poj 2352 Stars(线段树)-优快云博客

本文链接：https://blog.youkuaiyun.com/keshuai19940722/article/details/40403503

本文介绍了解决POJ2352 Stars问题的一种方法，使用线段树进行单点修改和区间查询，实现计算每个点左下角有多少个点，并输出相应计数。

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题目大意：给定N个点，然后计算每个点左下角有多少个点，然后输出左下角点个数为x的点有多少个。

解题思路：线段树维护，单点修改区间查询。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int maxn = 32000;
const int maxm = 15000;

#define lson(x) ((x)<<1)
#define rson(x) (((x)<<1)|1)
int lc[maxn << 2], rc[maxn << 2], c[maxn << 2], s[maxn << 2];

inline void pushup(int u) {
    s[u] = s[lson(u)] + s[rson(u)];
}

void build (int u, int l, int r) {
    lc[u] = l;
    rc[u] = r;
    s[u] = 0;

    if (l == r)
        return ;

    int mid = (l + r) >> 1;
    build(lson(u), l, mid);
    build(rson(u), mid + 1, r);
    pushup(u);
}

void modify(int u, int x, int w) {
    if (x == lc[u] && rc[u] == x) {
        s[u] += w;
        return;
    }

    int mid = (lc[u] + rc[u]) >> 1;
    if (x <= mid)
        modify(lson(u), x, w);
    else
        modify(rson(u), x, w);
    pushup(u);
}

int query(int u, int l, int r) {
    if (l <= lc[u] && rc[u] <= r)
        return s[u];

    int mid = (lc[u] + rc[u]) >> 1, ret = 0;
    if (l <= mid)
        ret += query(lson(u), l, r);
    if (r > mid)
        ret += query(rson(u), l, r);
    pushup(u);
    return ret;
}

int N, ans[maxm + 5];
struct point {
    int x, y;

    friend bool operator < (const point& a, const point& b) {
        if (a.x != b.x)
            return a.x < b.x;
        return a.y < b.y;
    }
}p[maxm + 5];

int main () {
    while (scanf("%d", &N) == 1) {
        build (1, 0, 32000);
        memset(ans, 0, sizeof(ans));

        for (int i = 0; i < N; i++)
            scanf("%d%d", &p[i].x, &p[i].y);
        sort(p, p + N);

        for (int i = 0; i < N; i++) {
            int k = query(1, 0, p[i].y);
            ans[k]++;
            modify(1, p[i].y, 1);
        }
        for (int i = 0; i < N; i++)
            printf("%d\n", ans[i]);
    }
    return 0;
}