HDU2830--Matrix Swapping II

Problem Description

Given an N * M matrix with each entry equal to 0 or 1. We can find some rectangles in the matrix whose entries are all 1, and we define the maximum area of such rectangle as this matrix’s goodness.

We can swap any two columns any times, and we are to make the goodness of the matrix as large as possible.

Input

There are several test cases in the input. The first line of each test case contains two integers N and M (1 ≤ N,M ≤ 1000). Then N lines follow, each contains M numbers (0 or 1), indicating the N * M matrix

Output

Output one line for each test case, indicating the maximum possible goodness.

Sample Input

3 41011100100013 4101010010001

Sample Output

42Note: Huge Input, scanf() is recommended.

/*此题和那个最大矩形面积其实是一样的只不过这题多了个排序。*/#include <iostream>#include <cstdio>#include <algorithm>using namespace std;struct REC{int l,r,h;}rec[1000008];//l是左边第一比他低的，r是右边第一个比他低的int h[1008][1008];int hh[1008];inline int max(int a,int b){return a>b?a:b;}bool cmp(int a,int b){return a>b?1:0;}int main(){int r,c;while(scanf("%d%d",&r,&c)==2){int maxs=0;getchar();for(int i=1;i<=r;i++){for(int j=1;j<=c;j++){h[i][j]=getchar()=='1'?h[i-1][j]+1:0;hh[j]=h[i][j];}getchar();sort(hh+1,hh+c+1,cmp);for(int j=1;j<=c;j++){rec[j].h=hh[j];}for(int j=1;j<=c;j++){if(j==1){rec[j].l=0;rec[0].h=-1;continue;}if(rec[j].h>rec[j-1].h){rec[j].l=j-1;continue;}if(rec[j].h==rec[j-1].h){rec[j].l=rec[j-1].l;continue;}if(rec[j].h<rec[j-1].h){int k=j-1;while(rec[k].h>=rec[j].h){k=rec[k].l;}rec[j].l=k;continue;}}//从左往右找一遍for(int j=c;j>=1;j--){if(j==c){rec[j].r=c+1;rec[c+1].h=-1;continue;}if(rec[j].h>rec[j+1].h){rec[j].r=j+1;continue;}if(rec[j].h==rec[j+1].h){rec[j].r=rec[j+1].r;continue;}if(rec[j].h<rec[j+1].h){int k=j+1;while(rec[j].h<=rec[k].h){k=rec[k].r;}rec[j].r=k;}}//现在已经每个矩形左边第一比他低和右边第一个比他低的都知道了//接下来是求最大面积for(int j=1;j<=c;j++){maxs=max(maxs,rec[j].h*(rec[j].r-rec[j].l-1));}}printf("%d\n",maxs);}return 0;}