2018牛客多校训练-----Maximum Mode

链接：https://www.nowcoder.com/acm/contest/142/G
来源：牛客网
 

题目描述

The mode of an integer sequence is the value that appears most often. Chiaki has n integers a1,a2,...,an. She woud like to delete exactly m of them such that: the rest integers have only one mode and the mode is maximum.

输入描述:

There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:
The first line contains two integers n and m (1 ≤ n ≤ 105, 0 ≤ m < n) -- the length of the sequence and the number of integers to delete.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109) denoting the sequence.
It is guaranteed that the sum of all n does not exceed 106.

输出描述:

For each test case, output an integer denoting the only maximum mode, or -1 if Chiaki cannot achieve it.

示例1

输入

5
5 0
2 2 3 3 4
5 1
2 2 3 3 4
5 2
2 2 3 3 4
5 3
2 2 3 3 4
5 4
2 2 3 3 4

输出

-1
3
3
3
4

mode的定义：一个序列中出现最多次数的数值。删掉m个数之后，使mode尽可能大。

 思路：

 以某个数x当做最后答案唯一的限制条件是当前序列没有比x个数多的数。枚举每一个序列中出现的数值为答案，看能否删除比当前答案出现次数多的数，如果能，作为备选答案，比较答案最大值。m>要删除的数，要求多删的数随便删就可以了。

#include<bits/stdc++.h>
using namespace std;

const int N=1e5+100;

int a[N],d[N];

struct Node
{
    int num;
    int value;

    bool operator <(const Node&other) const
    {
        return num>other.num;
    }

}node[N];

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int n,m,i,j;
        scanf("%d%d",&n,&m);
        for(i=1;i<=n;i++)
            scanf("%d",&a[i]);
        sort(a+1,a+n+1);
        int cnt=1;
        memset(node,0,sizeof(node));
        node[1].num=1;node[1].value=a[1];
        for(i=2;i<=n;i++)
        {
            if(a[i-1]!=a[i])
            {
                node[++cnt].num=1;
                node[cnt].value=a[i];
            }
            else
            {
                node[cnt].num++;
            }
        }
        sort(node+1,node+n+1);
        int t=1;
        for(i=1;i<=cnt;i++)
        {
            if(node[i-1].num==node[i].num&&i>=1)
                d[node[i].num]++;
            else
                d[node[i].num]=1;
        }
        int del=0,tt=-1;
        int ans=-1;
        for(i=1;i<=cnt;i++)
        {
            if(node[i-1].num>node[i].num&&i>=1)
            {
                del+=(node[i-1].num-node[i].num)*d[node[i-1].num];
                d[node[i].num]+=d[node[i-1].num];
            }
            tt=del+d[node[i].num]-1;
            if(m>=tt)
                ans=max(ans,node[i].value);
        }
        printf("%d\n",ans);
    }
    return 0;
}