树状数组的修改+查询

Color the ball

Time Limit: 9000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 29305    Accepted Submission(s): 14264

Problem Description

N个气球排成一排，从左到右依次编号为1,2,3....N.每次给定2个整数a b(a <= b),lele便为骑上他的“小飞鸽"牌电动车从气球a开始到气球b依次给每个气球涂一次颜色。但是N次以后lele已经忘记了第I个气球已经涂过几次颜色了，你能帮他算出每个气球被涂过几次颜色吗？

 

 

Input

每个测试实例第一行为一个整数N,(N <= 100000).接下来的N行，每行包括2个整数a b(1 <= a <= b <= N)。
当N = 0，输入结束。

 

 

Output

每个测试实例输出一行，包括N个整数，第I个数代表第I个气球总共被涂色的次数。

 

 

Sample Input

3 1 1 2 2 3 3 3 1 1 1 2 1 3 0

 

 

Sample Output

1 1 1 3 2 1

 

 

Author

8600

 

 

Source

 

 

//hdu  1556
//区间修改、单点查询

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <set>
#include <map>
using namespace std;
#define ll long long
#define  lowbit(x) x&(-x)
const int N=1e5+9;
ll  a[N],c[N];
ll  n,x,y;
void  update(ll x,ll  num){
    while(x<=n){
        c[x]+=num;
        x+=lowbit(x);
    }
}
ll  getsum(ll x){
    ll sum = 0;
    while(x>0){
        sum+=c[x];
        x-=lowbit(x);
    }
    return sum;
}
int  main()
{
    while(~scanf("%lld",&n),n){
        for(ll  i = 1 ;i <=n+9;i++){
            a[i]=0;
            c[i] =0;
        }
        for(ll i =0;i<n;i++){
            scanf("%lld%lld",&x,&y);
            update(x,1);
            update(y+1,-1);
        }
        for(ll  i =1;i<=n;i++){
            printf("%lld%c",getsum(i),i==n?'\n':' ');
        }
    }
    return  0;
}

 

//POJ   2155

Matrix

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 33412		Accepted: 12101

Description

Given an N*N matrix A, whose elements are either 0 or 1. A[i, j] means the number in the i-th row and j-th column. Initially we have A[i, j] = 0 (1 <= i, j <= N). 

We can change the matrix in the following way. Given a rectangle whose upper-left corner is (x1, y1) and lower-right corner is (x2, y2), we change all the elements in the rectangle by using "not" operation (if it is a '0' then change it into '1' otherwise change it into '0'). To maintain the information of the matrix, you are asked to write a program to receive and execute two kinds of instructions. 

1. C x1 y1 x2 y2 (1 <= x1 <= x2 <= n, 1 <= y1 <= y2 <= n) changes the matrix by using the rectangle whose upper-left corner is (x1, y1) and lower-right corner is (x2, y2). 
2. Q x y (1 <= x, y <= n) querys A[x, y]. 

Input

The first line of the input is an integer X (X <= 10) representing the number of test cases. The following X blocks each represents a test case. 

The first line of each block contains two numbers N and T (2 <= N <= 1000, 1 <= T <= 50000) representing the size of the matrix and the number of the instructions. The following T lines each represents an instruction having the format "Q x y" or "C x1 y1 x2 y2", which has been described above. 

Output

For each querying output one line, which has an integer representing A[x, y]. 

There is a blank line between every two continuous test cases. 

Sample Input

1
2 10
C 2 1 2 2
Q 2 2
C 2 1 2 1
Q 1 1
C 1 1 2 1
C 1 2 1 2
C 1 1 2 2
Q 1 1
C 1 1 2 1
Q 2 1

Sample Output

1
0
0
1

Source

 

 

 // 区间修改、单点查询
 1 #include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <set>
#include <map>
using namespace std;
#define ll long long
#define  lowbit(x) x&(-x)
/*
一维的进行互补操作(1变成0,0变成1)，[a,b] 只需要a:+1,b+1:+1
求第K个数的值，只需要getsum(k)%2即可
二维的要用到容斥定理:(x1,y1),(x2,y2)
(x1,y1,1),(x1,y2+1,1),(x2+1,y1,1),(x2+1,y2+1,1)
求(x,y)的值,只需要getsum(x,y)%2即可。
*/
const int N = 1e3+9;
int x,n,t;
char  s[5];
int  a[N][N],c[N][N];
void init(int n){
    for(int i =0;i<=n;i++){
         for(int j =0;j<=n;j++){
             a[i][j] = 0;
             c[i][j] = 0;
         }    
    }
}
void  update(int  x,int y,int num){
    for(int  i =x;i<=n;i+=lowbit(i)){
        for(int  j =y;j<=n;j+=lowbit(j)){
            c[i][j]+=num;
        }
    }
}
int  getsum(int x,int y){
    int  sum = 0;
    for(int   i =x;i>0;i-=lowbit(i)){
        for(int  j =y;j>0;j-=lowbit(j)){
            sum += c[i][j];
        }
    }
    return sum ;
}
int main()
{
    scanf("%d",&x);
    while(x--){
        scanf("%d%d",&n,&t);
        init(n);
        int  x1,y1,x2,y2;
        for(int  i = 0;i<t;i++){
            scanf("%s",s);
            if(s[0]=='C'){
                scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                update(x1,y1,1);
                update(x1,y2+1,1);
                update(x2+1,y1,1);
                update(x2+1,y2+1,1);
            }
            else{
                scanf("%d%d",&x1,&y1);
                printf("%d\n",getsum(x1,y1)%2);
            }
        }
                if(x>0) printf("\n");
    }
    return 0;
}

 

 //POJ   1656

Counting Black

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 11630		Accepted: 7501

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

 

 

//单点修改+区间查询
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <set>
#include <map>
using namespace std;
#define ll long long
#define  lowbit(x) x&(-x)
const  int N  = 1e2+5;
int a[N][N],c[N][N];
int t;
char s[10];
int  x,y,l;
void  update(int  x,int y,int  num){
    for(int i =x;i<=100;i+=lowbit(i)){
        for(int j =y;j<=100;j+=lowbit(j)){
            c[i][j]+=num;
        }
    }
}
int  getsum(int  x,int y){
    int  sum = 0;
    for(int i =x;i>0;i-=lowbit(i)){
        for(int  j = y;j>0;j-=lowbit(j)){
            sum += c[i][j] ;
        }
    }
    return  sum ;
}
int  main()
{
    scanf("%d",&t);
    while(t--){
        scanf("%s",s);
        if(s[0]=='B'){
            scanf("%d%d%d",&x,&y,&l);
            for(int i =x;i<=x+l-1;i++){
                for(int  j =y;j<=y+l-1;j++){
                    if(!a[i][j]){
                        a[i][j] = 1;
                        update(i,j,1);
                    }
                }
            }
        }
    else if(s[0]=='W'){            
            scanf("%d%d%d",&x,&y,&l);
            for(int i = x;i <= x+l-1;i++){
                for(int  j = y;j <= y+l-1;j++){
                    if(a[i][j] == 1 ){
                        a[i][j] = 0;
                        update(i,j,-1);
                    }
                }
            }
        }
        else{
            scanf("%d%d%d",&x,&y,&l);
            printf("%d\n",getsum(x-1,y-1)+getsum(x+l-1,y+l-1)-getsum(x-1,y+l-1)-getsum(x+l-1,y-1));
        }
    }
    return   0;
}