POJ - 3268Silver Cow Party

Silver Cow Party Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 20881 Accepted: 9550

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 T_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, and  X  Lines 2.. M+1: Line  i+1 describes road  i with three space-separated integers:  A_i,  B_i, and  T_i. The described road runs from farm  A_i to farm  B_i, requiring  T_i 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

把边正向建一个图，在反向建一个图，两遍dj，正向建是求回去的最短路径，而反向建是因为用dj求每个点到单点的最短距离复杂度很高，反向建图后求终点到每个点的最短路径就可以了

#include<stdio.h> #include<iostream> #include<string.h> #include<math.h> #include<string> #include<algorithm> #include<queue> #include<vector> #include<map> #include<set> #define eps 1e-9 #define PI 3.141592653589793 #define bs 1000000007 #define bsize 256 #define MEM(a) memset(a,0,sizeof(a)) #define inf 0x3f3f3f3f #define rep(i,be,n) for(i=be;i<n;i++) typedef long long ll; using namespace std; int mp[2][1005][1005]; int book[1005]; int x,n,m; int dijis(int k) { int index,i,j,minn; MEM(book); book[x]=1; for(i=1;i<=n;i++) { minn=inf; for(j=1;j<=n;j++) { if(!book[j]&&mp[k][x][j]<minn) { index=j; minn=mp[k][x][j]; } } book[index]=1; for(j=1;j<=n;j++) { if(!book[j]&&mp[k][index][j]+minn<mp[k][x][j]) { mp[k][x][j]=mp[k][index][j]+minn; } } } } int main() { int i,j,a,b,t; memset(mp,inf,sizeof(mp)); scanf("%d %d %d",&n,&m,&x); while(m--) { scanf("%d %d %d",&a,&b,&t); if(mp[0][a][b]>t) { mp[0][a][b]=t; } if(mp[1][b][a]>t) { mp[1][b][a]=t; } } dijis(0); dijis(1); int ans=0; for(i=1;i<=n;i++) { if(i==x) continue; ans=max(ans,mp[0][x][i]+mp[1][x][i]); } printf("%d\n",ans); return 0; }