CF1140F - Extending Set of Points

题意：对于点集S，定义函数F(S)为对S不断扩展到不能扩展时S的点数。一次扩展定义为如果有一个平行于坐标轴的矩形的三个点在S中，则第四个点加入S。

动态在S中加点删点，每次操作完后求F(S)的值。

解：首先有个结论就是，把这些点用平行于坐标轴的线段连接起来，则E值为每个连通块的横坐标种数 * 纵坐标种数之和。

线段树分治 + 可回退化并查集，O(nlog²n)。

  1 #include <bits/stdc++.h>
  2 
  3 typedef long long LL;
  4 const int N = 300010;
  5 
  6 struct Node {
  7     int x, y;
  8     Node(int X, int Y) {
  9         x = X;
 10         y = Y;
 11     }
 12 };
 13 
 14 std::map<int, int> mp[N];
 15 int n, m;
 16 LL ans[N];
 17 std::vector<Node> v[N << 2];
 18 
 19 namespace ufs {
 20     
 21     struct opt {
 22         int x, y;
 23         opt(){}
 24         opt(int X, int Y) {
 25             x = X, y = Y;
 26         }
 27     }stk[N * 4]; int top;
 28     
 29     int fa[N * 2], siz[N * 2], s1[N * 2], s2[N * 2], clock;
 30     LL tot, stk2[N];
 31     inline void init() {
 32         for(int i = 1; i < N; i++) {
 33             fa[i] = i;
 34             fa[i + N] = i + N;
 35             siz[i] = siz[i + N] = 1;
 36             s1[i] = s2[i + N] = 1;
 37             s2[i] = s1[i + N] = 0;
 38         }
 39         top = clock = 0;
 40         return;
 41     }
 42     int find(int x) {
 43         if(x == fa[x]) return x;
 44         return find(fa[x]);
 45     }
 46     inline LL cal(int x) {
 47         return 1ll * s1[x] * s2[x];
 48     }
 49     inline void merge(int x, int y) {
 50         x = find(x);
 51         y = find(y);
 52         if(x == y) return;
 53         if(siz[x] < siz[y]) {
 54             std::swap(x, y);
 55         }
 56         stk2[++clock] = tot;
 57         tot -= cal(x) + cal(y);
 58         stk[++top] = opt(y, fa[y]);
 59         fa[y] = x;
 60         stk[++top] = opt(x, siz[x]);
 61         siz[x] += siz[y];
 62         stk[++top] = opt(x, s1[x]);
 63         s1[x] += s1[y];
 64         stk[++top] = opt(x, s2[x]);
 65         s2[x] += s2[y];
 66         tot += cal(x);
 67         return;
 68     }
 69     inline void erase(int t) {
 70         while(clock > t) {
 71             tot = stk2[clock--];
 72             s2[stk[top].x] = stk[top].y;
 73             top--;
 74             s1[stk[top].x] = stk[top].y;
 75             top--;
 76             siz[stk[top].x] = stk[top].y;
 77             top--;
 78             fa[stk[top].x] = stk[top].y;
 79             top--;
 80         }
 81         return;
 82     }
 83 }
 84 
 85 void add(int L, int R, Node x, int l, int r, int o) {
 86     if(L <= l && r <= R) {
 87         v[o].push_back(x);
 88         return;
 89     }
 90     int mid = (l + r) >> 1;
 91     if(L <= mid) {
 92         add(L, R, x, l, mid, o << 1);
 93     }
 94     if(mid < R) {
 95         add(L, R, x, mid + 1, r, o << 1 | 1);
 96     }
 97     return;
 98 }
 99 
100 void solve(int l, int r, int o) {
101     int now = ufs::clock;
102     for(int i = 0; i < (int)v[o].size(); i++) {
103         ufs::merge(v[o][i].x, v[o][i].y + N);
104         //printf("[%d %d] merge %d %d \n", l, r, v[o][i].x, v[o][i].y);
105     }
106     if(l == r) {
107         ans[r] = ufs::tot;
108         ufs::erase(now);
109         return;
110     }
111     int mid = (l + r) >> 1;
112     solve(l, mid, o << 1);
113     solve(mid + 1, r, o << 1 | 1);
114     ufs::erase(now);
115     return;
116 }
117 
118 int main() {
119     ufs::init();
120     scanf("%d", &n);
121     for(int i = 1, x, y; i <= n; i++) {
122         scanf("%d%d", &x, &y);
123         if(mp[x][y]) {
124             add(mp[x][y], i - 1, Node(x, y), 1, n, 1);
125             //printf("add : (%d %d) [%d %d] \n", x, y, mp[x][y], i - 1);
126             mp[x][y] = 0;
127         }
128         else {
129             mp[x][y] = i;
130         }
131     }
132     std::map<int, int>::iterator it;
133     for(int i = 1; i < N; i++) {
134         for(it = mp[i].begin(); it != mp[i].end(); it++) {
135             if(it->second) {
136                 add(it->second, n, Node(i, it->first), 1, n, 1);
137                 //printf("add : (%d %d) [%d %d] \n", i, it->first, it->second, n);
138             }
139         }
140     }
141     
142     solve(1, n, 1);
143     for(int i = 1; i <= n; i++) {
144         printf("%lld ", ans[i]);
145     }
146     return 0;
147 }