首页 > 代码库 > hdu 1856 More is better(并查集)

hdu 1856 More is better(并查集)

2024-11-02 09:28:39   214人阅读

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1856

More is better

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 327680/102400 K (Java/Others)
Total Submission(s): 18985    Accepted Submission(s): 6990


Problem Description
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.
 

Input
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)
 

Output
The output in one line contains exactly one integer equals to the maximum number of boys Mr Wang may keep.
 

Sample Input
4
1 2
3 4
5 6
1 6
4
1 2
3 4
5 6
7 8
 

Sample Output
4
2


Hint

A and B are friends(direct or indirect), B and C are friends(direct or indirect), 
then A and C are also friends(indirect).

 In the first sample {1,2,5,6} is the result.
In the second sample {1,2},{3,4},{5,6},{7,8} are four kinds of answers.


一定要细致看题目啊！

！！

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 327680/102400 K (Java/Others)

时间限制尽管在1s 可是内存放的非常宽啊！

。 所以什么数组存不下等等的顾虑都抛开吧，大胆的开数组！

【代码】

#include <cstdio>const int maxn= 10000000+10;int father[maxn];int num[maxn];int Find(int x){ //路径压缩迭代版本号    int root = x;    while(root!=father[root])         root = father[root];//肇东根节点    while(x!=root){ //将这棵树上的节点都指向根节点        int tmp = father[x];        father[x] = root;        x= tmp;    }    return root;}void Union(int a,int b){    int p=Find(a);    int q=Find(b);    if(p!=q){        father[p]=q;        num[q]+=num[p]; //将合并后的数量加到 父节点上    }}int main(){    int n;    while(scanf("%d",&n)!=EOF){        if(n==0){            puts("1");continue;        }        for(int i=1;i<=maxn;i++){            num[i]=1;            father[i]=i;        }        int a,b;        int Max=0;        for(int i=0;i<n;i++){            scanf("%d%d",&a,&b);            if(a>Max)                Max=a;            if(b>Max)                Max=b;            Union(a,b);        }        //printf("sd");        int max=0;         for(int i=1;i<=Max;i++){             if(num[i]>max)                 max=num[i];            // printf("bug\n");         }        printf("%d\n",max);    }    return 0;}


hdu 1856 More is better(并查集)


联系
我们