poj3268最短路Dijkstrra

题目大意：给出n个点和m条边，接着是m条边，代表从牛a到牛b需要花费c时间，现在所有牛要到牛x那里去参加聚会，并且所有牛参加聚会后还要回来，给你牛x，除了牛x之外的牛，他们都有一个参加聚会并且回来的最短时间，从这些最短时间里找出一个最大值输出

D - Silver Cow Party

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status

Description

One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow party to be held at farm #X (1 ≤ X ≤ N). A total of M (1 ≤ M ≤ 100,000) unidirectional (one-way roads connects pairs of farms; road i requires T_i (1 ≤ T_i ≤ 100) units of time to traverse.

Each cow must walk to the party and, when the party is over, return to her farm. Each cow is lazy and thus picks an optimal route with the shortest time. A cow's return route might be different from her original route to the party since roads are one-way.

Of all the cows, what is the longest amount of time a cow must spend walking to the party and back?

Input

Line 1: Three space-separated integers, respectively: N, M, and X
Lines 2.. M+1: Line i+1 describes road i with three space-separated integers: A_i, B_i, and T_i. The described road runs from farm A_i to farm B_i, requiring T_itime units to traverse.

Output

Line 1: One integer: the maximum of time any one cow must walk.

Sample Input

4 8 21 2 41 3 21 4 72 1 12 3 53 1 23 4 44 2 3

Sample Output

10

#include <iostream>#include <stdio.h>#define INF 0xfffffffconst int MAx = 100005;using namespace std;struct edge{   int from,to,cost;}es[MAx];int d[MAx];int ans[MAx];int V,E;void meshort(int s)///从s点到任意点最短距离{    for(int i=0;i<=V;i++)    d[i]=INF;    d[s]=0;    while(1)    {        bool updata=false;        for(int i=0;i<E;i++)        {            edge e=es[i];            if(d[e.from]!=INF&&d[e.to]>d[e.from]+e.cost)            {                d[e.to]=d[e.from]+e.cost;                updata=true;            }        }        if(!updata)        break;    }}int me_max(int a,int b){    if(a<b)    return b;    return a;}int main(){    int endd;    int ed,st,ct;    while(~scanf("%d%d%d",&V,&E,&endd))    {        for(int i=0;i<E;i++)        {            scanf("%d%d%d",&st,&ed,&ct);            es[i].from=st;            es[i].to=ed;            es[i].cost=ct;        }        meshort(endd);        for(int i=1;i<=V;i++)        ans[i]=d[i];        int max=0;        for(int i=1;i<=V;i++)        {            if(i!=endd)            {                meshort(i);                max=me_max(max,d[endd]+ans[i]);            }        }        printf("%d\n",max);    }    return 0;}