Silver Cow Party
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 15897 Accepted: 7238
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 Ti (1 ≤ Ti ≤ 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: Ai, Bi, and Ti. The described road runs from farm Ai to farm Bi, requiring Ti time units to traverse.
Output
Line 1: One integer: the maximum of time any one cow must walk.
Sample Input
4 8 2
1 2 4
1 3 2
1 4 7
2 1 1
2 3 5
3 1 2
3 4 4
4 2 3
Sample Output
10
Hint
Cow 4 proceeds directly to the party (3 units) and returns via farms 1 and 3 (7 units), for a total of 10 time units.
Source
USACO 2007 February Silver
还是用dijikstra做，不同的是要分方向，并且要同时找出双向的最短路，最后再求出所有来回路程中最大的输出。
//
// main.cpp
// 最短路练习-D-Silver Cow Party
//
// Created by 袁子涵 on 15/10/10.
// Copyright (c) 2015年 袁子涵. All rights reserved.
// 63ms 8612KB#include <iostream>
#include <string.h>
#include <math.h>
#include <algorithm>
#include <stdio.h>
#include <stdlib.h>
#define MAXN 1005
#define INF 0x7fffffff
using namespace std;
long long int N,M,X,go[MAXN],back[MAXN];
long long int cost[MAXN][MAXN];
bool vis1[MAXN],vis2[MAXN];void Dijkstra(long long int beg)
{for (int i=1; i<=N; i++) {go[i]=back[i]=INF;vis1[i]=vis2[i]=0;}go[beg]=back[beg]=0;for (int j=1; j<=N; j++) {long long int k1=-1,k2=-1;long long int min1=INF,min2=INF;for (int i=1; i<=N; i++) {if (!vis1[i] && go[i]<min1) {min1=go[i];k1=i;}if (!vis2[i] && back[i]<min2) {min2=back[i];k2=i;}}if (k1==-1 && k2==-1)break;vis1[k1]=vis2[k2]=1;for (int i=1; i<=N; i++) {if (!vis1[i] && go[k1]+cost[k1][i]<go[i])go[i]=go[k1]+cost[k1][i];if (!vis2[i] && back[k2]+cost[i][k2]<back[i])back[i]=back[k2]+cost[i][k2];}}
}int main(int argc, const char * argv[]) {cin >> N >> M >> X;long long int a,b,c,out=0;for (int i=1; i<=N; i++) {for (int j=1; j<=N; j++) {if (i==j)cost[i][j]=0;elsecost[i][j]=INF;}}for (long long int i=1; i<=M; i++) {scanf("%lld %lld %lld",&a,&b,&c);cost[a][b]=c;}Dijkstra(X);for (int i=1; i<=N; i++)out=max(go[i]+back[i],out);cout << out << endl;return 0;
}