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

2464 -- Brownie Points II

Problem - 1156

　　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：

 #include <cstdio>
 #include <iostream>
 #include <algorithm>
 #include <cstring>
 #include <vector>
 #include <map>

 using namespace std;

 const int N = ;
 inline int lowbit(int x) { return (-x) & x;}
 struct BIT {
     int c[N];
     void init() { memset(c, , sizeof(c));}
     void insert(int x) { x += ; for (int i = x; i < N; i += lowbit(i)) c[i]++;}
     int query(int x) { int ret = ; for ( ; x > ; x -= lowbit(x)) ret += c[x]; return ret;}
     int query(int l, int r) { if (l > r) return ; l += , r += ; return query(r) - query(l - );}
 } bit;

 typedef pair<int, int> PII;
 PII sc[N], pt[N], buf[N];
 int y[N];
 map<int, int> yid;
 const int INF = 0x7fffffff;

 int main() {
 //    freopen("in", "r", stdin);
     int n;
     while (~scanf("%d", &n) && n) {
         memset(sc, , sizeof(sc));
         for (int i = ; i < n; i++) {
             scanf("%d%d", &pt[i].first, &pt[i].second);
             y[i] = pt[i].second;
         }
         sort(y, y + n);
         sort(pt, pt + n);
         int m = unique(y, y + n) - y;
         for (int i = ; i < m; i++) yid[y[i]] = i;
         bit.init();
         for (int i = , top = -; i < n; i++) {
             if (i && pt[i - ].first != pt[i].first) {
                 while (~top) bit.insert(yid[buf[top--].second]);
 //                cout << "pop! " << i << endl;
             }
             buf[++top] = pt[i];
             sc[i].first += bit.query(, yid[pt[i].second] - );
             sc[i].second += bit.query(yid[pt[i].second] + , m - );
         }
 //        for (int i = 0; i < n; i++) cout << sc[i].first << ' '; cout << endl;
 //        for (int i = 0; i < n; i++) cout << sc[i].second << ' '; cout << endl;
         bit.init();
         for (int i = n - , top = -; i >= ; i--) {
             if (i != n -  && pt[i + ].first != pt[i].first) {
                 while (~top) bit.insert(yid[buf[top--].second]);
 //                cout << "pop~ " << i << endl;
             }
             buf[++top] = pt[i];
             sc[i].first += bit.query(yid[pt[i].second] + , m - );
             sc[i].second += bit.query(, yid[pt[i].second] - );
         }
 //        for (int i = 0; i < n; i++) cout << sc[i].first << ' '; cout << endl;
 //        for (int i = 0; i < n; i++) cout << sc[i].second << ' '; cout << endl;
         vector<int> tmp, best;
         int fi = , se, mx = -INF;
         sc[n] = PII(-, -);
         while (fi < n) {
             se = fi;
             while (pt[fi].first == pt[se].first) se++;
             int tmx = INF;
 //            cout << fi << ' ' << se << endl;
             for (int i = fi; i < se; i++) {
                 if (tmx > sc[i].first) {
                     tmx = sc[i].first;
                     tmp.clear();
                     tmp.push_back(sc[i].second);
                 } else if (tmx == sc[i].first) tmp.push_back(sc[i].second);
             }
             fi = se;
             if (mx < tmx) {
                 mx = tmx;
                 best = tmp;
             } else if (mx == tmx) {
                 for (int i = , sz = tmp.size(); i < sz; i++) {
                     best.push_back(tmp[i]);
                 }
             }
         }
 //        cout << mx << ' ' << best.size() << endl;
         sort(best.begin(), best.end());
         printf("Stan: %d; Ollie:", mx);
         int t = (int) (unique(best.begin(), best.end()) - best.begin());
         while (best.size() > t) best.pop_back();
         for (int i = , sz = best.size(); i < sz; i++) cout << ' ' << best[i]; cout << ';' << endl;
     }
     return ;
 }

——written by Lyon