题目链接：uva 10869 - Brownie Points II

题目大意：给定若干点，第一个人选中一个存在点的横坐标，第二个人选中该横坐标上的一点，以该点作原点建立坐标系，一、三象限的点属于第一个人，二、四象限属于第二个人，坐标轴上的不属于任何人。问说在第一个人获得点最多的情况下第二个人可能获得多少点。

解题思路：将所有点按照x坐标从小到大，y坐标从大到小排序，这样从左向右可以处理处每个点右上角的点的个数（用树状数组维护即可），然后从右向左处理即可得到左下角区域的点数。

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

#define lowbit(x) ((x)&(-x))

using namespace std;
const int maxn = 200000;

set<int> vec;
int ans;
int N, R, C, fenx[maxn+5], rec[maxn+5];
int cntx[maxn+5], cnty[maxn+5], have[maxn+5];

struct point {
    int x, y;
    int rx, ry;
}p[maxn+5];

inline bool sort_y (const point& a, const point& b) {
    return a.y < b.y;
}

inline bool sort_xby (const point& a, const point& b) {
    if (a.x != b.x)
        return a.x < b.x;
    return a.y > b.y;
}

inline bool sort_xsy (const point& a, const point& b) {
    if (a.x != b.x)
        return a.x > b.x;
    return a.y < b.y;
}

void add_treeArr (int x, int val) {
    while (x <= maxn) {
        fenx[x] += val;
        x += lowbit(x);
    }
}

int query_treeArr (int x) {
    int ret = 0;
    while (x) {
        ret += fenx[x];
        x -= lowbit(x);
    }
    return ret;
}

void init () {
    memset(cntx, 0, sizeof(cntx));
    memset(cnty, 0, sizeof(cnty));
    memset(fenx, 0, sizeof(fenx));

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

    sort(p, p + N, sort_y);
    C = p[0].ry = 1;
    add_treeArr(C, 1);
    for (int i = 1; i < N; i++) {
        if (p[i].y != p[i-1].y)
            C++;
        p[i].ry = C;
        add_treeArr(C, 1);
    }

    sort(p, p + N, sort_xby);
    R = p[0].rx = 0;
    for (int i = 1; i < N; i++) {
        if (p[i].x != p[i-1].x)
            R++;
        p[i].rx = R;
    }
}

void set_ans (int ret, set<int> v) {
    if (ret > ans) {
        ans = ret;
        vec.clear();
    }

    if (ret == ans) {
        for (set<int>::iterator i = v.begin(); i != v.end(); i++)
            vec.insert(N - ret - cntx[p[*i].rx] - cnty[p[*i].ry] + 1);
    }
}

void solve () {
    memcpy(rec, fenx, sizeof(fenx));

    for (int i = 0; i < N; i++) {
        have[i] = query_treeArr(C) - query_treeArr(p[i].ry);
        add_treeArr(p[i].ry, -1);
        cntx[p[i].rx]++;
        cnty[p[i].ry]++;
    }

    ans = 0;
    vec.clear();
    memcpy(fenx, rec, sizeof(rec));

    int pre = p[N-1].rx, ret = N;
    set<int> oll;
    for (int i = N-1; i >= 0; i--) {
        int tmp = have[i] + query_treeArr(p[i].ry - 1);
        add_treeArr(p[i].ry, -1);

        if (p[i].rx != pre) {
            set_ans(ret, oll);
            pre = p[i].rx;

            ret = N;
            oll.clear();
        }

        if (ret > tmp) {
            ret = tmp;
            oll.clear();
        }

        if (ret == tmp)
            oll.insert(i);
    }
    set_ans(ret, oll);

    printf("Stan: %d; Ollie:", ans);
    for (set<int>::iterator i = vec.begin(); i != vec.end(); i++)
        printf(" %d", *i);
    printf(";\n");
}

int main () {
    while (scanf("%d", &N) == 1 && N) {
        init();
        solve();
    }
    return 0;
}

uva 10869 - Brownie Points II(树状数组)