洛谷 P2935 [USACO09JAN]最好的地方Best Spot

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1(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.

输入输出格式

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

输入样例#1:
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 
输出样例#1:
10 

说明

As the problem statement

As the problem statement.

 

floyd

①不读题②手残

mdzz。。

屠龙宝刀点击就送

#include <cstring>
#include <ctype.h>
#include <cstdio>
#define N 505

int F[N],ans1=0x7fffffff,ans2,p,f,c,Map[N][N];
inline int min(int a,int b) {return a>b?b:a;}
inline void read(int &x)
{
    bool f=0;
    register char ch=getchar();
    for(x=0;!isdigit(ch);ch=getchar()) if(ch=='-') f=1;
    for(;isdigit(ch);ch=getchar()) x=x*10+ch-'0';
    x=f?-x:x;
}
int main()
{
    read(p);read(f);read(c);
    for(int i=1;i<=f;++i) read(F[i]);
    memset(Map,1,sizeof(Map));
    for(int i=1;i<=p;i++) Map[i][i]=0;
    for(int x,y,z;c--;)
    {
        read(x);read(y);read(z);
        Map[x][y]=Map[y][x]=min(Map[x][y],z);
    }
    for(int k=1;k<=p;++k)
     for(int i=1;i<=p;++i)
      for(int j=1;j<=p;++j)
       if(Map[i][j]>Map[i][k]+Map[k][j])
       Map[i][j]=Map[i][k]+Map[k][j];
    for(int i=1;i<=p;++i)
    {
        int ans=0;
        for(int j=1;j<=f;++j) ans+=Map[i][F[j]];
        if(ans<ans1) ans1=ans,ans2=i;
    }
    printf("%d\n",ans2);
    return 0;
}

 

posted @ 2017-08-19 14:55  杀猪状元  阅读(258)  评论(1编辑  收藏  举报