POJ 1195 Mobile phones （二維樹狀數組或線段樹）

偶然發現這題還沒A掉............速速解決了.............

樹狀數組和線段樹比較下，線段樹是在是太冗餘了，以後能用樹狀數組還是儘量用.........

#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <vector>
#include <set>
#include <queue>
#include <stack>
#include <climits>//形如INT_MAX一類的
#define MAX 1111
#define INF 0x7FFFFFFF
#define REP(i,s,t) for(int i=(s);i<=(t);++i)
#define ll long long
#define mem(a,b) memset(a,b,sizeof(a))
#define mp(a,b) make_pair(a,b)
#define L(x) x<<1
#define R(x) x<<1|1
# define eps 1e-5
//#pragma comment(linker, "/STACK:36777216") ///傳說中的外掛
using namespace std;
int c[MAX][MAX],n;

int lowbit(int x) {
    return x & (-x);
}

void update(int x1,int y1,int va) {
    for(int i=x1; i<=n+1; i += lowbit(i)) {
        for(int j=y1; j<=n+1; j += lowbit(j)) {
            c[i][j] += va;
        }
    }
}

int query(int x1,int y1) {
    int sum = 0;
    for(int i=x1; i>0; i -= lowbit(i)) {
        for(int j=y1; j>0; j -= lowbit(j)) {
            sum += c[i][j];
        }
    }
    return sum;
}
int main() {
    int op,x1,y1,x2,y2,va;
    scanf("%d%d",&op,&n);
    memset(c,0,sizeof(c));

    while(scanf("%d",&op) ) {
        if(op == 3) break;
        if(op == 1) {
            scanf("%d%d%d",&x1,&y1,&va);
            update(x1+1,y1+1,va);
        }
        if(op == 2) {
            scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
            printf("%d\n",query(x2+1,y2+1) - query(x1,y2+1) + query(x1,y1) - query(x2+1,y1));
        }
    }
    return 0;
}