点击打开题目链接

More is better

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 327680/102400 K (Java/Others) Total Submission(s): 26832    Accepted Submission(s): 9611 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.   Author lxlcrystal@TJU   Source HDU 2007 Programming Contest - Final   Recommend lcy   |   We have carefully selected several similar problems for you:   1232  1325  1879  1863  1875    Statistic |  Submit |  Discuss |  Note

并查集的裸题，题目大意：老师想让几个男同学来做一个项目，这些同学必须直接或间接的为朋友关系，求最多招几个男同学。

思路：这里集合的数量不确定，目的是求每个集合的元素数量，再加一个num数组记录每个集合的元素，初始为1，合并时相加即可。

试了一下不同输入方式的时间，第一行为cin加取消同步的情况，第二行为cin不加取消同步，第三种是scanf加万能头文件，第四种是scanf不加万能头文件。

可以看出第二种直接超时，最优的方案还是写文件名加scanf。

附上AC代码：

#include<bits/stdc++.h> using namespace std; const int maxn=10000000+5; int par[maxn],num[maxn]; int n,f1,f2; int maxd,mind,maxcnt; void init() { for(int i=0;i<maxn;i++) par[i]=i,num[i]=1; } int find(int a) { if(a==par[a]) return a; return par[a]=find(par[a]); } void unite(int a,int b) { a=find(a); b=find(b); if(a==b)return ; par[a]=b; num[b]+=num[a]; } int main() { while(~scanf("%d",&n)) { if(n==0){printf("1\n");continue;} init(); maxd=INT_MIN; mind=INT_MAX; maxcnt=0; for(int i=0;i<n;i++) { scanf("%d%d",&f1,&f2); maxd=max(maxd,max(f1,f2)); mind=min(mind,min(f1,f2)); unite(f1,f2); } for(int i=mind;i<=maxd;i++) { maxcnt=max(maxcnt,num[i]); } printf("%d\n",maxcnt); } return 0; }