poj3494Largest Submatrix of All 1’s

Largest Submatrix of All 1’s

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 9943		Accepted: 3538
Case Time Limit: 2000MS

Description

Given a m-by-n (0,1)-matrix, of all its submatrices of all 1’s which is the largest? By largest we mean that the submatrix has the most elements.

Input

The input contains multiple test cases. Each test case begins with m and n (1 ≤ m, n ≤ 2000) on line. Then come the elements of a (0,1)-matrix in row-major order on m lines each with n numbers. The input ends once EOF is met.

Output

For each test case, output one line containing the number of elements of the largest submatrix of all 1’s. If the given matrix is of all 0’s, output 0.

Sample Input

2 2
0 0
0 0
4 4
0 0 0 0
0 1 1 0
0 1 1 0
0 0 0 0

Sample Output

0
4

Source

POJ Founder Monthly Contest – 2008.01.31, xfxyjwf

 

先进行预处理，将每行连着的高度向下加，如

0 0 0 0                 0 0 0 0
0 1 1 0————>0  1 1 0
0 1 1 0                 0 2 2 0 

0 0 0 0                 0 0 0 0

之后转变为求最大矩形面积模板题；

 

 

 代码

#include<iostream>
#include<cstring>
#include<stack>
#include<cstdio>
using namespace std;
const int N = 2e3+50;

int mp[N][N], dp[N][N];
int main()
{
    int m,n;
    while(scanf("%d%d",&n,&m)!=EOF){
    int ans=0;
    memset(dp,0,sizeof(dp));
    //预处理
    for(int i=1;i<=n;i++){
    for(int j=1;j<=m;j++){
    scanf("%d",&mp[i][j]);
    if(mp[i][j]==1)
    dp[i][j]=dp[i-1][j]+1;
}
dp[i][0]=dp[i][m+1]=-1;
}
for(int i=1;i<n;++i){
stack<int> s;
int r[N];
s.push(m+1);
for(int j=m;j;--j){
while(dp[i][s.top()]>=dp[i][j]) s.pop();
r[j]=s.top();
s.push(j);
}
while(!s.empty())s.pop();
s.push(0);
for(int j=1;j<=m;++j){
while(dp[i][s.top()]>=dp[i][j])s.pop();
ans=max(ans,dp[i][j]*(r[j]-s.top()-1));
s.push(j);
}
}
printf("%d ",ans);
}
}

 

数组模拟

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
const int N = 2e3+50;
struct elem{
 int height;
 int count;
};
int mp[N][N], dp[N][N];
elem stack[N];
int top,tmp;
int main()
{
    int m,n,height,tot;
    while(scanf("%d%d",&n,&m)!=EOF){
    int ans=0;
    memset(dp,0,sizeof(dp));
    //预处理
    for(int i=1;i<=n;i++){
    for(int j=1;j<=m;j++){
    scanf("%d",&mp[i][j]);
    if(mp[i][j]==1)
    dp[i][j]=dp[i-1][j]+1;
}
dp[i][0]=dp[i][m+1]=-1;
}
//
//for(int i=1;i<=n;i++){
// for(int j=1;j<=m;j++){
//  cout<<dp[i][j];
// }
// cout<<endl;
//}
for(int i=1;i<=n;i++){
 top=0;
 for(int j=1;j<=m;++j){
  height=dp[i][j];
      tmp=0;
      while (top > 0 && stack[top - 1].height >= height){
                tot = stack[top - 1].height * (stack[top - 1].count + tmp);
    if (tot > ans) ans = tot;
    tmp += stack[top - 1].count;
    --top;
    }
   stack[top].height = height;
   stack[top].count = 1 + tmp;
   ++top;
    }
    tmp = 0;
    while (top > 0)
    {
    tot = stack[top - 1].height * (stack[top - 1].count + tmp);
    if (tot > ans) ans = tot;
//    cout<<ans<<endl;
    tmp += stack[top - 1].count;
    --top;
    } 
 }
printf("%d ",ans);
}
}