POJ 3268 Silver Cow Party （最短路dijkstra）

Silver Cow Party

题目链接：

http://acm.hust.edu.cn/vjudge/contest/122685#problem/D

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.


##题意：

n个人分别住在1~n个点，他们要去某个点聚会并且结束后返回家中.
求每个人的最短路程. (往返路径可以不一样).


##题解：

先用dijkstra求出终点到所有点的最短路径(返程).
再沿着反向路径跑一次dijkstra，求出所有点到终点的最短路径.


##代码：
``` cpp
#include
#include
#include
#include
#include
#define mid(a,b) ((a+b)>>1)
#define LL long long
#define maxn 1010
#define inf 0x3f3f3f3f
#define IN freopen("in.txt","r",stdin);
using namespace std;

int n,m;

int value[maxn][maxn];

int dis[maxn];

bool vis[maxn];

int dis2[maxn];

void dijkstra(int s) {

memset(vis, 0, sizeof(vis));

for(int i=1; i<=n; i++) dis[i] = inf;

dis[s] = 0;

for(int i=1; i<=n; i++) {
    int p, mindis = inf;
    for(int j=1; j<=n; j++) {
        if(!vis[j] && dis[j]<mindis)
            mindis = dis[p=j];
    }
    vis[p] = 1;
    for(int j=1; j<=n; j++) {
        if(dis[j] > dis[p]+value[p][j]) {
            dis[j] = dis[p] + value[p][j];
        }
    }
}

}

void dijkstra2(int s) {

memset(vis, 0, sizeof(vis));

for(int i=1; i<=n; i++) dis2[i] = inf;

dis2[s] = 0;

for(int i=1; i<=n; i++) {
    int p, mindis = inf;
    for(int j=1; j<=n; j++) {
        if(!vis[j] && dis2[j]<mindis)
            mindis = dis2[p=j];
    }
    vis[p] = 1;
    for(int j=1; j<=n; j++) {
        if(dis2[j] > dis2[p]+value[j][p]) {
            dis2[j] = dis2[p] + value[j][p];
        }
    }
}

}

int main(int argc, char const *argv[])

{

//IN;

int aim;
while(scanf("%d %d %d", &n,&m,&aim) != EOF)
{
    for(int i=1; i<=n; i++)
        for(int j=1; j<=n; j++)
            value[i][j] = inf;
    while(m--){
        int u,v,w; cin>>u>>v>>w;
        if(w < value[u][v]) value[u][v] = w;
    }

    dijkstra(aim);
    dijkstra2(aim);

    int ans = 0;
    for(int i=1; i<=n; i++) if(dis[i]!=inf && dis2[i]!=inf)
        ans = max(ans, dis[i]+dis2[i]);

    printf("%d\n", ans);
}

return 0;

}