HDU 5618：Jam's problem again（CDQ分治+树状数组处理三维偏序）

http://acm.hdu.edu.cn/showproblem.php?pid=5618

题意：……

思路：和NEUOJ那题一样的。重新写了遍理解了一下，算作处理三维偏序的模板了。

 #include <cstdio>
 #include <algorithm>
 #include <iostream>
 #include <cstring>
 using namespace std;
 #define INF 0x3f3f3f3f
 #define N 100010
 typedef long long LL;
 struct node {
     int x, y, z, id, f;
 } p[N], s[N];
 int ans[N], same[N], bit[N], gap;

 bool cmpx(const node &a, const node &b) { if(a.x != b.x) return a.x < b.x; if(a.y != b.y) return a.y < b.y; return a.z < b.z; }
 bool cmpy(const node &a, const node &b) { if(a.y != b.y) return a.y < b.y; if(a.x != b.x) return a.x < b.x; return a.z < b.z; }
 bool cmpid(const node &a, const node &b) { return a.id < b.id; }
 bool check(node a, node b) { return a.x == b.x && a.y == b.y && a.z == b.z; }

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

 void update(int x, int w) { while(x <= gap) bit[x] += w, x += lowbit(x); }

 int query(int x) { int ans = ; while(x) ans += bit[x], x -= lowbit(x); return ans; }

 void CDQ(int l, int r) {
     if(l == r) return ;
     int m = (l + r) >> ;
     CDQ(l, m); CDQ(m + , r);
     int cnt = ;
     // 标记是在左边还是右边
     for(int i = l; i <= r; i++) s[++cnt] = p[i], s[cnt].f = i <= m ?  : ;
     sort(s + , s +  + cnt, cmpy);
     for(int i = ; i <= cnt; i++)
         if(!s[i].f) update(s[i].z, );
         else ans[s[i].id] += query(s[i].z);
     for(int i = ; i <= cnt; i++)
         if(!s[i].f) update(s[i].z, -);
 }

 int main() {
     int t, n;
     scanf("%d", &t);
     while(t--) {
         scanf("%d", &n); gap = ;
         for(int i = ; i <= n; i++) {
             scanf("%d%d%d", &p[i].x, &p[i].y, &p[i].z);
             p[i].id = i; if(p[i].z > gap) gap = p[i].z;
         }
         sort(p + , p +  + n, cmpx);
         // 离散化
         for(int i = ; i <= n; ) {
             int j = i + ;
             while(j <= n && check(p[j], p[i])) j++;
             while(i < j) same[p[i++].id] = p[j-].id;
         }
         for(int i = ; i <= n; i++) p[i].x = i;
         memset(bit, , sizeof(bit));
         memset(ans, , sizeof(ans));
         CDQ(, n);
         sort(p + , p +  + n, cmpid);
         for(int i = ; i <= n; i++) printf("%d\n", ans[same[i]]);
     }
     return ;
 }