POJ

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 ofM (1 ≤ M ≤ 100,000) unidirectional (one-way roads connects pairs of farms; roadi requiresT_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, andX
Lines 2.. M+1: Line i+1 describes road i with three space-separated integers:A_i,B_i, andT_i. The described road runs from farmA_i to farmB_i, requiringT_i time 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

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：就是1~n点的牛要去x点参加聚会，单向路，来的路和回去的路不一样，让你求是每个牛来回最短路中的最大的

#include<iostream>#include<cstdio>#include<cstring>#include<queue>using namespace std;#define INF 99999999int map1[1005][1005],dis1[1005];int map2[1005][1005],dis2[1005];int n;int init() {    int i,j;    for(i=1; i<=n; i++) {        for(j=1; j<=n; j++) {            map1[i][j]=INF;            map2[i][j]=INF;        }        dis1[i]=dis2[i]=INF;    }}void spfa(int x,int Map[1005][1005],int dis[1005]) {    int book[1005],i,j;    memset(book,0,sizeof(book));    queue<int>q;    while(!q.empty()) q.pop();    i=x;    book[i]=1;    dis[i]=0;    q.push(i);    while(!q.empty()) {        i=q.front();        q.pop();        for(j=1; j<=n; j++) {            if(Map[i][j]!=INF && book[j]==0) {                if(dis[j]>Map[i][j]+dis[i]) {                    dis[j]=Map[i][j]+dis[i];//更新j点的最短距离                    q.push(j);                }            }        }    }}int main() {    int m,x;    int i,j;    int a,b,c;    while(scanf("%d%d%d",&n,&m,&x)!=EOF) {        init();//初始化        while(m--) {            scanf("%d%d%d",&a,&b,&c);            map1[a][b]=c;            map2[b][a]=c;//存成两个图，这样方便计算所有点到x点        }        spfa(x,map1,dis1);        spfa(x,map2,dis2);//都是从x点到其他点        int ans=0;        for(i=1;i<=n;i++){                ans=max(ans,dis1[i]+dis2[i]);        }        printf("%d\n",ans);    }}