最短路练习4/poj/3268 /Silver Cow Party

题目链接：http://poj.org/problem?id=3268

Silver Cow Party

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 21544		Accepted: 9834

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

Sample Output

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：路是单向的，现在所有的牛都要去x点，给出m条路，n个点，这每个点的牛去x点的最短路和从x回来的最短路之和是这头的要走的最短路，求这些牛中最多的走了多少路。

思路：最短路模板，求x到每个点的最短路，和每个点到x的最短路，之和最小值。

做了几道最短路的题，这个代码我终于一遍过了，有进步；

AC代码：

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
#include<map>
#include<string>
#define LL long long
using namespace std;
const int mod = 1e7+7;
const int inf = 0x3f3f3f3f;
const int maxn = 1e6 +10;
int n,m,x,a,b,c,nn,point,sum,ans;
int ma[1005][1005];//两点的距离
int dis[1005];//从i点去x点的最短距离
int kis[1005];//从x点回到i点最短距离
int vis[1005];//是否已经更新为最小值
int main()
{
    while(~scanf("%d%d%d",&n,&m,&x))
    {
        memset(ma,inf,sizeof(ma));
        for(int i=1;i<=n;i++)
            ma[i][i]=0;
        for(int i=0;i<m;i++)
        {
            scanf("%d%d%d",&a,&b,&c);
            ma[a][b]=c;
        }
        for(int i=1;i<=n;i++)
        {
            dis[i]=ma[i][x];//i到x
            kis[i]=ma[x][i];//x到i
        }
        memset(vis,0,sizeof(vis));
        nn=n;
         while(nn--)
         {
                 sum=inf,point=-1;
             for(int i=1;i<=n;i++)
             {
                 if(!vis[i]&&dis[i]<sum)
                 {
                     sum=dis[i];
                     point=i;
                 }
             }
             if(point!=-1)
             {
                 vis[point ]=1;
                 for(int i=1;i<=n;i++)
                 {
                     if(!vis[i]&&dis[point]+ma[i][point]<dis[i])
                        dis[i]=dis[point]+ma[i][point];
                 }
             }
         }
         memset(vis,0,sizeof(vis));
         nn=n;
         while(nn--)
         {
                 sum=inf,point=-1;
             for(int i=1;i<=n;i++)
             {
                 if(!vis[i]&&kis[i]<sum)
                 {
                     sum=kis[i];
                     point=i;
                 }
             }
             if(point!=-1)
             {
                 vis[point ]=1;
                 for(int i=1;i<=n;i++)
                 {
                     if(!vis[i]&&kis[point]+ma[point][i]<kis[i])
                        kis[i]=kis[point]+ma[point][i];
                 }
             }
         }
         ans=0;
         for(int i=1;i<=n;i++)
         if(kis[i]+dis[i]>ans)
            ans=kis[i]+dis[i];
         printf("%d\n",ans);
    }
}

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最短路练习4/poj/3268 /Silver Cow Party

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