DLUTOJ 1033 Matrix

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Time Limit: 2 Sec Memory Limit: 128 MB

Description

We often use the matrix to analyze reality models. There are lots of algorithm about matrix in Linear Algebra like matrix multiplication, matrix determinant and matrix inversion, etc.

Recently, I should use matrix to do structural mechanics analysis. The element in the matrix indicating the mechanical properties of each unit in the structure. Stable sub-structure means a part with same mechanical properties. I want to find the largest stable sub-struture as it has good engineering applications. Reflected in the matrix, the problem above equals to find the largest sub-matrix whose members have the same value.

To accomplish the task perfectly, I wish you can help me to design a good algorithm to solve this problem.

Input

There are multiple test cases.

The first line contains two integers N and M, indicating the size of this N * M matrix A.

The next N line, each line containing M integers. The j-th integer in the i-th line means the element A(i, j).

1 <= N, M <= 800
1 <= A(i, j) <= 1000

Output

For each test, output the size of the largest sub-matrix satisfied the requests.

Sample Input

3 3

1 1 1

1 2 1

1 1 1

2 2

1 2

3 4

4 4

1 1 1 2

1 3 3 2

5 3 3 2

6 6 6 7

Sample Output

3 1 4

HINT

Source

2013大连市赛

Solution

单调栈

Implementation

#include <cstdio>
#include <stack>
using namespace std;
typedef long long LL;

const int N(800+5);

int h[N], L[N], R[N], a[N][N];

stack<int> st;
//[L[i], R[i])
int mono_stack(int l, int r){
    for(; st.size(); st.pop());
    for(int i=l; i<r; i++){
        for(; !st.empty()&&h[st.top()]>=h[i]; st.pop());
        if(st.empty()) L[i]=l;
        else L[i]=st.top()+1;
        st.push(i);
    }
    for(; st.size(); st.pop());
    for(int i=r-1; i>=l; i--){
        for(; !st.empty() && h[st.top()]>=h[i]; st.pop());
        if(st.empty()) R[i]=r;
        else R[i]=st.top();
        st.push(i);
    }
    int res=0;
    for(int i=l; i<r; i++)
        res=max(res, h[i]*(R[i]-L[i]));
    return res;
}

void solve(int n, int m){
    int res=0;
    for(int i=0; i<n; i++){
        if(i==0) for(int j=0; j<m; j++) h[j]=1;
        else for(int j=0; j<m; j++)
            if(a[i][j]==a[i-1][j]) h[j]++; 
            else h[j]=1;
        //two-pointers
        for(int l=0, r; l<m; l=r){
            for(r=l+1; r<m && a[i][r]==a[i][l]; r++);
            res=max(res, mono_stack(l, r));
        }
    }
    printf("%d\n", res);
}

int main(){
    for(int n, m; ~scanf("%d%d", &n, &m); ){
        for(int i=0; i<n; i++)
            for(int j=0; j<m; j++)
                scanf("%d", a[i]+j);
        solve(n, m);
    }
    return 0;
}

DLUTOJ 1033 Matrix

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原文地址：http://www.cnblogs.com/Patt/p/5067912.html