poj-1656-Counting Black-(树状数组)

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

5BLACK 1 1 2BLACK 2 2 2TEST 1 1 3WHITE 2 1 1TEST 1 1 3

 

Sample Output

76

题目给出n组数据，每组形式为x1,y1,l，每组数据形成一个矩阵（x1,y1)为左上角，（x1+l-1，y1+l-1）为右上角，给出3种操作，1.把矩阵内方块全部变白，2.全部变黑，3.统计黑块个数，

题目可以用树状数组做，但是看到这么小的数据，还是暴力解更合适，运行时间也说明了一切

代码（树状数组）（625 MS ，4708 KB）：

#include<iostream>#include<string>#include<cstdio>#include<algorithm>#include<cmath>#include<iomanip>#include<queue>#include<cstring>#include<map>#include<vector>using namespace std;typedef long long ll;#define M 105int tree[M][M],vis[M][M];inline int lowbit(int i){    return i&(-i);}void add(int x,int y,int v){    int i,j;    for(i=x;i<M;i+=lowbit(i))        for(j=y;j<M;j+=lowbit(j))    {        tree[i][j]+=v;    }}int sum(int x,int y){    int i,j,res=0;    for(i=x;i>0;i-=lowbit(i))        for(j=y;j>0;j-=lowbit(j))    {        res+=tree[i][j];    }    return res;}int main(){    int T,x1,y1,x2,y2,l,i,j;    char str[10];    scanf("%d",&T);    while(T--)    {        getchar();        scanf("%s",str);        scanf("%d%d%d",&x1,&y1,&l);        x2=x1+l-1;        y2=y1+l-1;        if(str[0]=='B')        {            for(i=x1;i<=x2;i++)                for(j=y1;j<=y2;j++)            {                if(vis[i][j]==0) {add(i,j,1);vis[i][j]=1;}            }        }        else if(str[0]=='W')        {            for(i=x1;i<=x2;i++)                for(j=y1;j<=y2;j++)            {                if(vis[i][j]==1) {add(i,j,-1); vis[i][j]=0;}            }        }        else        {            printf("%d\n",sum(x2,y2)+sum(x1-1,y1-1)-sum(x2,y1-1)-sum(x1-1,y2));        }    }    return 0;}

代码（0 MS ，748 KB）：

#include<iostream>#include<string>#include<cstdio>#include<algorithm>#include<cmath>#include<iomanip>#include<queue>#include<cstring>#include<map>#include<vector>using namespace std;typedef long long ll;#define M 105int tree[M][M];int main(){    int n,i,j,x1,y1,l,x2,y2;    char str[10];    scanf("%d",&n);    while(n--)    {        getchar();        scanf("%s",str);        scanf("%d%d%d",&x1,&y1,&l);        x2=x1+l-1;        y2=y1+l-1;        if(str[0]=='B')        {            for(i=x1;i<=x2;i++)                for(j=y1;j<=y2;j++)            {                tree[i][j]=1;            }        }        else if(str[0]=='W')        {            for(i=x1;i<=x2;i++)                for(j=y1;j<=y2;j++)            {                tree[i][j]=0;            }        }        else        {            int num=0;            for(i=x1;i<=x2;i++)                for(j=y1;j<=y2;j++)            {                if(tree[i][j]==1) num++;            }            printf("%d\n",num);        }    }    return 0;}