hdu 1081 To The Max 基础dp

To The Max Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Problem Description 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.   Input 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].   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 分析：通过转换成最大字段和 第一步:dp[i][j]=a[1][j]+ …… +a[i][j] 第二步:sum+=dp[j][k]-dp[k][i-1]; 第三步:求max和sum的较大值 代码： #include<stdio.h> #include<string.h> int a[102][102],dp[102][102]; int main() { int i,j,k,n,max,sum; while(scanf("%d",&n)!=EOF) { memset(dp,0,sizeof(dp)); for(i=1;i<=n;i++) for(j=1;j<=n;j++) { scanf("%d",&a[i][j]); dp[i][j]=dp[i][j-1]+a[i][j]; } int max=-9999; for(i=1;i<=n;i++) //控制行 for(j=i;j<=n;j++) //控制列 { int sum=0; for(k=1;k<=n;k++) //控制连续的层数 { sum=sum>0?sum:0; sum+=dp[k][j]-dp[k][i-1]; max=max>sum?max:sum; } } printf("%d\n",max); } return 0; }