Brownie Points II
Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 207 Accepted Submission(s): 77Problem DescriptionStan 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.
InputInput 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).
OutputFor 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 Input11 3 2 3 3 3 4 3 6 2 -2 1 -3 0 0 -3 -3 -3 -2 -3 -4 3 -7 0
Sample OutputStan: 7; Ollie: 2 3;
1 /*************************************************************************
2 > File Name: A.cpp
3 > Author: Stomach_ache
4 > Mail: sudaweitong@gmail.com
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 }
原文：http://www.cnblogs.com/Stomach-ache/p/3871751.html