UVA10869 - Brownie Points II(线段树)

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UVA10869 - Brownie Points II(线段树)

题目链接

题目大意：平面上有n个点，Stan和Ollie在玩游戏，游戏规则是：Stan先画一条竖直的线作为y轴，条件是必须要经过这个平面上的某一点，而ollie则画x轴，但是要在Stany画的y轴上经过的点中任意选择一点来作为原点画x轴。然后这个平面就被划分为4个象限，轴上的点都不算，1，3象限的点的个数就是Stan的得分，2，4就是ollie的得分。问Stan每画一条y轴，每条轴上都有个得分的最小值，求这些最小值中的最大值，并且输出这时ollie的得分。

解题思路：将y轴离散化建树，希望每次枚举一个点就可以得到这个点的Stan的得分。从左往右，从上往下处理每个点，通过线段树可以得到y高度上的点的个数，就是第一象限的得分。然后每次查询完一个点就要将这个点在线段树中删掉。同理得到第三象限的得分。这题还要输出ollie的得分，并且要按顺序，用set来存储。

代码：

#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(线段树)

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原文地址：http://blog.csdn.net/u012997373/article/details/40427885