題目鏈接：http://codeforces.com/gym/101964/problem/B

題意：q 種操作，①在(x,y)處加一個點，②加一個矩陣{(x1,y1),(x2,y2)}，問每次操作後點在矩陣中或矩陣邊界上的對數有多少。

題解：裸的CDQ分治，考慮對點和矩陣分別進行CDQ分治，因為x,y <= 1e9，要將其中一個坐標離散化。要註意先加點和先加矩陣的兩種情況下，矩陣的差分邊界是不同的。

  1 #include <bits/stdc++.h>
  2 using namespace std;
  3 #define ll long long
  4 #define ull unsigned long long
  5
 
 #define mst(a,b) memset((a),(b),sizeof(a))
  6 #define mp(a,b) make_pair(a,b)
  7 #define pi acos(-1)
  8 #define pii pair<int,int>
  9 #define pb push_back
 10 const int INF = 0x3f3f3f3f;
 11 const double eps = 1e-6;
 12 const int MAXN = 1e5 + 10;
 13 const int MAXM = 1e8 + 10;
 14 const ll mod = 1e9 + 7
 
;
 15 
 16 struct node {
 17     int op, x1, y1, x2, y2, id;
 18     bool operator < (const node &rhs) {
 19         if(x1 != rhs.x1)
 20             return x1 < rhs.x1;
 21         return id < rhs.id;
 22     }
 23 } a[MAXN], b[MAXN << 2], c[MAXN << 2];
 24 
 25 int tot;
 26
 
 
 27 void addnode(int op, int x1, int y1, int id) {
 28     ++tot;
 29     b[tot].op = op, b[tot].x1 = x1, b[tot].y1 = y1, b[tot].id = id;
 30 }
 31 
 32 int y[MAXN << 1], ysz = 0;
 33 ll ans[MAXN];
 34 
 35 int lowbit(int x) {
 36     return x & (-x);
 37 }
 38 
 39 ll bit[MAXN << 1];
 40 
 41 void add(int y, int val) {
 42     while(y <= ysz) {
 43         bit[y] += val;
 44         y += lowbit(y);
 45     }
 46 }
 47 
 48 ll query(int y) {
 49     ll sum = 0;
 50     while(y) {
 51         sum += bit[y];
 52         y -= lowbit(y);
 53     }
 54     return sum;
 55 }
 56 
 57 void cdq(int l, int r) {
 58     if(l == r)
 59         return ;
 60     int mid = (l + r) >> 1;
 61     cdq(l, mid);
 62     cdq(mid + 1, r);
 63     int len = 0;
 64     for(int i = l; i <= mid; i++)
 65         if(b[i].id == 0)
 66             c[++len] = b[i];
 67     for(int i = mid + 1; i <= r; i++)
 68         if(b[i].id)
 69             c[++len] = b[i];
 70     if(len == 0 || c[1].id || c[len].id == 0)
 71         return ;
 72     sort(c + 1, c + 1 + len);
 73     for(int i = 1; i <= len; i++) {
 74         if(c[i].id == 0)
 75             add(c[i].y1, c[i].op);
 76         else
 77             ans[c[i].id] += 1ll * c[i].op * query(c[i].y1);
 78     }
 79     for(int i = 1; i <= len; i++)
 80         if(c[i].id == 0)
 81             add(c[i].y1, -c[i].op);
 82 }
 83 
 84 int main() {
 85 #ifdef local
 86     freopen("data.txt", "r", stdin);
 87 //    freopen("data.txt", "w", stdout);
 88 #endif
 89     mst(ans, 0);
 90     int q;
 91     scanf("%d", &q);
 92     for(int i = 1; i <= q; i++) {
 93         scanf("%d%d%d", &a[i].op, &a[i].x1, &a[i].y1);
 94         y[++ysz] = a[i].y1;
 95         if(a[i].op == 2) {
 96             scanf("%d%d", &a[i].x2, &a[i].y2);
 97             y[++ysz] = a[i].y2;
 98         }
 99     }
100     sort(y + 1, y + 1 + ysz);
101     ysz = unique(y + 1, y + 1 + ysz) - y - 1;
102     for(int i = 1; i <= q; i++) {
103         a[i].y1 = lower_bound(y + 1, y + 1 + ysz, a[i].y1) - y;
104         if(a[i].op == 2)
105             a[i].y2 = lower_bound(y + 1, y + 1 + ysz, a[i].y2) - y;
106     }
107     tot = 0;
108     for(int i = 1; i <= q; i++) {
109         if(a[i].op == 1) {
110             addnode(1, a[i].x1, a[i].y1, 0);
111         } else {
112             addnode(1, a[i].x1 - 1, a[i].y1 - 1, i);
113             addnode(-1, a[i].x2, a[i].y1 - 1, i);
114             addnode(-1, a[i].x1 - 1, a[i].y2, i);
115             addnode(1, a[i].x2, a[i].y2, i);
116         }
117     }
118     cdq(1, tot);
119     tot = 0;
120     for(int i = 1; i <= q; i++) {
121         if(a[i].op == 1) {
122             addnode(1, a[i].x1, a[i].y1, i);
123         } else {
124             addnode(1, a[i].x1, a[i].y1, 0);
125             addnode(-1, a[i].x2 + 1, a[i].y1, 0);
126             addnode(-1, a[i].x1, a[i].y2 + 1, 0);
127             addnode(1, a[i].x2 + 1, a[i].y2 + 1, 0);
128         }
129     }
130     cdq(1, tot);
131     for(int i = 1; i <= q; i++) {
132         ans[i] += ans[i - 1];
133         printf("%lld\n", ans[i]);
134     }
135     return 0;
136 }

SEERC 2018 B. Broken Watch （CDQ分治）