C. New Year and Domino
time limit per test
3 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
They say "years are like dominoes, tumbling one after the other". But would a year fit into a grid? I don‘t think so.
Limak is a little polar bear who loves to play. He has recently got a rectangular grid with h rows and w columns. Each cell is a square, either empty (denoted by ‘.‘) or forbidden (denoted by ‘#‘). Rows are numbered 1 through h from top to bottom. Columns are numbered 1through w from left to right.
Also, Limak has a single domino. He wants to put it somewhere in a grid. A domino will occupy exactly two adjacent cells, located either in one row or in one column. Both adjacent cells must be empty and must be inside a grid.
Limak needs more fun and thus he is going to consider some queries. In each query he chooses some rectangle and wonders, how many way are there to put a single domino inside of the chosen rectangle?
Input
The first line of the input contains two integers h and w (1?≤?h,?w?≤?500) – the number of rows and the number of columns, respectively.
The next h lines describe a grid. Each line contains a string of the length w. Each character is either ‘.‘ or ‘#‘ — denoting an empty or forbidden cell, respectively.
The next line contains a single integer q (1?≤?q?≤?100?000) — the number of queries.
Each of the next q lines contains four integers r1i, c1i, r2i, c2i (1?≤?r1i?≤?r2i?≤?h,?1?≤?c1i?≤?c2i?≤?w) — the i-th query. Numbers r1i and c1i denote the row and the column (respectively) of the upper left cell of the rectangle. Numbers r2i and c2i denote the row and the column (respectively) of the bottom right cell of the rectangle.
Output
Print q integers, i-th should be equal to the number of ways to put a single domino inside the i-th rectangle.
Examples
input
Copy
5 8....#..#.#......##.#....##..#.##........41 1 2 34 1 4 11 2 4 52 5 5 8
output
Copy
401015
input
Copy
7 39........................................###..###..#..###.....###..###..#..###....#..#.#..#..#.........#..#.#..#..#....###..#.#..#..###.....###..#.#..#..###..#....#.#..#....#.....#....#.#..#..#.#..###..###..#..###.....###..###..#..###........................................61 1 3 202 10 6 302 10 7 302 2 7 71 7 7 71 8 7 8
output
Copy
53891202302
Note
A red frame below corresponds to the first query of the first sample. A domino can be placed in 4 possible ways.
自闭的人生
#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#include <algorithm>
#include <set>
#include <iostream>
#include <map>
#include <stack>
#include <string>
#include <vector>
#define pi acos(-1.0)
#define eps 1e-6
#define fi first
#define se second
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define bug printf("******\n")
#define mem(a,b) memset(a,b,sizeof(a))
#define fuck(x) cout<<"["<<x<<"]"<<endl
#define f(a) a*a
#define sf(n) scanf("%d", &n)
#define sff(a,b) scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define sffff(a,b,c,d) scanf("%d %d %d %d", &a, &b, &c, &d)
#define pf printf
#define FRE(i,a,b) for(i = a; i <= b; i++)
#define FREE(i,a,b) for(i = a; i >= b; i--)
#define FRL(i,a,b) for(i = a; i < b; i++)
#define FRLL(i,a,b) for(i = a; i > b; i--)
#define FIN freopen("DATA.txt","r",stdin)
#define gcd(a,b) __gcd(a,b)
#define lowbit(x) x&-x
#pragma comment (linker,"/STACK:102400000,102400000")
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
const int maxn = 2e5 + 10;
int n, m, sum[505][505], y[505][505], x[505][505];
char mp[505][505];
int main() {
sff(n, m);
for (int i = 1 ; i <= n ; i++)
scanf("%s", mp[i] + 1);
for (int i = 1 ; i <= n ; i++) {
for (int j = 1 ; j <= m ; j++) {
if (mp[i][j] == ‘.‘ && mp[i + 1][j] == ‘.‘ ) {
sum[i][j]++;
y[i][j] = 1;
}
if (mp[i][j] == ‘.‘ && mp[i][j + 1] == ‘.‘ ) {
sum[i][j]++;
x[i][j] = 1;
}
sum[i][j] += sum[i - 1][j] + sum[i][j - 1] - sum[i - 1][j - 1];
}
}
int q;
sf(q);
while(q--) {
int x1, y1, x2, y2;
sffff(x1, y1, x2, y2);
int ans = sum[x2][y2] - sum[x1 - 1][y2] - sum[x2][y1 - 1] + sum[x1 - 1][y1 - 1];
for (int i = x1 ; i <= x2 ; i++) ans -= x[i][y2];
for (int i = y1 ; i <= y2 ; i++) ans -= y[x2][i];
printf("%d\n", ans);
}
return 0;
}
原文地址：https://www.cnblogs.com/qldabiaoge/p/9452027.html