POJ1050 To the Max(贪心)

描述Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1*1 or greater located within the whole array. The sum of a rectangle is the sum of all the elements in that rectangle. In this problem the sub-rectangle with the largest sum is referred to as the maximal sub-rectangle.
As an example, the maximal sub-rectangle of the array:

0 -2 -7 0
9 2 -6 2
-4 1 -4 1
-1 8 0 -2
is in the lower left corner:

9 2
-4 1
-1 8
and has a sum of 15.输入The input consists of an N * N array of integers. The input begins with a single positive integer N on a line by itself, indicating the size of the square two-dimensional array. This is followed by N^2 integers separated by whitespace (spaces and newlines). These are the N^2 integers of the array, presented in row-major order. That is, all numbers in the first row, left to right, then all numbers in the second row, left to right, etc. N may be as large as 500. The numbers in the array will be in the range [-127,127].输出Output the sum of the maximal sub-rectangle.样例输入

4
0 -2 -7 0 9 2 -6 2
-4 1 -4  1 -1

8  0 -2

样例输出

15

题意：找出最大子矩阵和
题解：先考虑一维下的做法：

void maxx(){
	int sum=-mod;
	rep(i,1,n){
		if(sum>0)sum+=col[i];
		else sum=col[i];
		if(ans<sum)ans=sum;
	}
}

那么在二维下，只需要将所有的行的状态都考虑进去，放入一个一维数组中即可

//#include<bits/stdc++.h>
#include<algorithm>
#include <iostream>
#include  <cstdlib>
#include  <cstring>
#include  <cassert>
#include   <cstdio>
#include   <vector>
#include   <string>
#include    <cmath>
#include    <queue>
#include    <stack>
#include      <set>
#include      <map>
using namespace std;
#define P(a,b,c) make_pair(a,make_pair(b,c))
#define rep(i,a,n) for (int i=a;i<=n;i++)
#define per(i,a,n) for (int i=n;i>=a;i--)
#define CLR(vis) memset(vis,0,sizeof(vis))
#define MST(vis,pos) memset(vis,pos,sizeof(vis))
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef pair<int,pair<int,int> >pii;
typedef long long ll;
const ll mod = 1000000007;
const int INF = 0x3f3f3f3f;
ll gcd(ll a, ll b) { return b ? gcd(b, a%b) : a; }

const int maxn = 1e3+5;
int ma[maxn][maxn];
int col[maxn]; 
int n,ans=0;
void maxx(){
	int sum=-mod;
	rep(i,1,n){
		if(sum>0)sum+=col[i];
		else sum=col[i];
		if(ans<sum)ans=sum;
	}
}
int main(){
	scanf("%d", &n);
	rep(i,1,n)
		rep(j,1,n)
			scanf("%d", &ma[i][j]);
	rep(i,1,n){
		memset(col,0,sizeof(col));
		rep(j,i,n){
			rep(k,1,n){
				col[k]+=ma[j][k];
			}
			maxx();
		}
	}
	cout<<ans;
	return 0;
}
//可看作一行，但是列举所有情况，即列举不同的行数的和