Codeforces 919C - Seat Arrangements

传送门：http://codeforces.com/contest/919/problem/C

给出一张n×m的座位表（有已占座位和空座位），请选择同一行（或列）内连续的k个座位。求选择的方法数。

Hack：首先，若k=1，则所有的空座位均可选，方法数即为空座位数。

对于某一行（或列）中连续的len个座位，选择的方法数为len-k+1。

于是，分别逐行、逐列地统计，求出答案。参考程序如下：

#include <stdio.h>
#define MAX_N 2000

char map[MAX_N][MAX_N];
int row[MAX_N][MAX_N], col[MAX_N][MAX_N];

int main(void)
{
    int n, m, k;
    scanf("%d%d%d", &n, &m, &k);
    for (int i = 0; i < n; i++) {
        getchar();
        for (int j = 0; j < m; j++)
            map[i][j] = getchar();
    }
    int ans = 0;
    if (k == 1) {
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++)
                if (map[i][j] == '.') ans++;
        }
        printf("%d\n", ans);
        return 0;
    }
    //pass by row
    for (int i = 0; i < n; i++) {
        int len = 0, cnt = 0;
        for (int j = 0; j < m; j++) {
            if (map[i][j] == '*') {
                if (len) row[i][cnt++] = len;
                len = 0;
            }
            else len++;
        }
        if (len) row[i][cnt++] = len;
    }
    //pass by column
    for (int j = 0; j < m; j++) {
        int len = 0, cnt = 0;
        for (int i = 0; i < n; i++) {
            if (map[i][j] == '*') {
                if (len) col[j][cnt++] = len;
                len = 0;
            }
            else len++;
        }
        if (len) col[j][cnt++] = len;
    }
    for (int i = 0; i < n; i++) {
        for (int j = 0; row[i][j] != 0; j++)
            if (row[i][j] >= k) ans += row[i][j] - k + 1;
    }
    for (int i = 0; i < m; i++) {
        for (int j = 0; col[i][j] != 0; j++)
            if (col[i][j] >= k) ans += col[i][j] - k + 1;
    }
    printf("%d\n", ans);
    return 0;
}