poj1050 To the Max

poj1050 To the Max

(​​ 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 Source Greater New York 2001

n^3的时间内我去处理这个问题 相当于就是最大子段和问题

就是我可以这么去想

| a11 …… a1i ……a1j ……a1n | | a21 …… a2i ……a2j ……a2n | | . . . . . . . | | . . . . . . . | | ar1 …… ari ……arj ……arn | | . . . . . . . | | . . . . . . . | | ak1 …… aki ……akj ……akn | | . . . . . . . | | an1 …… ani ……anj ……ann |

我把每一行中的同一列加起来 就是我可以n^2枚举 我在行中是哪一段 然后加起来变成一维的 然后求一下 最大子段和即可

#include#include#include#define N 11000using namespace std; inline char gc(){ static char now[1<<16],*S,*T; if (T==S) {T=(S=now)+fread(now,1,1<<16,stdin);if (T==S) return EOF;} return *S++;}inline int read(){ int x=0,f=1;char ch=gc(); while(ch<'0'||ch>'9') {if (ch=='-') f=-1;ch=gc();} while(ch<='9'&&ch>='0') x=x*10+ch-'0',ch=gc(); return x*f;}int n,s[110][110];int main(){ freopen("poj1050.in","r",stdin); n=read();int ans=-0x3f3f3f3f; for (int i=1;i<=n;++i){ int tmp=-0x3f3f3f3f; for (int j=1;j<=n;++j) { s[i][j]=read();if (tmp<0) tmp=s[i][j];else tmp+=s[i][j];ans=max(ans,tmp); } } for (int i=1;i<=n;++i){ for (int j=i+1;j<=n;++j){ int tmp=-0x3f3f3f3f; for (int k=1;k<=n;++k){ s[i][k]+=s[j][k]; if (tmp<0) tmp=s[i][k];else tmp+=s[i][k];ans=max(ans,tmp); } } }printf("%d",ans); return 0;}

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