牛客网多校赛第七场J--Sudoku Subrectangle

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

时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 32768K，其他语言65536K
Special Judge, 64bit IO Format: %lld

题目描述

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 (x1, y1, x2, y2), where (x1, y1) and (x2, y2) 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.

 

数据量不是很大 可以枚举

但是纯暴力的枚举可能会T 因为是52*52*n*m

感觉还是挺考验思维的...反正我不是很会敲...看了题解觉得好巧妙

可以优化到52*n*m

首先预处理出每个位置距离上一个相同字母的距离    dp

然后枚举每一个点 每次往左走一格看看能多多少个长方形【与高度有关】，这个高度是不可能变大的

 

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<stack>
#include<queue>
#include<map>
#include<set>
#define inf 0x3f3f3f3f
using namespace std;

int n, m;
const int maxn = 1005;
char grid[maxn][maxn];
int pos[maxn], L[maxn][maxn], U[maxn][maxn], len[maxn];
//pos 上一次该字母出现的位子 LU分别表示距离相同字母最近的距离

int main()
{
	while(scanf("%d%d", &n, &m) != EOF){
        for(int i = 1; i <= n; i++){
            scanf("%s", grid[i] + 1);
        }


        for(int i = 1; i <= n; i++){
            memset(pos, 0, sizeof(pos));
            for(int j = 1; j <= m; j++){
                L[i][j] = min(L[i][j - 1] + 1, j - pos[grid[i][j]]);
                pos[grid[i][j]] = j;
            }
        }

        for(int j = 1; j <= m; j++){
            memset(pos, 0, sizeof(pos));
            for(int i = 1; i <= n; i++){
                U[i][j] = min(U[i - 1][j] + 1, i - pos[grid[i][j]]);
                pos[grid[i][j]] = i;
            }
        }

        long long ans = 0;
        for(int j = 1; j <= m; j++){
            memset(len, 0, sizeof(len));
            for(int i = 1; i <= n; i++){
                for(int k = 0; k < L[i][j]; k++){
                    len[k] = min(len[k] + 1, U[i][j - k]);
                    if(k) len[k] = min(len[k], len[k - 1]);
                    ans += len[k];
                }
                for(int k = L[i][j]; k < 54; k++) len[k] = 0;
            }
        }
        printf("%lld\n", ans);
	}
	return 0;
}

        for(int j = 1; j <= m; j++){
            memset(pos, 0, sizeof(pos));
            for(int i = 1; i <= n; i++){
                U[i][j] = min(U[i - 1][j] + 1, i - pos[grid[i][j]]);
                pos[grid[i][j]] = i;
            }
        }

        long long ans = 0;
        for(int j = 1; j <= m; j++){
            memset(len, 0, sizeof(len));//len表示当前列可以向上的长度
            for(int i = 1; i <= n; i++){
                for(int k = 0; k < L[i][j]; k++){
                    len[k] = min(len[k] + 1, U[i][j - k]);
                    if(k) len[k] = min(len[k], len[k - 1]);
                    ans += len[k];
                }
                for(int k = L[i][j]; k < 54; k++) len[k] = 0;
            }
        }
        printf("%lld\n", ans);
	}
	return 0;
}