HDU 1081 To The Max (DP) 扩展最大子列和,求最大子矩阵和
2011-11-24 14:16
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To The Max
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3768 Accepted Submission(s): 1798
[align=left]Problem Description[/align]
Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1 x 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.
[align=left]Input[/align]
The input consists of an N x 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].
[align=left]Output[/align]
Output the sum of the maximal sub-rectangle.
[align=left]Sample Input[/align]
4 0 -2 -7 0 9 2 -6 2 -4 1 -4 1 -1 8 0 -2
[align=left]Sample Output[/align]
15
[align=left]Source[/align]
Greater New York 2001
#include<stdio.h> #include<string.h> #define MAXN 110 int map[110][110]; /*//数组a,长度为n,的最大连续子段和*/ int MaxSubArray(int *a,int n) { int Max=-65535; int i,tmp=0; for(i=0;i<n;i++) { if(tmp>0)tmp+=a[i]; else tmp=a[i]; if(tmp>Max) Max=tmp; } return Max; } /*//求 n 行,m 列的矩阵的最大子矩阵和 */ int MaxSubMatrix(int n,int m) { int Max=-65535; int i,j,k; int sum; int b[MAXN]; for(i=0;i<n;i++) { memset(b,0,sizeof(b)); for(j=i;j<n;j++) { for(k=0;k<m;k++) b[k]+=map[j][k]; sum=MaxSubArray(b,m); if(sum>Max)Max=sum; } } return Max; } int main() { int n; int i,j; while(scanf("%d",&n)!=EOF) { for(i=0;i<n;i++) for(j=0;j<n;j++) scanf("%d",&map[i][j]); printf("%d\n",MaxSubMatrix(n,n)); } return 0; }
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