Counting Black
Time Limit: 1000MS | Memory Limit: 10000K |
Total Submissions: 8499 | Accepted: 5469 |
Description
There is a board with 100 * 100 grids as shown below. The left-top gird is denoted as (1, 1) and the right-bottom grid is (100, 100).
We may apply three commands to the board:
1. WHITE x, y, L // Paint a white square on the board,
// the square is defined by left-top grid (x, y)
// and right-bottom grid (x+L-1, y+L-1)
2. BLACK x, y, L // Paint a black square on the board,
// the square is defined by left-top grid (x, y)
// and right-bottom grid (x+L-1, y+L-1)
3. TEST x, y, L // Ask for the number of black grids
// in the square (x, y)- (x+L-1, y+L-1)
In the beginning, all the grids on the board are white. We apply a series of commands to the board. Your task is to write a program to give the numbers of black grids within a required region when a TEST command is applied.
Input
The first line of the input is an integer t (1 <= t <= 100), representing the number of commands. In each of the following lines, there is a command. Assume all the commands are legal which means that they won't try to paint/test the grids outside the board.
Output
For each TEST command, print a line with the number of black grids in the required region.
Sample Input
5
BLACK 1 1 2
BLACK 2 2 2
TEST 1 1 3
WHITE 2 1 1
TEST 1 1 3
Sample Output
7
6
Source
POJ Monthly--2004.05.15 Liu [email protected]
有一個100*100的方格,對其進行N次操作,操作類型分為BLACK、WHITE、TEST三種。如果是BLACK,就将(x, y)到(x+l-1,y+l-1)之間的區域變成黑色,如果是WHITE,則讓其為白色,如果是TEST,就計算這個區域内有多少個黑色方塊,并且輸出。
由于資料給的弱,是以用的方法是暴力模拟
注意: 1)如果是高資料,這種方法會TLE,可以使用二維線段樹進行塗色功能。
代碼(1AC):
#include <cstdio>
#include <cstdlib>
#include <cstring>
int count[110][110];
int main(void){
int casenum ,ii;
char oprator[10];
int x, y, l;
int i, j;
int num, value;
while (scanf("%d", &casenum) != EOF){
getchar();
memset(count, 0, sizeof(count));
for (ii = 0; ii < casenum; ii++){
scanf("%s%d%d%d", oprator, &x, &y, &l);
getchar();
if (!strcmp(oprator, "TEST")){
value = -1;
num = 0;
}
else if (!strcmp(oprator, "WHITE")){
value = 0;
}
else if (!strcmp(oprator, "BLACK")){
value = 1;
}
for (i = x; i <= x + l - 1; i++){
for (j = y; j <= y + l - 1; j++){
if (value == -1){
if (count[i][j] == 1){
num ++;
}
}
else{
count[i][j] = value;
}
}
}
if (value == -1){
printf("%d\n", num);
}
}
}
return 0;
}