Uva10869-Brownie points II (line segment tree)

Uva10869-Brownie points II (line segment tree)

Question Link

There are N points on the plane. Stan and Ollie are playing games. The rule is: Stan first draws a vertical line as the Y axis, the condition is that a point on the plane must pass through, while Ollie draws the X axis, but any point must be selected from the point on the Y axis of stany to draw the X axis as the origin. Then the plane is divided into four quadrants. The vertices on the axis are not counted. The number of vertices in the quadrants is the Stan score, and the 2, 4 is the Ollie score. Ask Stan every time he draws a Y axis, each axis has a minimum score, calculate the maximum among these minimum values, and output the Ollie score at this time.

Solution: divide the Y axis into discrete values. We hope that we can get the Stan score for this point every time we enumerate a point. From left to right, process each vertex from top to bottom, and use the line segment tree to obtain the number of vertices on the Y height, that is, the score in the First quadrant. After each query, the online segment tree of the vertex is deleted. Similarly, the third quadrant score is obtained. This question also outputs the Ollie score, which should be stored in order using set.

Code:

#include <cstdio>#include <cstring>#include <vector>#include <map>#include <set>#include <algorithm>#include <iostream>using namespace std;const int maxn = 2e5;#define lson(x) (x<<1)#define rson(x) ((x<<1) | 1)int n;vector<int> pos;map<int, int>x, y;set<int> score, vec;struct Point {    int x, y, score;}p[maxn];int cmp_x (const Point &a, const Point &b) {    if (a.x == b.x)        return a.y > b.y;    return a.x < b.x;}struct Node {    int l, r, v, addv;    void set (int l, int r, int addv, int v) {        this->l = l;        this->r = r;        this->v = v;        this->addv = addv;    }}node[4 * maxn + 5];void add_node (int u, int addv) {    node[u].addv += addv;    node[u].v += (node[u].r - node[u].l + 1) * addv;}void pushdown (int u) {    if (node[u].addv) {        add_node(lson(u), node[u].addv);        add_node(rson(u), node[u].addv);    }}void pushup (int u) {    node[u].set (node[lson(u)].l, node[rson(u)].r, 0, node[lson(u)].v + node[rson(u)].v);}void build (int u, int l, int r) {    if (l == r) {        node[u].set (l, r, 0, y[pos[l]]);        return ;    }    int m = (l + r)>>1;    build (lson(u), l, m);    build (rson(u), m + 1, r);    pushup(u);}void update (int u, int l, int r, int add) {    if (node[u].l >= l && node[u].r <= r) {        add_node(u, add);        return;    }    pushdown(u);    int m = (node[u].l + node[u].r)>>1;    if (l <= m)        update (lson(u), l, r, add);    if (r > m)        update (rson(u), l, r, add);    pushup(u);}int query (int u, int l, int r) {    if (node[u].l >= l && node[u].r <= r)        return node[u].v;    pushdown(u);    int m = (node[u].l + node[u].r)>>1;    int ans = 0;    if (l <= m)        ans += query(lson(u), l, r);    if (r > m)        ans += query(rson(u), l, r);    pushup(u);    return ans;}void init () {    x.clear();    y.clear();    pos.clear();    for (int i = 0; i < n; i++) {        scanf ("%d%d", &p[i].x, &p[i].y);        p[i].score = 0;        x[p[i].x]++;        y[p[i].y]++;        pos.push_back(p[i].y);    }    sort (pos.begin(), pos.end());    sort (p, p + n, cmp_x);    pos.erase(unique (pos.begin(), pos.end()), pos.end());}void solve_rtop () {    int x;    build(1, 0, (int)pos.size() - 1);    for (int i = 0; i < n; i++) {        x = lower_bound(pos.begin(), pos.end(), p[i].y) - pos.begin();        if (x + 1 <= pos.size() - 1) {            p[i].score += query(1, x + 1, pos.size() - 1);                    //printf ("%d %d %d\n", p[i].x, p[i].y, x);        }        update (1, x, x, -1);            }    }void solve_lboutom () {    int x;        build(1, 0, (int)pos.size() - 1);    for (int i = n - 1; i >= 0; i--) {        x = lower_bound(pos.begin(), pos.end(), p[i].y) - pos.begin();        if (x - 1 >= 0) {            p[i].score += query (1, 0, x - 1);            //printf ("%d %d %d\n", p[i].x, p[i].y, p[i].score);        }        update (1, x, x, -1);    }}void add_ans (int tmp) {    for (set<int>::iterator it = vec.begin(); it != vec.end(); it++)        score.insert(n - tmp - x[p[*it].x] - y[p[*it].y] + 1);    }void solve () {    init();    solve_lboutom();    solve_rtop();    score.clear();    vec.clear();    int ans = -1;    int tmp = p[0].score;                int pre = p[0].x;    for (int i = 0; i < n; i++)    {        if (pre == p[i].x) {            if (tmp == p[i].score)                vec.insert(i);            else if (tmp > p[i].score) {                vec.clear();                vec.insert(i);                    tmp = p[i].score;            }        } else {            if (ans <= tmp) {                if (ans < tmp)                     score.clear();                add_ans(tmp);            }            ans = max (ans, tmp);            tmp = p[i].score;            pre = p[i].x;            vec.clear();            vec.insert(i);        }    }    if (ans <= tmp) {        if (ans < tmp)             score.clear();        add_ans(tmp);    }    ans = max (ans, tmp);    printf("Stan: %d; Ollie:", ans);    for (set<int>::iterator it = score.begin(); it != score.end(); it++)        printf(" %d", *it);    printf(";\n");}int main () {    while (scanf ("%d", &n) && n) {            solve();    }    return 0;}

Uva10869-Brownie points II (line segment tree)