POJ 1195 (Mobile phones) 二维树状数组

来源：互联网发布：java 文件md5校验编辑：程序博客网时间：2024/04/25 09:43

题目链接：http://poj.org/problem?id=1195

题意：给出一个矩阵及一些操作，求一些子矩阵的和。

思路：二维树状数组裸题，要注意下标从0开始，所以可以在输入后对每个坐标值都加1。

#include<cstdio>#include<cstring>#include<string>#include<cctype>#include<iostream>#include<set>#include<map>#include<cmath>#include<sstream>#include<vector>#include<stack>#include<queue>#include<bitset>#include<algorithm>#define fin(a) freopen("a.txt","r",stdin)#define fout(a) freopen("a.txt","w",stdout)typedef long long ll;using namespace std;const int maxn = 1050;int a[maxn][maxn];int lowbit(int x) {  return x & -x;}void add(int x, int y, int v, int n) {  for(int i = x; i <= n; i += lowbit(i))    for(int j = y; j <= n; j += lowbit(j))      a[i][j] += v;}ll sum(int x, int y) {  ll ans = 0;  for(int i = x; i >= 1; i -= lowbit(i))    for(int j = y; j >= 1; j -= lowbit(j))      ans += (ll)a[i][j];  return ans;}int main() {  int op, n;  scanf("%d%d", &op, &n);  while(scanf("%d", &op) && op != 3) {    if(op == 1) {      int x, y, z;      scanf("%d%d%d", &x, &y, &z); x++, y++;      add(x, y, z, n);    }    else {      int x1, y1, x2, y2;      scanf("%d%d%d%d", &x1, &y1, &x2, &y2);      x1++, y1++, x2++, y2++;      ll ans = sum(x2, y2) - sum(x1-1, y2) - sum(x2, y1-1) + sum(x1-1, y1-1);      printf("%lld\n", ans);    }  }  return 0;}

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