Codeforces 420 C. Bug in Code

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码代码的猿猿的AC之路

最新推荐文章于 2024-06-13 23:08:19 发布

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在一家名为F公司的组织中，发现了一项严重错误。公司领导决定在会议上找出问题的源头，即谁是造成代码错误的人。共有n名开发者参与了讨论，每个人声称他们知道错误是由另外两个人中的一个造成的。为了确保至少p个人同意选择的嫌疑人，公司领导需要找出所有可能的嫌疑人组合。本文将详细介绍如何计算所有可能的嫌疑人组合数量。

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Recently a serious bug has been found in the FOS code. The head of the F company wants to find the culprit and punish him. For that, he set up an organizational meeting, the issue is: who's bugged the code? Each of the n coders on the meeting said: 'I know for sure that either x or y did it!'

The head of the company decided to choose two suspects and invite them to his office. Naturally, he should consider the coders' opinions. That's why the head wants to make such a choice that at least p of n coders agreed with it. A coder agrees with the choice of two suspects if at least one of the two people that he named at the meeting was chosen as a suspect. In how many ways can the head of F choose two suspects?

Note that even if some coder was chosen as a suspect, he can agree with the head's choice if he named the other chosen coder at the meeting.

Input

The first line contains integers n and p (3 ≤ n ≤ 3·105; 0 ≤ p ≤ n) — the number of coders in the F company and the minimum number of agreed people.

Each of the next n lines contains two integers xi, yi (1 ≤ xi, yi ≤ n) — the numbers of coders named by the i-th coder. It is guaranteed that xi ≠ i, yi ≠ i, xi ≠ yi.

Output

Print a single integer –– the number of possible two-suspect sets. Note that the order of the suspects doesn't matter, that is, sets (1, 2)и (2, 1) are considered identical.

Sample test(s)

input
4 2
2 3
1 4
1 4
2 1

output
6

input
8 6
5 6
5 7
5 8
6 2
2 1
7 3
1 3
1 4

output
1

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>

using namespace std;

const int maxn=330000;
typedef long long int LL;

map<pair<int,int>, int> tu;
int agree[maxn];

int main()
{
    int n,p;
    scanf("%d%d",&n,&p);
    for(int i=1;i<=n;i++)
    {
        int u,v;
        scanf("%d%d",&u,&v);
        if(u>v) swap(u,v);
        agree[u]++; agree[v]++;
        tu[make_pair(u,v)]++;
    }
    LL ans=0;
    map<pair<int,int> ,int>::iterator it;
    for(it=tu.begin();it!=tu.end();it++)
    {
        int u=it->first.first,v=it->first.second;
        int res=agree[u]+agree[v];
        if(res>=p&&res-it->second<p)
            ans--;
    }
    int now=n;
    sort(agree+1,agree+1+n);
    for(int i=1;i<=n;i++)
    {
        now=max(i,now);
        while(i<now&&agree[i]+agree[now]>=p) now--;
        ans+=n-now;
    }
    printf("%I64d\n",ans);
    return 0;
}