POJ 1195 Mobile Phones 二维树状数组 

 1 #include<iostream>
 2 #include<math.h>
 3 #include<string.h>
 4 #include<queue>
 5 #include<algorithm>
 6 #include<cstdio>
 7 #define rep(i,a,n) for(i=a;i<=n;i++)
 8 #define per(i,a,n) for(i=n;i>=a;i--)
 9 #define scd1(a) scanf("%d",&a)
10 #define scd2(a,b) scanf("%d%d",&a,&b)
11 #define ll long long
12 #define mem(a,b) memset(a,b,sizeof a);
13 using namespace std;
14 
15 const int maxn=1024;
16 int T,cnt=0,N,M;
17 int mt[maxn+5][maxn+5];
18 int ist;
19 
20 int lowbit(int num){
21     return num&(-num);
22 }
23 
24 void add(int x, int y, int d){
25     while(x<=N){
26         int j=y;
27         while(j<=N){
28             mt[x][j]+=d;j+=lowbit(j);
29         }x+=lowbit(x);
30     }
31 }
32 
33 int sum(int x, int y){
34     int ret=0;
35     while(x>0){
36         int j=y;
37         while(j>0){
38             ret+=mt[x][j];j-=lowbit(j);
39         }
40         x-=lowbit(x);
41     }
42     return ret;
43 }
44 
45 int main(){
46     while(~scd2(ist,N)){
47         mem(mt,0);
48         while(~scd1(ist)&&ist!=3){
49             if(ist==1){
50                 int X,Y,A;
51                 scd2(X,Y);scd1(A);X++;Y++;
52                 add(X,Y,A);
53             }
54             if(ist==2){
55                 int L,B,R,T;
56                 scd2(L,B);scd2(R,T);L++;B++;R++;T++;
57                 int tmp=sum(R,T);
58                 tmp+=sum(L-1,B-1);
59                 tmp-=sum(L-1,T);
60                 tmp-=sum(R,B-1);
61                 printf("%d\n",tmp);
62             }
63         }
64     }
65 }
View Code

题意描述: 给出一个矩阵, 每次修改一个点的值, 每次质询要求回答一个子矩阵的数值和. 

还是裸的二维树状数组, 它的对偶题是POJ2155 Matrix 题解链接:http://www.cnblogs.com/LiXinze/p/7353436.html

posted @ 2017-08-13 14:37  Li_Xinze  阅读(123)  评论(0)  编辑  收藏  举报