To the Max(poj1050,最大连续和)
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/*http://poj.org/problem?id=1050
To the Max
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 36957 Accepted: 19470
Description
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.
Input
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 100. The numbers in the array will be in the range [-127,127].
Output
Output the sum of the maximal sub-rectangle.
Sample Input
4
0 -2 -7 0 9 2 -6 2
-4 1 -4 1 -1
8 0 -2
Sample Output
15
解析:
题意:
给出一个矩阵,找出一个最大和矩形区域;
思路:
把二维转化为一维;
如何转化?就是把连续的行对应位置相加后形成一维,然后按照一维的做法求连续最大和,
依次枚举每一个连续行,然后更新最大值
192 KB 16 ms C++ 729 B
*/
#include<stdio.h>#include<stdlib.h>#include<string.h>#include<math.h>#include<algorithm>#include <iostream>using namespace std;const int maxn=100+10;int arr[maxn][maxn],s[maxn],n;int getpmax(){ int ans,sum; ans=sum=s[1];for(int i=2;i<=n;i++){if(sum<0)sum=0;sum+=s[i];if(sum>ans)ans=sum;}return ans;}int main(){int i,j,ans,k,temp; while(scanf("%d",&n)!=EOF) { for(i=1;i<=n;i++) for(j=1;j<=n;j++) scanf("%d",&arr[i][j]); ans=-1270002;for(i=1;i<=n;i++) { memset(s,0,sizeof(s)); for(j=i;j<=n;j++) { for(k=1;k<=n;k++) { s[k]+=arr[j][k];//对应位置相加,变成一维 } temp=getpmax(); if(ans<temp) ans=temp; } }printf("%d\n",ans); } return 0;}
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