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.

思路

這道題用到了最短路的思想。這道題的要求也是單向的路徑，在編寫代碼的時候需要注意一下。

#include <iostream> #include <stdio.h> #include <cstring>using namespace std; int n,m,x,minn,u,ans=0; int ai,bi,ti; int f[1005][1005]; int dis[1005],book[1005],way[1005]; const int inf=99999999;void distant(int x) {//book[]數組用來記錄查詢過的點for(int i=1;i<=n;i++)book[i]=0;//重置book數組book[x]=1;for(int i=1;i<=n;i++)//記錄點i到x的距離dis[i]=f[x][i];for(int i=2;i<=n;i++){minn=inf;for(int j=1;j<=n;j++)//兩個農場之間取最短的路徑{if(!book[j]&&dis[j]<minn){minn=dis[j];u=j;}}book[u]=1;for(int j=1;j<=n;j++){if(!book[j]&&dis[j]>dis[u]+f[u][j])dis[j]=dis[u]+f[u][j];}} }int main() {while(scanf("%d %d %d",&n,&m,&x)!=EOF){for(int i=0;i<=n;i++)//這里i的范圍要取小于n，如果取小于1005z則會runtime error{//重置數組for(int j=0;j<=n;j++){if(i==j){f[i][j]=0;}else{f[i][j]=inf;}}}for(int i=0;i<m;i++){scanf("%d %d %d",&ai,&bi,&ti);if(f[ai][bi]>ti)f[ai][bi]=ti;}distant(x);for(int i=1;i<=n;i++)way[i]=dis[i];//記錄distant函數之后的dis的數據for(int i=1;i<=n;i++){for(int j=i+1;j<=n;j++){int tem;tem=f[j][i];f[j][i]=f[i][j];f[i][j]=tem;}}distant(x);for(int i=1;i<=n;i++){ans=max(ans,way[i]+dis[i]);}printf("%d\n",ans);}return 0; }

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