Friendship

时间：2015-07-07 17:03:29 阅读：165 评论：0 收藏：0 [点我收藏+]

标签：

A - Friendship

Time Limit:3000MS Memory Limit:32768KB 64bit IO Format:%lld & %llu

Description

A friend is like a flower,
a rose to be exact,
Or maybe like a brand new gate
that never comes unlatched.

A friend is like an owl,
both beautiful and wise.
Or perhaps a friend is like a ghost,
whose spirit never dies.

A friend is like a heart that goes
strong until the end.
Where would we be in this world
if we didn‘t have a friend?

- By Emma Guest

Now you‘ve grown up, it‘s time to make friends. The friends you make in university are the friends you make for life. You will be proud if you have many friends.

Input

There are multiple test cases for this problem.

Each test case starts with a line containing two integers N, M (1 <= N <= 100‘000, 1 <= M <= 200‘000), representing that there are totally N persons (indexed from 1 to N) and M operations, then M lines with the form "M a b" (without quotation) or "Q a" (without quotation) follow. The operation "M a b" means that person a and b make friends with each other, though they may be already friends, while "Q a" means a query operation.

Friendship is transitivity, which means if a and b, b and c are friends then a and c are also friends. In the initial, you have no friends except yourself, when you are freshman, you know nobody, right? So in such case you have only one friend.

Output

For each test case, output "Case #:" first where "#" is the number of the case which starts from 1, then for each query operation "Q a", output a single line with the number of person a‘s friends.

Separate two consecutive test cases with a blank line, but Do NOT output an extra blank line after the last one.

Sample Input

3 5
M 1 2
Q 1
Q 3
M 2 3
Q 2
5 10
M 3 2
Q 4
M 1 2
Q 4
M 3 2
Q 1
M 3 1
Q 5
M 4 2
Q 4

Sample Output

Case 1:
2
1
3

Case 2:
1
1
3
1
4

Notes

This problem has huge input and output data, please use ‘scanf()‘ and ‘printf()‘ instead of ‘cin‘ and ‘cout‘ to avoid time limit exceed.

/*
Author:2486
Memory: 952 KB		Time: 170 MS
Language: C++ (g++ 4.7.2)		Result: Accepted
*/
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn=100000+5;
int par[maxn],sum[maxn];
int n,m,x,y;
char op[5];
void init(int x) {
    for(int i=0; i<=x; i++) {
        par[i]=i;
        sum[i]=1;
    }
}
int find(int x) {
    return par[x]==x?x:par[x]=find(par[x]);
}
bool same(int x,int y) {
    return find(x)==find(y);
}
void unite(int x,int y) {
    x=find(x);
    y=find(y);
    if(x==y)return;
    par[x]=y;
    sum[y]+=sum[x];
}
int main() {
    int cases=0;
    //freopen("D://imput.txt","r",stdin);
    while(~scanf("%d%d",&n,&m)) {
        init(n);
        cases++;
        if(cases!=1)printf("\n");
        printf("Case %d:\n",cases);
        for(int i=0; i<m; i++) {
            scanf("%s",op);
            if(op[0]=='Q') {
                scanf("%d",&x);
                printf("%d\n",sum[find(x)]);
            } else {
                scanf("%d%d",&x,&y);
                unite(x,y);
            }
        }
    }
    return 0;
}

Friendship

标签：

原文地址：http://blog.csdn.net/qq_18661257/article/details/46790007

踩

(0)

评论一句话评论（0）

分享档案

更多>

2021年07月29日 (22)
2021年07月28日 (40)
2021年07月27日 (32)
2021年07月26日 (79)
2021年07月23日 (29)
2021年07月22日 (30)
2021年07月21日 (42)
2021年07月20日 (16)
2021年07月19日 (90)
2021年07月16日 (35)

周排行