uva 10869 - Brownie Points II(树状数组)

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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;}

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