zoukankan      html  css  js  c++  java
  • poj 3494 Largest Submatrix of All 1’s

    Largest Submatrix of All 1’s
    Time Limit: 5000MS   Memory Limit: 131072K
    Total Submissions: 7700   Accepted: 2779
    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 ≤ mn ≤ 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
    题意:找出最大子矩阵
    思路:单调栈,找出每一行每一个元素上的累计高度,再一行一行的考虑。
    AC代码:
    #define _CRT_SECURE_NO_DEPRECATE
    #include<iostream>
    #include<algorithm>
    #include<vector>
    #include<cstring>
    #include<string>
    #include<cmath>
    using namespace std;
    #define INF 0x3f3f3f3f
    #define N_MAX 2000+20
    typedef long long ll;
    int n,m,h[N_MAX][N_MAX],Map[N_MAX][N_MAX];
    int st[N_MAX],L[N_MAX],R[N_MAX];
    ll solve(const int *h,const int& n) {
        int t = 0;
        for (int i = 0; i < n;i++) {
            while (t > 0 && h[st[t - 1]] >= h[i])t--;
            L[i] = t == 0 ? 0 : st[t - 1] + 1;
            st[t++] = i;
        }
        t = 0;
        for (int i = n - 1; i >= 0;i--) {
            while (t > 0 && h[st[t - 1]] >= h[i])t--;
            R[i] = t == 0 ? n : st[t - 1];
            st[t++] = i;
        }
        ll res = 0;
        for (int i = 0; i < n;i++) {
            res = max(res, (ll)h[i] * (R[i] - L[i]));
        }
        return res;
    }
    
    void prep() {
        for (int i = 0; i < m; i++)h[0][i] = Map[0][i];
        for (int i = 1; i < n;i++) {
            for (int j = 0; j < m;j++) {
               if(Map[i][j])h[i][j] = Map[i][j] + h[i - 1][j];
               else h[i][j] = 0;
            }
        }
    
    }
    
    int main() {
        while (scanf("%d%d",&n,&m)!=EOF){
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < m; j++) {
                    scanf("%d",&Map[i][j]);
                }
            }
            prep();
            ll res = 0;
            for (int i = 0; i < n;i++) {
                res=max(res,solve(h[i],m));
            }
            printf("%lld
    ",res);
        }    
        return 0;
    }
  • 相关阅读:
    中国地区免费注册bitcointalk论坛教程
    Broken Keyboard (a.k.a. Beiju Text) 思路
    IE兼容性測試軟件
    HttpRunner接口自动化框架的使用
    在Linux服务器上安装Python3.7
    在Linux系统上安装Git
    在Linux系统上安装配置ant环境
    桥接模式:探索JDBC的接口
    Intelij Idea下的git使用
    SSM+Maven(教程二):Idea快速入门SSM+Maven框架。
  • 原文地址:https://www.cnblogs.com/ZefengYao/p/9021619.html
Copyright © 2011-2022 走看看