poj3668 spfa

来源：互联网发布：谢朝阳云计算编辑：程序博客网时间：2024/05/20 05:10
#include <iostream>#include <cstdio>#include <queue>#include <vector>#include <cstring>using namespace std;//求多点到一定点的距离是， 可以建立反图，后就转化为求定点到各点的最短路径问题。const int M = 1005;const int inf = 99999999;struct node {    int to;    int w;    node(int tto, int ww) : to(tto), w(ww){}};int n, m, x;int vist[M];vector<node>g[M], rg[M];int dist1[M], dist2[M];void spfa(int s, vector<node> gr[], int *dist) {    queue<int>que;    for(int i = 0; i <= n; i++)        dist[i] = inf;    memset(vist, 0, sizeof(vist));    vist[s] = 1;     que.push(s);     dist[s] = 0;     while(!que.empty()) {        int v = que.front();        que.pop();        vist[v] = 0;        for(int i = 0; i < (int)gr[v].size(); i++) {            node p = gr[v][i];            if(dist[p.to] > dist[v] + p.w) {                dist[p.to] = dist[v] + p.w;                if(!vist[p.to]) {                    vist[p.to] = 1;                    que.push(p.to);                }            }        }     }}int main(){    int a, b, t;   while(scanf("%d%d%d", &n, &m, &x) != EOF) {        for(int i = 0; i < m; i++) {            scanf("%d%d%d", &a, &b, &t);            g[a].push_back(node(b, t));            rg[b].push_back(node(a, t));        }        spfa(x, g, dist1);        spfa(x, rg, dist2);        int ans = 0;        for(int i = 1; i <= n; i++) {            if(dist1[i] != inf&&dist2[i] != inf)                ans = max(ans, dist1[i]+dist2[i]);        }        printf("%d\n", ans);   }    return 0;}
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