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九度 OJ 1444:More is better

2017-03-12 17:26 337 查看
题目描述:

Mr Wang wants some boys to help him with a project. Because the project is rather complex, the more boys come, the better it will be. Of course there are certain requirements.Mr Wang selected a room big enough to hold the boys.
The boy who are not been chosen has to leave the room immediately. There are 10000000 boys in the room numbered from 1 to 10000000 at the very beginning. After Mr Wang's selection any two of them who are still in this room should be friends (direct or indirect),
or there is only one boy left. Given all the direct friend-pairs, you should decide the best way.
输入:

The first line of the input contains an integer n (0 ≤ n ≤ 100 000) - the number of direct friend-pairs. The following n lines each contains a pair of numbers A and B separated by a single space that suggests A and B are direct
friends. (A ≠ B, 1 ≤ A, B ≤ 10000000)
输出:

The output in one line contains exactly one integer equals to the maximum number of boys Mr Wang may keep.
样例输入:
4
1 2
3 4
5 6
1 6
4
1 2
3 4
5 6
7 8

样例输出:
4
2

#include <stdio.h>
#define N 10000001
using namespace std;
int Tree
;

int findRoot(int x) {
if(Tree[x] == -1) return x;     //若x为根节点返回节点号x
else {
int tmp = findRoot(Tree[x]);    //递归找x的根节点
Tree[x] = tmp;                  //树的压缩,使得树高变矮
return tmp;
}
}
int sum
;  //用sum[i]保存以i为根节点的树的子节点个数

int main()
{
int n;
while(scanf("%d", &n)!= EOF) {
//初始化
for(int i=0;i<N;i++) {
Tree[i] = -1;
sum[i] = 1;     //初始每个节点为一个集合,只包含自己
}
int m = n;
while(m-- != 0) {
int a,b;
scanf("%d%d", &a,&b);
a = findRoot(a);
b = findRoot(b);
if(a != b) {         //若a,b不是一个集合内的,将其合并。
Tree[a] = b;     //将a的根设为b
sum[b] += sum[a];
}
}
int max = 1;
for(int i=0;i<N;i++){
if((Tree[i] == -1) && (sum[i]>max))
max = sum[i];
}
printf("%d\n", max);

}

return 0;
}
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