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  • 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
    ", 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;
    }
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  • 原文地址:https://www.cnblogs.com/Patt/p/5067912.html
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