Hard problem

1022: Hard problem

Time Limit: 1 Sec   Memory Limit: 128 MB
Submit: 43   Solved: 12

Description

The cat gets N mice from dreamone, and she can choose K mice from them as the order which is listed from 

left to right for dinner. But there is a limitation that the second mouse is no bigger than the first one, the third 

one is no bigger than the second one….the Kth one is no bigger than the (K-1) th one. Actually, there is always 

not a single method to choose the K mice from the N mice as the way described above; we can assume there 

is M ways. This time, the cat of dreamone’s has thought another hard problem:  

For each way, there is always a value for the Kth mouse. She wants to know the biggest value for the Kth 

mouse of all the M ways. Can you get it? Of course, not all of you have understood the idea, so there is an 

example blew: 

We can assume N=4, K=2. 

The N (N=4) numbers represented the N mice are given blew from left to right: 

4 6 5 4 

According to the rules above, we can get four ways for choosing K mice, such as: 

4 4;  6 5;  6 4;  5 4 

So the answer is 5.because the value 5 is the biggest one of the four ways for the Kth number.  

If the cat can not solve the problem as quickly as she can, she will feel very boring, so she turns to you, a 

topcoder of SWUST, for help. Can you help her? 

 

Input

The first line of input will be a positive integer indicating how many test cases will be included (T). Each of 

the next T cases will contain two parts: 

The first part: two integer N, K (1<=N<=10000, 1<=K<=10) 

The second part: N numbers (which is no larger than 1000000) represented the N mice from left to right. 

 

Output

For each test, you should output the biggest value for the Kth numbers of all the manners. If you can not find 

any way to choose the Kth mouse, just output “OH, NO” instead. 

 

Sample Input

2 

4 2 

4 6 5 4 

4 2 

1 2 3 4

Sample Output

5 

OH,NO

思路：dp问题，找好状态转移方程。

代码：

#include<iostream>
#include<stdio.h>
#include<cstring>
using namespace std;
#define MAX 10000
int line[MAX+10];
int num[11];
int dp[MAX+10][11];
int get_min(int a,int b)
{
    return a>b?b:a;
}
int get_max(int a,int b)
{
    return a>b?a:b;
}
int main()
{
    //freopen("E.in","r",stdin);
    //freopen("E1.out","w",stdout);
    int n,m,t,i,j,flag,ans;
    scanf("%d",&n);
    while(n--)
    {
        flag=0;ans=0;
        scanf("%d%d",&m,&t);
        for(i=0;i<m;i++)
            scanf("%d",&line[i]);
        if(t>m)
            printf("OH,NO\n");
        else
        {
            memset(num,-1,sizeof(num));
            memset(dp,-1,sizeof(dp));
            num[0]=line[0];
            for(i=0;i<m;i++)
            {
                dp[i][0]=line[i];
                int temp=get_min(i,t);
                for(j=temp;j>=1;j--)
                {
                    if(num[j-1]==-1)
                        continue;
                    else if(num[j-1]>=line[i])
                    {
                        dp[i][j]=line[i];
                        num[j]=get_max(num[j],line[i]);
                    }
                }
                num[0]=get_max(num[0],line[i]);
            }
            for(i=t-1;i<m;i++)
            {
                if(dp[i][t-1]>ans)
                {
                    ans=dp[i][t-1];
                    flag=1;
                }
            }
            if(flag)
                printf("%d\n",ans);
            else
                printf("OH,NO\n");
        }
    }
    return 0;
}

FROM：暑假第三场