Best Spot S（Floyd）&最短路(Floyd）

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 #1复制

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 #1复制

10

#include<bits/stdc++.h>
using namespace std;
int a[510][510],b[510],n,f,m,inf=1e9;
int main(){
	cin>>n>>f>>m;
	for(int i=1;i<=n;i++)
	for(int j=1;j<=n;j++){
		if(i==j)
		a[i][j]=0;
		else
		a[i][j]=inf;
	}
	for(int i=0;i<f;i++)
	cin>>b[i];
	for(int i=0;i<m;i++){
		int x,y,m;
		cin>>x>>y>>m;
		a[x][y]=m;
		a[y][x]=m;
	}
	for(int k=1;k<=n;k++)
	for(int i=1;i<=n;i++)
	for(int j=1;j<=n;j++)
	if(a[i][j]>a[i][k]+a[k][j])
	a[i][j]=a[i][k]+a[k][j];
	int ans,minn=inf,sum;
	for(int i=1;i<=n;i++){
		sum=0;
		for(int j=0;j<f;j++)
		sum+=a[i][b[j]];
		if(sum<minn){
		minn=sum;
		ans=i;	
		}
	}
	cout<<ans<<endl;
	return 0; 
} 

#include<iostream>
using namespace std;
int a[110][110],n,m,inf=1e9;
int main(){
	while(~scanf("%d %d",&n,&m)){
		if(n==0&&m==0)
		break;
		for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++){
			if(i==j)
			a[i][j]=0;
			else
			a[i][j]=inf;
		}
		for(int i=0;i<m;i++){
			int x,y,z;
			scanf("%d %d %d",&x,&y,&z);
			a[x][y]=z;
			a[y][x]=z;
		}
		for(int k=1;k<=n;k++)
		for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++){
			if(a[i][j]>a[i][k]+a[k][j])
			a[i][j]=a[i][k]+a[k][j]; 
		}
		printf("%d\n",a[1][n]); 
	} 
	return 0;
}