HDOJ-1156 Brownie Points II 线段树/树状数组(模板)

http://acm.hdu.edu.cn/showproblem.php?pid=1156

在一张二位坐标系中,给定n个点的坐标,玩一个划线游戏(线必须穿过点),Stan先手画一条垂直的线,然后Ollie画一条水平的线(要求要穿过Stan那条线所穿过的某个点)。划分后,左上和右下点数是Ollie 的得分,左下和右上是Stan的得分。求Stan在保证最低得分(即不论Ollie后手怎么划,Stan最少能的的分数)最高,并给出基于符合的先手划法,Ollie后手的各种划线的得分(需要去重),升序输出。

这里可以发现,每次划分等同于枚举以某个点为原点,求一次局面,不同的局面分类到x轴坐标中,每个坐标中的局面集合求最小值,所有x轴坐标分类求最大值即求出了最低最大划法。

#include <iostream>
#include <cstring>
#include <string>
#include <queue>
#include <vector>
#include <set>
#include <stack>
#include <cmath>
#include <cstdio>
#include <map>
#include <algorithm>
using namespace std;
struct node
{
int x, y;
}p[];
bool cmpx(node a, node b)
{
if (a.x == b.x) return a.y < b.y;
return a.x < b.x;
}
const int N = ;
map<int, int> mpx, mpy;
int st[N];
int tre[N];
int ans[N];
int lowbit(int k)
{
return k&-k;
}
int query(int k)
{
int rec = ;
while (k)
{
rec += tre[k];
k -= lowbit(k);
}
return rec;
}
void add(int k)
{
while (k <= N)
{
tre[k]++;
k += lowbit(k);
}
}
bool cmp(node a, int b)
{
return a.x < b;
}
int main() {
cin.sync_with_stdio(false);
int n;
while (cin>>n)
{
if (n == ) break;
mpx.clear(), mpy.clear();
queue<node> wq;
fill(tre, tre + N, );
fill(ans, ans + N, );
for (int i = ; i < n; i++)
cin >> p[i].x>>p[i].y,mpx[p[i].x]++,mpy[p[i].y]++,st[i]=p[i].y;
sort(st, st + n);
int len = unique(st, st + n) - st;
sort(p, p + n, cmpx); for (int i = ; i < n; i++)
{
while (!wq.empty())
{
node ad = wq.front();
if (ad.x < p[i].x) add(upper_bound(st, st + len, ad.y) - st + ),wq.pop();
else break;
}
int py = upper_bound(st, st + len, p[i].y)-st+;
ans[i] += query(py-);
wq.push(p[i]);
}
while (!wq.empty()) wq.pop();
fill(tre, tre + , );
for (int i = n-; i >= ; i--)
{
while (!wq.empty())
{
node ad = wq.front();
if (ad.x > p[i].x) add(len+st-upper_bound(st, st + len, ad.y) + ),wq.pop();
else break;
}
int py = len + st - upper_bound(st, st + len, p[i].y) + ;
ans[i] += query(py-);
wq.push(p[i]);
}
map<int, int> getMi;
for (int i = ; i < n; i++)
{
if (getMi.find(p[i].x) == getMi.end()) getMi[p[i].x] = ans[i];
else getMi[p[i].x] = min(getMi[p[i].x], ans[i]);
}
map<int, int>::iterator mxit = getMi.end();
for (map<int, int>::iterator it = getMi.begin(); it != getMi.end(); it++)
{
if (mxit == getMi.end())
mxit = it;
else if (mxit->second < it->second)
mxit = it;
}
cout << "Stan: " << mxit->second << "; Ollie:";
set<int> rpy;
for (int i = ; i < n; i++)
if (getMi[p[i].x]==ans[i]&&ans[i] == mxit->second)
rpy.insert(n - ans[i] - mpx[p[i].x] - mpy[p[i].y] + ); for (set<int>::iterator it = rpy.begin(); it != rpy.end(); it++)
{
cout << ' ' << *it;
}
cout << ";" << endl;
}
return ;
}
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