USACO 2009 Jan Silver 1.Best Spot SPFA

题目描述

Bessie, always wishing to optimize her life, has realized that she really enjoys visiting F (1 <= F <= P) favorite pastures F_i of the P (1 <= P <= 500; 1 <= F_i <= P) total pastures (conveniently

numbered 1..P) that compose Farmer John's holdings.

Bessie knows that she can navigate the C (1 <= C <= 8,000) bidirectional cowpaths (conveniently numbered 1..C) that connect various pastures to travel to any pasture on the entire farm. Associated with each path P_i is a time T_i (1 <= T_i <= 892) to traverse that path (in either direction) and two path endpoints a_i and b_i (1 <= a_i <= P; 1 <= b_i <= P).

Bessie wants to find the number of the best pasture to sleep in so that when she awakes, the average time to travel to any of her F favorite pastures is minimized.

By way of example, consider a farm laid out as the map below shows, where *'d pasture numbers are favorites. The bracketed numbers are times to traverse the cowpaths.

            1*--[4]--2--[2]--3
                     |       |
                    [3]     [4]
                     |       |
                     4--[3]--5--[1]---6---[6]---7--[7]--8*
                     |       |        |         |
                    [3]     [2]      [1]       [3]
                     |       |        |         |
                    13*      9--[3]--10*--[1]--11*--[3]--12*

The following table shows distances for potential 'best place' of pastures 4, 5, 6, 7, 9, 10, 11, and 12:

      * * * * * * Favorites * * * * * *
 Potential      Pasture Pasture Pasture Pasture Pasture Pasture     Average
Best Pasture       1       8      10      11      12      13        Distance
------------      --      --      --      --      --      --      -----------
    4              7      16       5       6       9       3      46/6 = 7.67
    5             10      13       2       3       6       6      40/6 = 6.67
    6             11      12       1       2       5       7      38/6 = 6.33
    7             16       7       4       3       6      12      48/6 = 8.00
    9             12      14       3       4       7       8      48/6 = 8.00
   10             12      11       0       1       4       8      36/6 = 6.00 ** BEST
   11             13      10       1       0       3       9      36/6 = 6.00
   12             16      13       4       3       0      12      48/6 = 8.00

Thus, presuming these choices were the best ones (a program would have to check all of them somehow), the best place to sleep is pasture 10.

约翰拥有P(1<=P<=500)个牧场.贝茜特别喜欢其中的F个.所有的牧场 由C(1 < C<=8000)条双向路连接，第i路连接着ai,bi，需要Ti(1<=Ti< 892)单 位时间来通过.

作为一只总想优化自己生活方式的奶牛，贝茜喜欢自己某一天醒来，到达所有那F个她喜欢的 牧场的平均需时最小.那她前一天应该睡在哪个牧场呢？请帮助贝茜找到这个最佳牧场.

此可见，牧场10到所有贝茜喜欢的牧场的平均距离最小，为最佳牧场.

输入格式

* Line 1: Three space-separated integers: P, F, and C

* Lines 2..F+1: Line i+2 contains a single integer: F_i

* Lines F+2..C+F+1: Line i+F+1 describes cowpath i with three

space-separated integers: a_i, b_i, and T_i

输出格式

* Line 1: A single line with a single integer that is the best pasture in which to sleep. If more than one pasture is best, choose the smallest one.

输入输出样例

输入

13 6 15

11

13

10

12

8

1

2 4 3

7 11 3

10 11 1

4 13 3

9 10 3

2 3 2

3 5 4

5 9 2

6 7 6

5 6 1

1 2 4

4 5 3

11 12 3

6 10 1

7 8 7

输出

10

分析：从题目分析可知，可以枚举1~N的所有顶点到F个牧场的权值和，找到一个顶点使其到F个牧场的权值和最小，顶点的编号！！即为答案。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<queue>
#define M 16100
#define N 510
using namespace std;
queue<int>q;
int nxt[M],head[N],to[M],val[M],tot;
int dis[N];
bool check[N];
void Add(int x,int y,int z)
{
    val[++tot]=z;
    to[tot]=y;
    nxt[tot]=head[x];
    head[x]=tot;
}
void SPFA(int S)
{
    memset(dis,0x3f3f,sizeof(dis));
    dis[S]=0;
    q.push(S);
    check[S]=1;
    while(!q.empty())
    {
        int k=q.front(),y;
        q.pop();
        check[k]=0;
        for(int i=head[k];i;i=nxt[i])
        {
            if(dis[y=to[i]]>dis[k]+val[i])
            {
                dis[y]=dis[k]+val[i];
                if(check[y]==0)
                {
                    q.push(y);
                    check[y]=1;
                }
            }
        }
    }
}
int f[N];
int main()
{
    int n,m,F;
    scanf("%d%d%d",&n,&F,&m);
    for(int i=1;i<=F;i++)
        scanf("%d",&f[i]);
    for(int i=1;i<=m;i++)
    {
        int x,y,z;
        scanf("%d%d%d",&x,&y,&z);
        Add(x,y,z);
        Add(y,x,z);
    }
    int ans=2147483647;
    int idx=0;
    for(int i=1;i<=n;i++)
    {
        SPFA(i);
        int t=0;
        for(int j=1;j<=F;j++)
        {
            t+=dis[f[j]];
        }
        if(ans>t)
        {
            ans=t;
            idx=i;
        }
    }
    printf("%d\n",idx);
    return 0;
}