POJ_1050_ToTheMax

To the Max

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 43794 Accepted: 23206

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

40 -2 -7 0 9 2 -6 2-4 1 -4  1 -18  0 -2

Sample Output

15

Source

Greater New York 2001

第一次写二维的树状数组，完全自己从一维推过来的

给自己点个赞。确实像有一本书上那样说的，二维树状数组不是一个多难的东西……

基本裸的二维树状数组

穷举下矩阵的范围就可以了，

用下容斥原理，把不需要的部分去掉

另外我把和都存储下来了，这样应该会省去一部分的运算

#include <iostream>#include <stdio.h>using namespace std;typedef long long LL;const int M=105;LL tree[M][M];LL su[M][M];inline int lowbit(int x){    return x&(-x);}void add(int x,int y,LL v){    for(int i=x;i<M;i+=lowbit(i))        for(int j=y;j<M;j+=lowbit(j))            tree[i][j]+=v;}LL getsum(int x,int y)                       //左上为1,1右下角为x,y的矩阵的和{    LL s=0;    for(int i=x;i;i-=lowbit(i))       for(int j=y;j;j-=lowbit(j))           s+=tree[i][j];    return s;}LL getsu(int stx,int edx,int sty,int edy)        //左上stx,sty右下edx,edy的矩阵的和{    return su[edx][edy]-su[stx-1][edy]-su[edx][sty-1]+su[stx-1][sty-1];}int main(){    int n;    int num;    LL ans=-1e9;    //freopen("1.in","r",stdin);    scanf("%d",&n);    for(int i=1;i<=n;i++)        for(int j=1;j<=n;j++)        {            scanf("%d",&num);            add(i,j,num);        }    //cout<<"1"<<endl;    for(int i=1;i<=n;i++)        for(int j=1;j<=n;j++)            su[i][j]=getsum(i,j);    for(int stx=1;stx<=n;stx++)        for(int edx=stx;edx<=n;edx++)            for(int sty=1;sty<=n;sty++)                for(int edy=sty;edy<=n;edy++)                    ans=max(ans,getsu(stx,edx,sty,edy));    printf("%I64d\n",ans);    return 0;}