zoukankan      html  css  js  c++  java
  • 【POJ2155】【二维树状数组】Matrix

    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

    POJ Monthly,Lou Tiancheng
    【分析】
    算是真正明白二维树状数组了。
    维护的时候只要更改矩阵的四个端点就行了。呵呵呵..
     1 #include <iostream>
     2 #include <cstdio>
     3 #include <algorithm>
     4 #include <cstring>
     5 #include <vector>
     6 #include <utility>
     7 #include <iomanip>
     8 #include <string>
     9 #include <cmath>
    10 #include <map>
    11 
    12 const int MAXN = 1000 + 10; 
    13 const int MAX = 32000 + 10; 
    14 using namespace std;
    15 int n, m;//m为操作次数 
    16 int C[MAXN][MAXN];
    17 
    18 int lowbit(int x){return x&-x;}
    19 /*int sum(int x, int y){
    20     int cnt = 0, tmp;
    21     while (x > 0){
    22           tmp = y;
    23           while (tmp > 0){
    24                 cnt += C[x][tmp];
    25                 tmp -= lowbit(tmp);
    26           }
    27           x -= lowbit(x);
    28     } 
    29     return cnt;
    30 }
    31 void add(int x, int y, int val){
    32      int tmp;
    33      while (x <= 1000){
    34            tmp = y;
    35            while (tmp <= 1000){
    36                  C[x][tmp] += val;
    37                  tmp += lowbit(tmp);
    38            }
    39            x += lowbit(x);
    40      }
    41      return;
    42 }*/
    43 void add(int x,int y)  {  
    44     int i,k;  
    45     for(i=x; i<=n; i+=lowbit(i))  
    46         for(k=y; k<=n; k+=lowbit(k))  
    47             C[i][k]++;  
    48 }  
    49 int sum(int x,int y)  {  
    50     int i,k,cnt = 0;  
    51     for(i=x; i>0; i-=lowbit(i))  
    52         for(k=y; k>0; k-=lowbit(k))  
    53             cnt += C[i][k];  
    54     return cnt;  
    55 }  
    56 
    57 void work(){
    58      scanf("%d%d", &n, &m);
    59      for (int i = 1; i <= m; i++){
    60          char str[2];
    61          scanf("%s", str);
    62          if (str[0] == 'Q'){
    63             int x, y;
    64             scanf("%d%d", &x, &y);
    65             //x++;y++;
    66             printf("%d
    ", sum(x, y)%2);
    67          }else if (str[0] == 'C'){
    68                int x1, y1, x2, y2;
    69                scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
    70                x1++;y1++;x2++;y2++;
    71                add(x2, y2);
    72                add(x2, y1 - 1);
    73                add(x1 - 1, y2);
    74                add(x1 - 1, y1 - 1);
    75          }
    76      }
    77 }
    78 
    79 int main(){
    80     int T;
    81     #ifdef LOCAL
    82     freopen("data.txt",  "r",  stdin);
    83     freopen("out.txt",  "w",  stdout); 
    84     #endif
    85     scanf("%d", &T);
    86     while (T--){ 
    87           memset(C, 0, sizeof(C));
    88           work();
    89           printf("
    ");
    90     }
    91     return 0;
    92 }
    View Code
  • 相关阅读:
    boost库:函数对象
    boost库:智能指针
    linux 查看和修改文件时间
    linux正则表达式
    UVA
    UVA
    UVA
    UVA
    UVA
    对JavaScript的认识?
  • 原文地址:https://www.cnblogs.com/hoskey/p/4320030.html
Copyright © 2011-2022 走看看