題意：給一個下圖這樣的矩陣，'.'表示空位置，‘#’表示牆，給一個木塊，長度爲2個單位，可橫放可豎放，問有幾種放法？
....#..# .#...... ##.#.... ##..#.## ........
解析：這是一個前綴和的問題，我們可將二維矩陣拆成兩個一維前綴和：行綴和與列綴和，行綴和即爲木塊在某一行放置的種數，列綴和同理；然後再分別一行行一列列的加起來，得出結果。
代碼如下：
#include<iostream>
#include<algorithm>
#include<stdlib.h>
#include<stdio.h>
using namespace std;
typedef long long ll;
const int maxn = 551;
int h, w;
int sum1[maxn][maxn];
int sum2[maxn][maxn];
char c[maxn][maxn];
int q;
int r1, c1, r2, c2;
int main()
{
scanf("%d %d", &h, &w);
getchar();//這裏的getchar()不要忘了，一開始沒清除緩存區，矩陣沒讀進來。。。
for (int i = 1; i <= h; i++) {
for (int j = 1; j <= w; j++) {
scanf("%c", &c[i][j]);
}
getchar();
}
memset(sum1, 0, sizeof(sum1));
memset(sum2, 0, sizeof(sum2));
for (int i = 1; i <= h; i++) {
for (int j = 1; j<w; j++) {
if (c[i][j] == '.'&&c[i][j + 1] == '.') {//計算行的前綴和
sum1[i][j+1] = sum1[i][j] + 1;
}
else {
sum1[i][j+1] = sum1[i][j];
}
}
}
/*for (int i = 1; i <= h; i++) {
for (int j = 1; j <= w; j++) {
printf("%d", sum1[i][j]);
}
printf("\n");
}*/
for (int i = 1; i <= w; i++) {
for (int j = 1; j<h; j++) {
if (c[j][i] == '.'&&c[j + 1][i] == '.') {//計算列的前綴和
sum2[j+1][i] = sum2[j][i] + 1;
}
else {
sum2[j+1][i] = sum2[j][i];
}
}
}
/*printf("\n\n");
for (int i = 1; i <= h; i++) {
for (int j = 1; j <= w; j++) {
printf("%d", sum2[i][j]);
}
printf("\n");
}*/
scanf("%d", &q);
int ans = 0;
while (q--) {
ans = 0;
scanf("%d%d%d%d", &r1, &c1, &r2, &c2);
for (int i = r1; i <= r2; i++) {
ans += (sum1[i][c2] - sum1[i][c1]);//一行行的和
}
for (int i = c1; i <= c2; i++) {
ans += (sum2[r2][i] - sum2[r1][i]);//一列列的和
}
printf("%d\n", ans);
}
return 0;
}