Hdu 1156

题目链接

Brownie Points II

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 207    Accepted Submission(s): 77

Problem Description

Stan and Ollie play the game of Odd Brownie Points. Some brownie points are located in the plane, at integer coordinates. Stan plays first and places a vertical line in the plane. The line must go through a brownie point and may cross many (with the same x-coordinate). Then Ollie places a horizontal line that must cross a brownie point already crossed by the vertical line. 
Those lines divide the plane into four quadrants. The quadrant containing points with arbitrarily large positive coordinates is the top-right quadrant. 

The players score according to the number of brownie points in the quadrants. If a brownie point is crossed by a line, it doesn't count. Stan gets a point for each (uncrossed) brownie point in the top-right and bottom-left quadrants. Ollie gets a point for each (uncrossed) brownie point in the top-left and bottom-right quadrants. 

Stan and Ollie each try to maximize his own score. When Stan plays, he considers the responses, and chooses a line which maximizes his smallest-possible score. 

 

Input

Input contains a number of test cases. The data of each test case appear on a sequence of input lines. The first line of each test case contains a positive odd integer 1 < n < 200000 which is the number of brownie points. Each of the following n lines contains two integers, the horizontal (x) and vertical (y) coordinates of a brownie point. No two brownie points occupy the same place. The input ends with a line containing 0 (instead of the n of a test). 

 

Output

For each input test, print a line of output in the format shown below. The first number is the largest score which Stan can assure for himself. The remaining numbers are the possible (high) scores of Ollie, in increasing order.

 

Sample Input

11 3 2 3 3 3 4 3 6 2 -2 1 -3 0 0 -3 -3 -3 -2 -3 -4 3 -7 0

 

Sample Output

Stan: 7; Ollie: 2 3;

Accepted Code:

 1 /*************************************************************************  2  > File Name: A.cpp  3  > Author: Stomach_ache  4  > Mail: [email protected]  5  > Created Time: 2014年07月27日 星期日 14时45分32秒  6  > Propose:  7  ************************************************************************/

 8 

 9 #include <cmath>

 10 #include <string>

 11 #include <vector>

 12 #include <cstdio>

 13 #include <fstream>

 14 #include <cstring>

 15 #include <iostream>

 16 #include <algorithm>

 17 using namespace std;  18 

 19 const int maxn = 200002;  20 //l维护垂直线左侧的点，r维护垂直线右侧的点

 21 int l[maxn], r[maxn];  22 //每一条垂直于x轴的直线信息

 23 struct Line {  24       int x, y;  25     friend bool operator < (Line a, Line b) {  26           return a.x < b.x;  27  }  28 }line[maxn];  29 //保存所有y轴坐标

 30 int y[maxn];  31 int n, w;  32 

 33 //BIT

 34 int lowbit(int x) {  35       return x & -x;  36 }  37 

 38 void add(int t[], int x, int v) {  39       while (x <= w) {  40           t[x] += v;  41         x += lowbit(x);  42  }  43 }  44 

 45 int sum(int t[], int x) {  46       int res = 0;  47     while (x > 0) {  48           res += t[x];  49         x -= lowbit(x);  50  }  51     return res;  52 }  53 

 54 int main(void) {  55 #ifndef ONLINE_JUDGE  56       freopen("in.txt", "r", stdin);  57 #endif

 58       while (~scanf("%d", &n) && n) {  59           for (int i = 0; i < n; i++) {  60               scanf("%d %d", &line[i].x, &line[i].y);  61             y[i] = line[i].y;  62  }  63         //y轴坐标离散化

 64         sort(y, y + n);  65         w = unique(y, y + n) - y;  66         //按x轴坐标从小到大排序

 67         sort(line, line + n);  68         //初始化BIT数组

 69         memset(l, 0, sizeof(l));  70         memset(r, 0, sizeof(r));  71         //把所有点加入右侧的BIT

 72         for (int i = 0; i < n; i++) add(r, lower_bound(y, y + w, line[i].y)+1-y, 1);  73         //Stan是其可以获得的最大的最小值  74         //st保存重复x坐标出现的起点

 75         int Stan = -1, st = 0;  76         //保存Ollie可能的结果

 77         vector<int> Ollie;  78         for (int i = 1; i <= n; i++) {  79               if (i == n || line[i].x != line[i-1].x) {  80               //把重复的点从右侧BIT中删除

 81                   for (int j = st; j < i; j++) add(r, lower_bound(y, y + w, line[j].y)+1-y, -1);  82                 int stan = -1, ollie = -1;  83                 //扫描ｘ坐标重复的点，枚举平行于x轴的直线

 84                 for (int j = st; j < i; j++) {  85                       int f = lower_bound(y, y + w, line[j].y) + 1 - y;  86                       int s = sum(l, f-1) + sum(r, w) - sum(r, f);  87                     int o = sum(l, w) - sum(l, f) + sum(r, f-1);  88                     //为了使ollie最大

 89                     if (o > ollie) {  90                           ollie = o;  91                         stan = s;  92                     } else if (o == ollie) {  93                           stan = min(stan, s);  94  }  95  }  96                 //更新最大的最小值

 97                 if (stan > Stan) {  98                       Stan = stan;  99  Ollie.clear(); 100  Ollie.push_back(ollie); 101                 } else if (stan == Stan) { 102  Ollie.push_back(ollie); 103  } 104                 //把重复的点加入左侧的BIT

105                   for (int j = st; j < i; j++) add(l, lower_bound(y, y + w, line[j].y)+1-y, 1); 106                 st = i; 107  } 108  } 109         //注意要将Ollie的结果去重

110  sort(Ollie.begin(), Ollie.end()); 111         int len = unique(Ollie.begin(), Ollie.end()) - Ollie.begin(); 112         printf("Stan: %d; Ollie:", Stan); 113         for (int i = 0; i < len; i++) printf(" %d", Ollie[i]); 114         puts(";"); 115  } 116 

117     return 0; 118 }