【POJ1195】【二维树状数组】Mobile phones
Description
Write a program, which receives these reports and answers queries about the current total number of active mobile phones in any rectangle-shaped area.
Input
The values will always be in range, so there is no need to check them. In particular, if A is negative, it can be assumed that it will not reduce the square value below zero. The indexing starts at 0, e.g. for a table of size 4 * 4, we have 0 <= X <= 3 and 0 <= Y <= 3.
Table size: 1 * 1 <= S * S <= 1024 * 1024
Cell value V at any time: 0 <= V <= 32767
Update amount: -32768 <= A <= 32767
No of instructions in input: 3 <= U <= 60002
Maximum number of phones in the whole table: M= 2^30
Output
Sample Input
0 4
1 1 2 3
2 0 0 2 2
1 1 1 2
1 1 2 -1
2 1 1 2 3
3
Sample Output
3 4
【分析】
不说了,最基本的二维树状数组
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <vector>
#include <utility>
#include <iomanip>
#include <string>
#include <cmath>
#include <map> const int MAXN = * + ;
const int N=;
using namespace std; int n, array[N][N]; int lowbit(int x){return x & (-x);}
void add(int i, int j, int w) {
int tmp;
while(i <= n){
tmp=j;
while(tmp <= n){
array[i][tmp] += w;
tmp += lowbit(tmp);
}
i += lowbit(i);
}
} int sum(int i, int j){
int tmp, ans=;
while(i > ){
tmp=j;
while(tmp > ){
ans += array[i][tmp];
tmp -= lowbit(tmp);
}
i -= lowbit(i);
}
return ans;
} int main(){
#ifdef LOCAL
freopen("data.txt", "r", stdin);
freopen("out.txt", "w", stdout);
#endif
int op, x1, y1, x2, y2, w;
memset(array, , sizeof(array));
while( ~scanf("%d", &op) ){
if(op == ){
scanf("%d", &n);
n++;
}else if(op == ){
scanf("%d%d%d", &x1, &y1, &w);
add(x1 + , y1 + , w);
}else if(op == ){
scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
int ans=sum(x2 + , y2 + ) - sum(x1, y2 + ) - sum(x2 + , y1) + sum(x1, y1);
printf("%d\n", ans);
}else if(op == )
break;
}
return ;
}
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