Educational Codeforces Round 30 C. Strange Game On Matrix

C. Strange Game On Matrix

Problem Statement

    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:

    1.Initially Ivan’s score is 0;
    2.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

Example 1
    Input
        4 3 2
        0 1 0
        1 0 1
        0 1 0
        1 1 1
    Output
        4 1
Example 2
    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 .

题意

    给出一个n行m列的01矩阵，你可以选矩阵中的几个1把它改成0，操作完之后，对于每一列他会找到最上面的一个1，并从他开始，包含他往下一共k个，里面有几个一就是他的得分。然后一共m列加起来就是他最终的分数。问你最后可以拿到的最大分和最少把几个1改成0。

思路

    这题只要贪心就行了..对于每一个位置，我们先求出从这个位置开始向下的k个位置，或者到n的地方中的1的个数，然后对每一列我们求出这一列最多的一出现在哪里，然后将他前面的1都给消去就行了。

Code

#pragma GCC optimize(3)
#include
using namespace std;
typedef long long ll;
inline void readInt(int &x) {
    x=0;int f=1;char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
    x*=f;
}
inline void readLong(ll &x) {
    x=0;int f=1;char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
    x*=f;
}
/*================Header Template==============*/
int n,m,k,arr[105][105],mxlen[105],ind[105];
int main() {
    readInt(n);
    readInt(m);
    readInt(k);
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
            readInt(arr[i][j]);
    for(int i=1;i<=m;i++) {
        for(int j=1;j<=n;j++) {
            if(arr[j][i]==0)
                continue;
            int len=0;
            for(int p=j;p<=j-1+min(k,n-j+1);p++)
                len+=arr[p][i];
            if(mxlen[i]int ans=0,now=0;
    for(int i=1;i<=m;i++) {
        ans+=mxlen[i];
        for(int j=1;jif(arr[j][i]==1)
                now++;
    }
    printf("%d %d\n",ans,now);
    return 0;
}