zoukankan      html  css  js  c++  java
  • POJ3494Largest Submatrix of All 1’s[单调栈]

    Largest Submatrix of All 1’s
    Time Limit: 5000MS   Memory Limit: 131072K
    Total Submissions: 5883   Accepted: 2217
    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

    Source


    最大全1子矩阵

    对于每一行,维护一个全1高度(tot)递减栈就行了
    注意0时tot[j]=-1,相当于一个高度为0的
    最后依旧要弹出栈里的,如果没有上一行连sample都错
    注意l已经包含了这个点,最后+n-st[top].pos不需要再+1
    //
    //  main.cpp
    //  poj3494
    //
    //  Created by Candy on 10/5/16.
    //  Copyright © 2016 Candy. All rights reserved.
    //
    
    #include <iostream>
    #include <cstdio>
    #include <algorithm>
    #include <cstring>
    using namespace std;
    typedef long long ll;
    const int N=2e3+5;
    inline int read(){
        char c=getchar();int x=0,f=1;
        while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
        while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
        return x;
    }
    int n,m,tot[N],a,ans=0;
    struct data{
        int h,l,pos;
    }st[N];
    int top=0;
    int main(int argc, const char * argv[]) {
        while(scanf("%d%d",&n,&m)!=EOF){
            ans=0;
            memset(tot,0,sizeof(tot));
            for(int i=1;i<=n;i++){
                top=0;
                for(int j=1;j<=m;j++){
                    a=read();
                    if(!a) tot[j]=-1;
                    data t;
                    t.h=++tot[j];t.l=1;t.pos=j;
                    int right=0;
                    while(top&&st[top].h>=t.h){
                        ans=max(ans,(st[top].l+right)*st[top].h);
                        //printf("%d %d %d  %d %d %d
    ",i,j,ans,st[top].l,right,st[top].h);
                        right+=st[top].l; t.l+=st[top].l;
                        top--;
                    }
                    st[++top]=t;//printf("top %d
    ",top);
                }//printf("toptop %d
    ",top);
                while(top){
                    ans=max(ans,st[top].h*(st[top].l +n-st[top].pos));
                    //printf("p %d %d %d %d
    ",st[top].h,st[top].pos,st[top].l,ans);
                    top--;
                }
            }
            printf("%d
    ",ans);
        }
        
        return 0;
    }
     
  • 相关阅读:
    sql优化的几种方法
    webService使用
    Springboot 项目pom.xml 配置文件以及启动类的一些配置
    WebSocket消息推送(群发和指定到个人)
    有n级台阶,每次爬1或者2级台阶,用程序计算有多少种爬法
    对两个字符串进行比较,取出两个字符串中一样部分的长度
    java 日志 log4j
    进制转换
    java 调用第三方动态链接库
    如何使用 OpenFileDialog 组件
  • 原文地址:https://www.cnblogs.com/candy99/p/5933271.html
Copyright © 2011-2022 走看看