POJ 1195 Mobile phones【树状数组二维】

题目大意

更新：一个矩阵的一个元素
求和：求一个子矩阵元素和

学习了二维的树状数组
和一维一样 
只需要记住，对于矩阵A
C[i][j]的表示的是 
矩阵A[i][j]向左上分别延伸lowbit(i)和lowbit(j)个单位形成的子矩阵的和

#include <stdio.h>
#include <iostream>

using namespace std;

#pragma warning(disable:4996)

#define debug(x)  cout<<#x<<": "<<(x)<<endl;
#define low(x) ((x)&(-x))

typedef long long ll;

//ll a[1025][1025];
ll c[1025][1025];

ll s;

bool update(ll x, ll y, ll A) {
  for (ll i=x; i <= s; i += low(i)) {
    for (ll j=y; j <= s; j += low(j)) {
      c[i][j] += A;
    }
  }
  return true;
}

ll getS(ll x, ll y) {
  ll ret = 0;
  for (ll i = x; i > 0 ; i -= low(i)) {
    for (ll j = y; j > 0; j -= low(j)) {
      ret += c[i][j];
      
    }
  }
  return ret;
}

int main() {

  //freopen("../in1.txt","r",stdin);
  ll t;
  memset(c, 0, sizeof(c));
  while (cin >> t) {

    if (t == 0) {
      cin >> s;
    }
    else if (t == 1) {
      ll x, y, A;
      cin >> x >> y >> A;
      update(x + 1, y + 1, A);
    }
    else if (t == 2) {
      ll x1, y1, x2, y2;
      cin >> x1 >> y1 >> x2 >> y2;
      //debug(getS(x2 + 1, y2 + 1))
      ll s = getS(x2+1,y2+1) - getS(x2+1,y1) - getS(x1,y2+1) + getS(x1,y1) ;
      cout << s << endl;
    }
  }
  
  return 0;
}

更新a[1][1]示例：
图片链接,侵删
​​​https://www.docin.com/p-403698374.html​​