Problem : [Usaco2009 Jan]Best Spot
Time Limit: 1 Sec Memory Limit: 128 MB[Submit][Status][Web Board]
Description
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
Input
之后C行每行输入三个整数ai,bi,Ti,描述一条路.
Output
Sample Input
5 5
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
Sample Output
10
HINT
#include<stdio.h>
#include<string.h>
class BestSpot {
public:
int g[501][501],p,f,c,l[501];
void init() {
scanf("%d %d %d",&p,&f,&c);
memset(g,0x3f,sizeof(g));
for(i=1; i<=f; i++)
scanf("%d",&l[i]);
for(i=1; i<=c; i++)
scanf("%d %d %d",&u,&v,&w),g[u][v]=g[v][u]=w;
for(i=1; i<=p; i++)
g[i][i]=0;
}
void work() {
for(k=1; k<=p; k++)
for(i=1; i<=p; i++)
for(j=1; j<=p; j++)
if(g[i][k]+g[k][j]<g[i][j])
g[i][j]=g[i][k]+g[k][j];
}
void print() {
for(i=1; i<=p; i++) {
sum=0;
for(j=1; j<=f; j++)
if(g[i][l[j]]!=0x3f3f3f3f)sum+=g[i][l[j]];
if(sum<mn)
mn=sum,ans=i;
}
printf("%d",ans);
}
private:
int i,j,k,u,v,w,sum,ans,mn=0x7fffffff;
};
int main() {
BestSpot Bessie;
Bessie.init();
Bessie.work();
Bessie.print();
}