zoukankan      html  css  js  c++  java
  • 牛客第七场 Sudoku Subrectangles

    链接:https://www.nowcoder.com/acm/contest/145/J
    来源:牛客网

    You have a n * m grid of characters, where each character is an English letter (lowercase or uppercase, which means there are a total of 52 different possible letters).

    A nonempty subrectangle of the grid is called sudoku-like if for any row or column in the subrectangle, all the cells in it have distinct characters.

    How many sudoku-like subrectangles of the grid are there?

    输入描述:

    The first line of input contains two space-separated integers n, m (1 ≤ n, m ≤ 1000).

    The next n lines contain m characters each, denoting the characters of the grid. Each character is an English letter (which can be either uppercase or lowercase).

    输出描述:

    Output a single integer, the number of sudoku-like subrectangles.

    示例1

    输入

    复制
    2 3
    AaA
    caa

    输出

    复制
    11

    说明

    For simplicity, denote the j-th character on the i-th row as (i, j).

    For sample 1, there are 11 sudoku-like subrectangles. Denote a subrectangle
    by (x
    1
    , y
    1
    , x
    2
    , y
    2
    ), where (x
    1
    , y
    1
    ) and (x
    2
    , y
    2
    ) are the upper-left and lower-right coordinates of the subrectangle.

    The sudoku-like subrectangles are (1, 1, 1, 1), (1, 2, 1, 2), (1, 3, 1, 3), (2, 1, 2, 1), (2, 2, 2, 2), (2, 3, 2, 3), (1, 1, 1, 2), (1, 2, 1, 3), (2, 1, 2, 2), (1, 1, 2, 1), (1, 3, 2, 3).
    示例2

    输入

    复制
    4 5
    abcde
    fGhij
    klmno
    pqrst

    输出

    复制
    150

    说明

    For sample 2, the grid has 150 nonempty subrectangles, and all of them are sudoku-like.


    预处理每一个点向下想右最大的延伸长度,然后再遍历一遍所有点,枚举所有点是否成立的情况
    AC代码:
    #include <map>
    #include <set>
    #include <stack>
    #include <cmath>
    #include <queue>
    #include <cstdio>
    #include <vector>
    #include <string>
    #include <bitset>
    #include <cstring>
    #include <iomanip>
    #include <iostream>
    #include <algorithm>
    #define ls (r<<1)
    #define rs (r<<1|1)
    #define debug(a) cout << #a << " " << a << endl
    using namespace std;
    typedef long long ll;
    const ll maxn = 1e3+10;
    const ll mod = 1e9+7;
    const double pi = acos(-1.0);
    const double eps = 1e-8;
    ll n, m, lmax[maxn][maxn], pos[maxn];
    ll umax[maxn][maxn], len[maxn];
    char s[maxn][maxn];
    int main() {
        ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
        scanf("%lld%lld",&n,&m);
        for( ll i = 1; i <= n; i ++ ) {
            scanf("%s",s[i]+1);
        }
        for( ll i = 1; i <= n; i ++ ) {
            memset(pos,0,sizeof(pos));
            for( ll j = 1; j <= m; j ++ ) {
                lmax[i][j] = min(lmax[i][j-1]+1,j-pos[s[i][j]]);
                pos[s[i][j]] = j;
            }
        }
        for( ll j = 1; j <= m; j ++ ) {
            memset(pos,0,sizeof(pos));
            for( ll i = 1; i <= n; i ++ ) {
                umax[i][j] = min(umax[i-1][j]+1,i-pos[s[i][j]]);
                pos[s[i][j]] = i;
            }
        }
        ll ans = 0;
        for( ll r = 1; r <= m; r ++ ) {
            memset(len,0,sizeof(len));
            for( ll d = 1; d <= n; d ++ ) {
                for( ll i = 0; i < lmax[d][r]; i ++ ) {
                    len[i] = min(umax[d][r-i],len[i]+1);
                    if(i) {
                        len[i] = min(len[i],len[i-1]);
                    }
                    ans += len[i];
                }
                for( ll i = lmax[d][r]; i <= 54; i ++ ) {
                    len[i] = 0;
                }
            }
        }
        printf("%lld
    ",ans);
        return 0;
    }
    

      

     
    彼时当年少,莫负好时光。
  • 相关阅读:
    Bootstrap入门
    CSS3动画详解(图文教程)
    CSS3属性详解(图文教程)
    CSS3选择器详解
    HTML5详解
    jQuery动画详解
    jQuery的介绍和选择器详解
    html 出现粒子线条,鼠标移动会以鼠标为中心吸附的特效之canvas-nest.js插件
    div 内容宽度自适应、超出后换行
    layui layui.open弹窗后按enter键不停弹窗问题的解决
  • 原文地址:https://www.cnblogs.com/l609929321/p/9454143.html
Copyright © 2011-2022 走看看