Educational Codeforces Round 30 C

Strange Game On Matrix

题意:给出n*m的0 1矩阵,在每一列第一个1后面找k-1个数相加,可以将1修改为0,求和的最大值并且修改的最少次数

思路:暴力过去n^3,或者求列的前缀n^2

AC代码:

#include "iostream"
#include "iomanip"
#include "string.h"
#include "stack"
#include "queue"
#include "string"
#include "vector"
#include "set"
#include "map"
#include "algorithm"
#include "stdio.h"
#include "math.h"
#pragma comment(linker, "/STACK:102400000,102400000")
#define bug(x) cout<<x<<" "<<"UUUUU"<<endl;
#define mem(a,x) memset(a,x,sizeof(a))
#define step(x) fixed<< setprecision(x)<<
#define mp(x,y) make_pair(x,y)
#define pb(x) push_back(x)
#define ll long long
#define endl ("\n")
#define ft first
#define sd second
#define lrt (rt<<1)
#define rrt (rt<<1|1)
using namespace std;
const ll mod=1e9+7;
const ll INF = 1e18+1LL;
const int inf = 1e9+1e8;
const double PI=acos(-1.0);
const int N=1e5+100;

int M[105][105],n,m,k,ans,mi;
int main(){
    ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
    cin>>n>>m>>k;
    for(int i=1; i<=n; ++i){
        for(int j=1; j<=m; ++j){
            cin>>M[i][j];
        }
    }
    for(int j=1; j<=m; ++j){
        int x=0, v=0, q=0;
        for(int i=1; i<=n; ++i){
            if(M[i][j]==1){
                int u=0;
                for(int t=i; t<=n && t<i+k; ++t){
                    if(M[t][j]==1) u++;
                }
                if(u>x){
                    x=u,  q=v;
                }
                v++;
            }
        }
        ans+=x, mi+=q;
    }
    cout<<ans<<" "<<mi<<endl;
    return 0;
}

 

posted on 2017-10-14 12:15  lazzzy  阅读(132)  评论(0编辑  收藏  举报

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