Light OJ 1266 - Points in Rectangle

题目

Link
就是查询矩形内有多少个点。

分析

二维树状数组维护就好了,。

Code

#include 

const int maxn = 1000 + 131;
struct Point {
    int x, y;
};
int Num[maxn][maxn];
bool Vis[maxn][maxn];

int lowbit(int x) { return x&(-x); }
int Sum(int x, int y) {
    int ret = 0;
    int xx = x;
    while(xx) {
        int yy = y;
        while(yy)
        {
            ret += Num[xx][yy];
            yy -= lowbit(yy);
        }
        xx  -= lowbit(xx);
    }
    return ret;
}
void Add(int x,int y, int val) {
    int xx = x;
    while(xx < maxn)
    {
        int yy = y;
        while(yy < maxn)
        {
            Num[xx][yy] += val;
            yy += lowbit(yy);
        }
        xx += lowbit(xx);
    }
}

int main() {
    int T;
    scanf("%d",&T);
    for(int kase = 1; kase <= T; kase++)
    {
        printf("Case %d:\n",kase);
        memset(Num, 0, sizeof(Num));
        memset(Vis, false, sizeof(Vis));
        int Q, tmp;
        Point a, b;
        scanf("%d",&Q);
        while(Q--)
        {
            scanf("%d",&tmp);
            if(tmp == 0)
            {
                scanf("%d%d",&a.x,&a.y);
                if(!Vis[a.x][a.y])
                    Add(a.x+1,a.y+1,1), Vis[a.x][a.y] = true;
            }
            else {
                scanf("%d%d%d%d",&a.x,&a.y,&b.x,&b.y);
                int Ans = Sum(b.x+1,b.y+1) - Sum(a.x,b.y+1) - Sum(b.x+1,a.y) + Sum(a.x,a.y);
                printf("%d\n",Ans);
            }
        }
    }
    return 0;
}

你可能感兴趣的:(Light OJ 1266 - Points in Rectangle)