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：

输出样例#1：

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=;

    register char ch=getchar();

    for(x=;!isdigit(ch);ch=getchar()) if(ch=='-') f=;

    for(;isdigit(ch);ch=getchar()) x=x*+ch-'';

    x=f?-x:x;

}

int main()

{

    read(p);read(f);read(c);

    for(int i=;i<=f;++i) read(F[i]);

    memset(Map,,sizeof(Map));

    for(int i=;i<=p;i++) Map[i][i]=;

    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=;k<=p;++k)

     for(int i=;i<=p;++i)

      for(int j=;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=;i<=p;++i)

    {

        int ans=;

        for(int j=;j<=f;++j) ans+=Map[i][F[j]];

        if(ans<ans1) ans1=ans,ans2=i;

    }

    printf("%d\n",ans2);

    return ;

}