Codeforces 873C Strange Game On Matrix【贪心】水题

最新推荐文章于 2024-06-14 10:34:43 发布

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最新推荐文章于 2024-06-14 10:34:43 发布

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分类专栏：水题文章标签： Codeforces 873C

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本文介绍了一个基于矩阵的游戏，玩家需通过转换矩阵元素来最大化得分，并确保转换次数最少。文章提供了问题解决思路及AC代码。

Ivan is playing a strange game.

He has a matrix a with n rows and m columns. Each element of the matrix is equal to either 0 or 1. Rows and columns are 1-indexed. Ivan can replace any number of ones in this matrix with zeroes. After that, his score in the game will be calculated as follows:

Initially Ivan's score is 0;
In each column, Ivan will find the topmost 1 (that is, if the current column is j, then he will find minimum i such that ai, j = 1). If there are no 1's in the column, this column is skipped;
Ivan will look at the next min(k, n - i + 1) elements in this column (starting from the element he found) and count the number of 1's among these elements. This number will be added to his score.

Of course, Ivan wants to maximize his score in this strange game. Also he doesn't want to change many elements, so he will replace the minimum possible number of ones with zeroes. Help him to determine the maximum possible score he can get and the minimum possible number of replacements required to achieve that score.

Input

The first line contains three integer numbers n, m and k (1 ≤ k ≤ n ≤ 100, 1 ≤ m ≤ 100).

Then n lines follow, i-th of them contains m integer numbers — the elements of i-th row of matrix a. Each number is either 0 or 1.

Output

Print two numbers: the maximum possible score Ivan can get and the minimum number of replacements required to get this score.

Examples

input
4 3 2
0 1 0
1 0 1
0 1 0
1 1 1

output
4 1

input
3 2 1
1 0
0 1
0 0

output
2 0

Note

In the first example Ivan will replace the element a1, 2.

题目大意：

给出一个N*M的矩阵，我们可以使得矩阵中任意一个1变成0.计算价值的方法是：找到每一列的第一个1，然后计算这个1以及其下边小于等于k的连续段落里边有多少个1，然后加到答案中，现在我们希望使得结果尽可能的大，而且使得变换的次数尽可能的小，输出最大值以及变换的最小次数。

思路：

每一列都是一个子问题，那么对应每一列我们枚举到1的时候，判断删除其前边所有的1是否更优即可。

Ac代码：

#include<stdio.h>
#include<string.h>
using namespace std;
int a[150][150];
int main()
{
    int n,m,k;
    while(~scanf("%d%d%d",&n,&m,&k))
    {
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=m;j++)
            {
                scanf("%d",&a[i][j]);
            }
        }
        int ans1=0,ans2=0;
        for(int j=1;j<=m;j++)
        {
            int pre=0;
            int maxn=0;
            int del=0;
            for(int i=1;i<=n;i++)
            {
                int cnt=0;
                for(int l=i;l<=n&&l-i+1<=k;l++)
                {
                    if(a[l][j]==1)cnt++;
                }
                if(maxn<cnt)
                {
                    maxn=cnt;
                    del=pre;
                }
                pre+=a[i][j];
            }
            ans1+=maxn;
            ans2+=del;
        }
        printf("%d %d\n",ans1,ans2);
    }
}