标签:
http://codevs.cn/problem/2630/
Solution
预处理f[i][j],代表第j列前i行的代价
枚举上下界,然后做最大子段和,g[i]代表选到第i列的代价,
g[k]=(g[k-1]<0?0:g[k-1])+f[j][k]-f[i-1][k]
复杂度O(n^3)
Notice
注意赋初值
// <2630.cpp> - Tue Oct 18 20:36:24 2016 // This file is made by YJinpeng,created by XuYike‘s black technology automatically. // Copyright (C) 2016 ChangJun High School, Inc. // I don‘t know what this program is. #include <iostream> #include <cstring> #include <cstdio> #include <cmath> using namespace std; const int MAXN=410; int g[MAXN],f[MAXN][MAXN]; int main() { freopen("2630.in","r",stdin); freopen("2630.out","w",stdout); int n,m,ans=0;scanf("%d%d",&n,&m); for(int i=1;i<=n;i++) for(int j=1;j<=m;j++){ char ch=getchar(); while(ch!=‘0‘&&ch!=‘1‘)ch=getchar(); f[i][j]=f[i-1][j]+(ch==‘0‘?-1:1); } for(int i=1;i<=n;i++) for(int j=i;j<=n;j++) for(int k=1;k<=m;k++) g[k]=(g[k-1]<0?0:g[k-1])+f[j][k]-f[i-1][k],ans=max(g[k],ans); printf("%d\n",ans); return 0; }
标签:
原文地址:http://www.cnblogs.com/YJinpeng/p/5975095.html