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
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
10
说明
As the problem statement
As the problem statement.
思路：先跑一遍floyed，然后暴力枚举即可。
#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #define MAXN 510 using namespace std; int P,F,C,ans; int maxn=0x7f7f7f7f; int f[MAXN],map[MAXN][MAXN]; int main(){ scanf("%d%d%d",&P,&F,&C); for(int i=1;i<=F;i++) scanf("%d",&f[i]); for(int i=1;i<=P;i++) for(int j=1;j<=P;j++){ map[i][j]=999999; if(i==j) map[i][j]=0; } for(int i=1;i<=C;i++){ int x,y,z; scanf("%d%d%d",&x,&y,&z); map[x][y]=map[y][x]=z; } for(int k=1;k<=P;k++) for(int i=1;i<=P;i++) for(int j=1;j<=P;j++) if(i!=k&&k!=j&&i!=j&&map[i][k]+map[k][j]<map[i][j]) map[i][j]=map[i][k]+map[k][j]; for(int i=1;i<=P;i++){ long long sum=0; for(int j=1;j<=F;j++) sum+=map[i][f[j]]; if(sum<maxn){ maxn=sum; ans=i; } } cout<<ans; }