POJ1195 Mobile phones 二维树状数组的应用

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这题可以用线段树离散化做，用二维树状数组做了一下，不懂得可以看一下这篇文章：http://www.java3z.com/cwbwebhome/article/article1/1369.html?id=4804

题意：

给你一个s*s的正方形区域，先输入一个x，若x==0,则再输入一个s，若x==1,则输入x,y,a，表示矩阵中(x,y)这点的值加上a，若x==2，输入l,b,r,t,代表以左上角的点(l,b)右下角的点(r,t),求这一片矩形内的矩阵元素之和，若x==3则结束此次程序

就是最基础的二维树状数组的应用，修改单点的值，并且快速求区间和

#include<iostream>#include<cstdio>#include<list>#include<algorithm>#include<cstring>#include<string>#include<stack>#include<map>#include<vector>#include<cmath>#include<memory.h>#include<set>#include<cctype>#define ll long long#define LL __int64#define eps 1e-8//const ll INF=9999999999999;#define inf 0xfffffffusing namespace std;//vector<pair<int,int> > G;//typedef pair<int,int> P;//vector<pair<int,int>> ::iterator iter;////map<ll,int>mp;//map<ll,int>::iterator p;int n;int c[1000 + 35][1000 + 35];void clear() {memset(c,0,sizeof(c));}int lowbit(int x) {return x&(-x);}void add(int x,int y,int value) {int i = y;while(x <= n) {y = i;while(y <= n) {c[x][y] += value;y += lowbit(y);}x += lowbit(x);}}int get_sum(int x,int y) {int sum=0,i = y;while(x > 0) {y = i;while(y > 0) {sum += c[x][y];y -= lowbit(y);}x -= lowbit(x);}return sum;}int main() {int x;while(true) {scanf("%d",&x);if(x == 0) {scanf("%d",&n);n++;clear();}else if(x == 1) {int x,y,a;scanf("%d %d %d",&x,&y,&a);add(++x,++y,a);//矩阵行列标由0开始要处理掉}else if(x == 2) {int l,b,r,t;scanf("%d %d %d %d",&l,&b,&r,&t);l++,b++,r++,t++;//处理行列标int ans = 0;//int a1 = get_sum(r,b-1);//int a2 = get_sum(l-1,b-1);//int a3 = get_sum(r,t);//int a4 = get_sum(l-1,t);ans = get_sum(r,t) - get_sum(l-1,t) - get_sum(r,b-1) + get_sum(l-1,b-1);printf("%d\n",ans);}else break;}return 0;}

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