Educational Codeforces Round 30 C
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; }