　　hdu分类线段树的题。题意是，给出一堆点的位置，stan和ollie玩游戏，stan通过其中一个点画垂线，ollie通过在垂线上的点画一条平行线。他们的得分是stan在一三象限中有多少点，ollie在二四象限中共有多少点。在平行线和垂线的点不算任何人的。

　　开始的时候以为暴力二维线段树，结果发现点的数目也太庞大了。后来想了一想，只是统计这么几个区域，逐个插点就可以的嘛。用一个树状数组存下y坐标中相应位置的点的个数，向前向后做两次差点即可！

　　错了一次忘记将答案unique，然后就AC了。

代码及debug的数据如下：

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <algorithm>
 4 #include <cstring>
 5 #include <vector>
 6 #include <map>
 7 
 8 using namespace std;
 9 
10 const int N = 222222;
11 inline int lowbit(int x) { return (-x) & x;}
12 struct BIT {
13     int c[N];
14     void init() { memset(c, 0, sizeof(c));}
15     void insert(int x) { x += 10; for (int i = x; i < N; i += lowbit(i)) c[i]++;}
16     int query(int x) { int ret = 0; for ( ; x > 0; x -= lowbit(x)) ret += c[x]; return ret;}
17     int query(int l, int r) { if (l > r) return 0; l += 10, r += 10; return query(r) - query(l - 1);}
18 } bit;
19 
20 typedef pair<int, int> PII;
21 PII sc[N], pt[N], buf[N];
22 int y[N];
23 map<int, int> yid;
24 const int INF = 0x7fffffff;
25 
26 int main() {
27 //    freopen("in", "r", stdin);
28     int n;
29     while (~scanf("%d", &n) && n) {
30         memset(sc, 0, sizeof(sc));
31         for (int i = 0; i < n; i++) {
32             scanf("%d%d", &pt[i].first, &pt[i].second);
33             y[i] = pt[i].second;
34         }
35         sort(y, y + n);
36         sort(pt, pt + n);
37         int m = unique(y, y + n) - y;
38         for (int i = 0; i < m; i++) yid[y[i]] = i;
39         bit.init();
40         for (int i = 0, top = -1; i < n; i++) {
41             if (i && pt[i - 1].first != pt[i].first) {
42                 while (~top) bit.insert(yid[buf[top--].second]);
43 //                cout << "pop! " << i << endl;
44             }
45             buf[++top] = pt[i];
46             sc[i].first += bit.query(0, yid[pt[i].second] - 1);
47             sc[i].second += bit.query(yid[pt[i].second] + 1, m - 1);
48         }
49 //        for (int i = 0; i < n; i++) cout << sc[i].first << ' '; cout << endl;
50 //        for (int i = 0; i < n; i++) cout << sc[i].second << ' '; cout << endl;
51         bit.init();
52         for (int i = n - 1, top = -1; i >= 0; i--) {
53             if (i != n - 1 && pt[i + 1].first != pt[i].first) {
54                 while (~top) bit.insert(yid[buf[top--].second]);
55 //                cout << "pop~ " << i << endl;
56             }
57             buf[++top] = pt[i];
58             sc[i].first += bit.query(yid[pt[i].second] + 1, m - 1);
59             sc[i].second += bit.query(0, yid[pt[i].second] - 1);
60         }
61 //        for (int i = 0; i < n; i++) cout << sc[i].first << ' '; cout << endl;
62 //        for (int i = 0; i < n; i++) cout << sc[i].second << ' '; cout << endl;
63         vector<int> tmp, best;
64         int fi = 0, se, mx = -INF;
65         sc[n] = PII(-1, -1);
66         while (fi < n) {
67             se = fi;
68             while (pt[fi].first == pt[se].first) se++;
69             int tmx = INF;
70 //            cout << fi << ' ' << se << endl;
71             for (int i = fi; i < se; i++) {
72                 if (tmx > sc[i].first) {
73                     tmx = sc[i].first;
74                     tmp.clear();
75                     tmp.push_back(sc[i].second);
76                 } else if (tmx == sc[i].first) tmp.push_back(sc[i].second);
77             }
78             fi = se;
79             if (mx < tmx) {
80                 mx = tmx;
81                 best = tmp;
82             } else if (mx == tmx) {
83                 for (int i = 0, sz = tmp.size(); i < sz; i++) {
84                     best.push_back(tmp[i]);
85                 }
86             }
87         }
88 //        cout << mx << ' ' << best.size() << endl;
89         sort(best.begin(), best.end());
90         printf("Stan: %d; Ollie:", mx);
91         int t = (int) (unique(best.begin(), best.end()) - best.begin());
92         while (best.size() > t) best.pop_back();
93         for (int i = 0, sz = best.size(); i < sz; i++) cout << ' ' << best[i]; cout << ';' << endl;
94     }
95     return 0;
96 }

View Code

——written by Lyon

posted @ 2013-07-05 16:20 LyonLys 阅读(441) 评论(0) 收藏举报

刷新页面返回顶部

hdu 1156 && poj 2464 Brownie Points II (BIT)