SEERC 2018 B. Broken Watch （CDQ分治）

题目链接：http://codeforces.com/gym/101964/problem/B

题意：q 种操作，①在(x,y)处加一个点，②加一个矩阵{(x1,y1),(x2,y2)}，问每次操作后点在矩阵中或矩阵边界上的对数有多少。

题解：裸的CDQ分治，考虑对点和矩阵分别进行CDQ分治，因为x,y <= 1e9，要将其中一个坐标离散化。要注意先加点和先加矩阵的两种情况下，矩阵的差分边界是不同的。

 #include <bits/stdc++.h>
 using namespace std;
 #define ll long long
 #define ull unsigned long long
 #define mst(a,b) memset((a),(b),sizeof(a))
 #define mp(a,b) make_pair(a,b)
 #define pi acos(-1)
 #define pii pair<int,int>
 #define pb push_back
 const int INF = 0x3f3f3f3f;
 const double eps = 1e-;
 const int MAXN = 1e5 + ;
 const int MAXM = 1e8 + ;
 const ll mod = 1e9 + ;

 struct node {
     int op, x1, y1, x2, y2, id;
     bool operator < (const node &rhs) {
         if(x1 != rhs.x1)
             return x1 < rhs.x1;
         return id < rhs.id;
     }
 } a[MAXN], b[MAXN << ], c[MAXN << ];

 int tot;

 void addnode(int op, int x1, int y1, int id) {
     ++tot;
     b[tot].op = op, b[tot].x1 = x1, b[tot].y1 = y1, b[tot].id = id;
 }

 int y[MAXN << ], ysz = ;
 ll ans[MAXN];

 int lowbit(int x) {
     return x & (-x);
 }

 ll bit[MAXN << ];

 void add(int y, int val) {
     while(y <= ysz) {
         bit[y] += val;
         y += lowbit(y);
     }
 }

 ll query(int y) {
     ll sum = ;
     while(y) {
         sum += bit[y];
         y -= lowbit(y);
     }
     return sum;
 }

 void cdq(int l, int r) {
     if(l == r)
         return ;
     int mid = (l + r) >> ;
     cdq(l, mid);
     cdq(mid + , r);
     int len = ;
     for(int i = l; i <= mid; i++)
         if(b[i].id == )
             c[++len] = b[i];
     for(int i = mid + ; i <= r; i++)
         if(b[i].id)
             c[++len] = b[i];
     if(len ==  || c[].id || c[len].id == )
         return ;
     sort(c + , c +  + len);
     for(int i = ; i <= len; i++) {
         if(c[i].id == )
             add(c[i].y1, c[i].op);
         else
             ans[c[i].id] += 1ll * c[i].op * query(c[i].y1);
     }
     for(int i = ; i <= len; i++)
         if(c[i].id == )
             add(c[i].y1, -c[i].op);
 }

 int main() {
 #ifdef local
     freopen("data.txt", "r", stdin);
 //    freopen("data.txt", "w", stdout);
 #endif
     mst(ans, );
     int q;
     scanf("%d", &q);
     for(int i = ; i <= q; i++) {
         scanf("%d%d%d", &a[i].op, &a[i].x1, &a[i].y1);
         y[++ysz] = a[i].y1;
         if(a[i].op == ) {
             scanf("%d%d", &a[i].x2, &a[i].y2);
             y[++ysz] = a[i].y2;
         }
     }
     sort(y + , y +  + ysz);
     ysz = unique(y + , y +  + ysz) - y - ;
     for(int i = ; i <= q; i++) {
         a[i].y1 = lower_bound(y + , y +  + ysz, a[i].y1) - y;
         if(a[i].op == )
             a[i].y2 = lower_bound(y + , y +  + ysz, a[i].y2) - y;
     }
     tot = ;
     for(int i = ; i <= q; i++) {
         if(a[i].op == ) {
             addnode(, a[i].x1, a[i].y1, );
         } else {
             addnode(, a[i].x1 - , a[i].y1 - , i);
             addnode(-, a[i].x2, a[i].y1 - , i);
             addnode(-, a[i].x1 - , a[i].y2, i);
             addnode(, a[i].x2, a[i].y2, i);
         }
     }
     cdq(, tot);
     tot = ;
     for(int i = ; i <= q; i++) {
         if(a[i].op == ) {
             addnode(, a[i].x1, a[i].y1, i);
         } else {
             addnode(, a[i].x1, a[i].y1, );
             addnode(-, a[i].x2 + , a[i].y1, );
             addnode(-, a[i].x1, a[i].y2 + , );
             addnode(, a[i].x2 + , a[i].y2 + , );
         }
     }
     cdq(, tot);
     for(int i = ; i <= q; i++) {
         ans[i] += ans[i - ];
         printf("%lld\n", ans[i]);
     }
     return ;
 }