Silver Cow Party

    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. 

这个题关键是能想到把边的两边互换，其他到是基础的最短路径，用spfa很好写。

代码：

#include<stdio.h>
#include<iostream>
#include<queue>
#include<cstring>
#include<cmath>
#include<vector>

#define INF 0x3f3f3f3f

using namespace std;

int N,M,X;
int data[1005][1005];
int come[1005];
int back[1005];
bool book[1005];

void SpfaB(){
	queue<int>Q;
	Q.push(X);
	book[X] = true;
	while(!Q.empty()){
		int mid = Q.front();
		Q.pop();
		book[mid] = false;
		for(int i=1; i<=N ; i++){
			if(back[mid]+data[mid][i]<back[i]){
				back[i] = back[mid]+data[mid][i];
				if(!book[i]){
					Q.push(i);
					book[i] = true;
				}
			}
		}
	}
}

void SpfaC(){
	queue<int>Q;
	Q.push(X);
	book[X] = true;
	while(!Q.empty()){
		int mid = Q.front();
		Q.pop();
		book[mid] = false;
		for(int i=1; i<=N ; i++){
			if(come[mid]+data[i][mid]<come[i]){
				come[i] = come[mid]+data[i][mid];
				if(!book[i]){
					Q.push(i);
					book[i] = true;
				}
			}
		}
	}
}

int main(){
	scanf("%d %d %d",&N,&M,&X);
	memset(data,INF,sizeof(data));
	for(int i=0 ; i<1005 ; i++){
		data[i][i] = 0;
	}
	for(int i=1 ; i<=M ; i++){
		int a,b,c;
		scanf("%d %d %d",&a,&b,&c);
		data[a][b] = c;
	}
	memset(come,INF,sizeof(come));
	come[X] = 0;
	memset(back,INF,sizeof(back));
	back[X] = 0;
	SpfaB();
	SpfaC();
	int max = 0;
	for(int i=1; i<=N ; i++){
		int mid = come[i]+back[i];
		if(mid>max)max = mid;
	}
	printf("%d
",max);
	return 0;
}