Problem B
Reading books
(Input File: book.in / Standard Output)
In the summer vacation, LRJ wants to improve himself in computer science. So he finds out N books of computer science in the school library. The books are numbered from 0 to N-1.
To finish reading the i-th book, it takes LRJ time[i] minutes. But some books are similar in the content. If the i-th book and the j-th book are similar, then if LRJ has finished reading the i-th book, it will take him only minutes to finish reading the j-th book. Of course if LRJ has finished reading the j-th book, it will take him only minutes to finish reading the i-th book. Now you are asked to tell LRJ the minimal total time to finish reading all the N books.
Input:
The first line contains two integers N (0<=N<=100) and M (0<=M<=N*(N-1)/2). N is the total number of books. M is the number of pairs which are similar.
Then the following N lines describe time[0], time[1], … , time[N-1] (1<=time[i]<=105).
Next comes M lines, each contains two integer (i, j), indicating that the i-th book and the j-th book are similar.
Input is ended by N=0 and M=0.
Output:
For each test case, just output the minimal total time on a single line.
Sample input:
2 1
6
10
0 1
3 2
1
2
3
0 1
1 2
3 1
2
4
6
0 1
0 0
Sample output:
11
3
10
Hints:
For the first test case, if LRJ read the books in the order (0, 1), then the total time = 6+10/2=11; If in the order (1, 0), then the total time =10+ 6/2=13.
用矩阵将全部相似的书都标记起来，然后通过dfs去算和就可以
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
int map[105][105],val[105],n,m,sum,minn;
int vis[105],ans;
void dfs(int x)
{
vis[x] = 1;
sum+=val[x]/2;
if(minn == -1 || val[x]<minn)
minn = val[x];
for(int i = 0; i<n; i++)
{
if(!vis[i] && map[x][i])
dfs(i);
}
}
int main()
{
int i,x,y;
while(~scanf("%d%d",&n,&m),n+m)
{
memset(map,0,sizeof(map));
memset(vis,0,sizeof(vis));
ans = 0;
for(i = 0; i<n; i++)
scanf("%d",&val[i]);
for(i = 0; i<m; i++)
{
scanf("%d%d",&x,&y);
map[x][y] = map[y][x] = 1;
}
for(i = 0; i<n; i++)
{
minn = -1;
if(vis[i])
continue;
sum = 0;
dfs(i);
ans = ans+sum-minn/2+minn;
}
printf("%d
",ans);
}
return 0;
}