; i + 1; i = next[i]) {            if (relax(u, p[i].v, p[i].c) && !vis[p[i].v]) {                if ((++cnt[p[i].v]) > n)                    return true;                vis[p[i].v] = 1;                if (!que.empty()) {                    if (dis[p[i].v] > dis[que.front()])                        que.push_back(p[i].v);                    else                        que.push_front(p[i].v);                }                else                    que.push_back(p[i].v);            }        }    }    return false;}int main(){    //freopen("F:\\test.txt", "r", stdin);    int cas, u, v, c;    scanf("%d", &cas);    while (cas--) {        init();//        for (int i = 1; i <= n; i++) {  //通过由0点向每个点加一条边，权值为0//            addnode(0, i, 0);//                //addnode(i, 0, 0);//        }        scanf("%d%d%d", &n, &m, &w);        while (m--) {            scanf("%d%d%d", &u, &v, &c);            addnode(u, v, c);            addnode(v, u, c);        }        while (w--) {            scanf("%d%d%d", &u, &v, &c);            addnode(u, v, -c);        }        if (spfa())            puts("YES");        else            puts("NO");    }    return 0;}