bzoj 1818 [CQOI 2010] 内部白点 - 扫描线 - 树状数组

题目传送门

　　快速的列车

　　慢速的列车

题目大意

　　一个无限大的方格图内有$n$个黑点。问有多少个位置上下左右至少有一个黑点或本来是黑点。

　　扫描线是显然的。

　　考虑一下横着的线段，取它两个端点，横坐标小的地方放一个+1，大的地方放一个-1事件。

　　然后扫描，扫到的横着的线段更新，竖着的线段用树状数组求答案。

　　然后考虑这一列上原来存在的黑点有没有被统计，如果没有就加上。

Code

 /**
  * bzoj
  * Problem#1818
  * Accepted
  * Time: 1824ms
  * Memory: 9776k
  */
 #include <algorithm>
 #include <iostream>
 #include <cstdlib>
 #include <cstring>
 #include <cstdio>
 #include <vector>
 #ifndef WIN32
 #define Auto "%lld"
 #else
 #define Auto "%I64d"
 #endif
 using namespace std;
 #define smax(a, b) (a = max(a, b))
 #define smin(a, b) (a = min(a, b))
 typedef bool boolean;
 #define ll long long

 const int N = 1e5 + ;

 typedef class IndexedTree {
     public:
         int s;
         int* ar;

         IndexedTree() {    }
         IndexedTree(int s):s(s) {
             ar = new int[(s + )];
             memset(ar, , sizeof(int) * (s + ));
         }

         void add(int idx, int val) {
             for ( ; idx <= s; idx += (idx & (-idx)))
                 ar[idx] += val;
         }

         int query(int idx) {
             int rt = ;
             for ( ; idx; idx -= (idx & (-idx)))
                 rt += ar[idx];
             return rt;
         }
 }IndexedTree;

 typedef class Event {
     public:
         int x, y, val;

         Event(int x = , int y = , int val = ):x(x), y(y), val(val) {    }

         boolean operator < (Event b) const {
             return x < b.x;
         }
 }Event;

 int n;
 int xs[N], ys[N], bxs[N], bys[N];
 int ls[N], rs[N], us[N], ds[N];
 ll res = ;
 int tp = ;
 Event es[N << ];
 vector<int> vs[N];
 IndexedTree it;

 inline void init() {
     scanf("%d", &n);
     for (int i = ; i <= n; i++)
         scanf("%d%d", xs + i, ys + i);
 }

 inline void descrete(int* ar, int* br, int n) {
     memcpy(br, ar, sizeof(int) * (n + ));
     sort(br + , br + n + );
     for (int i = ; i <= n; i++)
         ar[i] = lower_bound(br + , br + n + , ar[i]) - br;
 }

 inline void solve() {
     descrete(xs, bxs, n);
     descrete(ys, bys, n);

     for (int i = ; i <= n; i++)
         ls[i] = ds[i] = N;
     for (int i = ; i <= n; i++)
         us[i] = rs[i] = ;

     for (int i = ; i <= n; i++) {
         smin(ls[ys[i]], xs[i]);
         smax(rs[ys[i]], xs[i]);
         smin(ds[xs[i]], ys[i]);
         smax(us[xs[i]], ys[i]);
     }

     for (int i = ; i <= n; i++)
         vs[xs[i]].push_back(ys[i]);

     for (int i = ; i <= n; i++)
         if (ls[i] < rs[i] - ) {
             es[++tp] = Event(ls[i], i, );
             es[++tp] = Event(rs[i], i, -);
         }
     sort(es + , es + tp + );

     int pe = ;
     it = IndexedTree(n);
     bxs[] = -1e9 - ;
     for (int i = ; i <= n; i++) {
         if (bxs[i] == bxs[i - ])    continue;
         while (pe <= tp && es[pe].x == i)
             it.add(es[pe].y, es[pe].val), pe++;
         if (ds[i] <= us[i])
             res += it.query(us[i]) - it.query(ds[i] - );//, cerr << bxs[i] << endl;
         for (int j = ; j < (signed) vs[i].size(); j++)
             if (it.query(vs[i][j]) - it.query(vs[i][j] - ) == )
                 res++;
 //        cerr << bxs[i] << " " << res << endl;
     }
     printf(Auto"\n", res);
 }

 int main() {
     init();
     solve();
     return ;
 }