POJ 3268 dijkstra() 정반대 행렬

dijkstra의 누드 문제, 정반 행렬을 각각 한 번씩 구하면 된다...제목:
Silver Cow Party
Time Limit: 2000MS
 
Memory Limit: 65536K
Total Submissions: 7988
 
Accepted: 3538
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

ac 코드:

#include <iostream>
#include <cstdio>
#include <string.h>
const int N=1010;
int nfarm,npath,pos;
int map1[N][N],map2[N][N],visted[N];
int dis1[N],dis2[N];
#define MAX 0x7fffffff
void dijkstra(int dis[N],int map[N][N]){
  memset(visted,0,sizeof(visted));
  for(int i=1;i<=nfarm;++i)
	  dis[i]=map[pos][i];
  dis[pos]=0;
  visted[pos]=1;
  int newpos=pos;
  for(int i=1;i<=nfarm;++i){
	  for(int j=1;j<=nfarm;++j){
		  if(!visted[j]&&map[newpos][j]!=MAX&&dis[newpos]+map[newpos][j]<dis[j]){
		    dis[j]=dis[newpos]+map[newpos][j];
		  }
	  }
	  int mmin=MAX;
	  for(int j=1;j<=nfarm;++j){
		  if(!visted[j]&&mmin>dis[j]){
		    mmin=dis[j];
			newpos=j;
		  }
	  }
	  visted[newpos]=1;
  }
}
int main(){
	//freopen("11.txt","r",stdin);
	while(~scanf("%d%d%d",&nfarm,&npath,&pos)){
	  int x,y,z;
	  for(int i=0;i<=nfarm;++i){
		  for(int j=0;j<=nfarm;++j){
		    map1[i][j]=MAX;
			map2[i][j]=MAX;
		  }
	  }
	  while(npath--){
	    scanf("%d%d%d",&x,&y,&z);
		map1[x][y]=z;
		map2[y][x]=z;
	  }
	  dijkstra(dis1,map1);
	  dijkstra(dis2,map2);
	  int mmax=0;
	 /* for(int i=1;i<=nfarm;++i){
	    printf("%d ",dis1[i]);
	  }*/
	  for(int i=1;i<=nfarm;++i){
	    if(dis1[i]+dis2[i]>mmax)
			mmax=dis1[i]+dis2[i];
	  }
	  printf("%d
",mmax);
	}
  return 0;
}