POJ3268

Silver Cow Party

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 14455		Accepted: 6511

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.

#include<stdio.h>
#define M 99999999
int map[1002][1002]={0},mapp[1002][1002]={0};
int d[1002],dt[1002];

void dijistra(int d[],int g[][1002],int n,int v)
{
	int i,j;
	int ss[1002];
	int min;
	int u;

	for(i=1;i<=n;i++)
	{
		d[i]=M;
		ss[i]=0;
	}
	d[v]=0;
	for(i=1;i<=n;i++)
	{
		min=M;
		for(j=1;j<=n;j++)
		{
			if(ss[j]==0&&d[j]<min)
			{
				min=d[j];
				u=j;
			}
		}
		ss[u]=1;
		for(j=1;j<=n;j++)
		{
			if(d[u]+g[u][j]<d[j])
			{
				d[j]=d[u]+g[u][j];
			}
		}
	}
}
int main()
{
	int i,j,n,edge,party,cost,u,v;
	int max;

	scanf("%d %d %d",&n,&edge,&party);
	for(i=1;i<=n;i++)
		for(j=1;j<=n;j++)
		{
			map[i][j]=M;
			mapp[i][j]=M;
		}
	for(i=0;i<=edge;i++)
	{
		scanf("%d%d",&u,&v);
		scanf("%d",&map[u][v]);
		mapp[v][u]=map[u][v];
	}
	dijistra(d,map,n,party);
	dijistra(dt,mapp,n,party);
	max=0;
	for(i=1;i<=n;i++)
	{
		if(max<d[i]+dt[i])
			max=d[i]+dt[i];
	}
	printf("%d",max);
	return 0;
}

不是很会，仿照的写的，感觉迪杰斯特拉还是不怎么好写，理解起来还好