uva 108 - Maximum Sum
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Maximum Sum
Background
A problem that is simple to solve in one dimension is often much more difficult to solve in more than one dimension. Consider satisfying a boolean expression in conjunctive normal form in which each conjunct consists of exactly 3 disjuncts. This problem (3-SAT) is NP-complete. The problem 2-SAT is solved quite efficiently, however. In contrast, some problems belong to the same complexity class regardless of the dimensionality of the problem.
The Problem
Given a 2-dimensional array of positive and negative integers, find the sub-rectangle with the largest sum. 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. A sub-rectangle is any contiguous sub-array of size or greater located within the whole array. As an example, the maximal sub-rectangle of the array:
is in the lower-left-hand corner:
and has the sum of 15.
Input and Output
The input consists of an array of integers. The input begins with a single positive integerN on a line by itself indicating the size of the square two dimensional array. This is followed by integers separated by white-space (newlines and spaces). These integers make up the array in row-major order (i.e., all numbers on the first row, left-to-right, then all numbers on 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].
The output is the sum of the maximal sub-rectangle.
Sample Input
40 -2 -7 0 9 2 -6 2-4 1 -4 1 -18 0 -2
Sample Output
15
一维的时候是最大连续子序列问题,二维求的是矩形,可以用捆绑法转换为一维的情况,只要枚举不同的组合方法
#include<stdio.h>#include<string.h>#define inf -9999999int b[101],n;int count(){ int i; int f=1,max=inf,sum=0; for (i=1;i<=n;i++) { if (b[i]>max) max=b[i]; if (b[i]>=0) f=0; } if (f) return max; for (i=1;i<=n;i++) { sum=sum+b[i]; if (sum>max) max=sum; if (sum<0) sum=0; } return max;}int main(){ int i,j,k,t,a[101][101],max,ans; scanf("%d",&n); for (i=1;i<=n;i++) for (j=1;j<=n;j++) scanf("%d",&a[i][j]); max=inf; for (k=0;k<n;k++) { for (i=1;i<=n;i++) { if (i+k<=n) { memset(b,0,sizeof(b)); for (j=i;j<=i+k;j++) for (t=1;t<=n;t++) b[t]+=a[j][t]; ans=count(); if (ans>max) max=ans; } } } printf("%d\n",max); return 0;}
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