Little Artem and Time Machine CodeForces-669 E(CDQ分治)

テーマリンク:http://codeforces.com/problemset/problem/669/E
n個の操作をあげます.全部で3種類です.
1、op=1、時間tに数を追加
2、op=2、時間tで1つの数を削除
3、op=3、クエリが時間tに現れる回数
3 Dバイアス問題は,CDQで解決できるか,2 Dツリー配列で直接やるか,動的議長ツリー(書くのが面倒)である.
 

#include 
#define ll long long
using namespace std;
const int maxn = 1e5 + 30;
struct node{
    int x, y, z, id, op;
    int val;
}q[maxn * 5], tmp[maxn * 5];
int ans[maxn * 5], num[maxn * 5];
bool cmp(node x, node y){
    if(x.x != y.x) return x.x < y.x;
    else if(x.y != y.y) return x.y < y.y;
    return x.z < y.z;
}
bool cmp1(node x, node y){
    return x.id < y.id;
}
void CDQ(int l, int r){
    if(r - l <= 1) {
        return;
    }
    int mid = l + r >> 1;
    CDQ(l, mid);
    CDQ(mid, r);
    int t1 = l, t2 = mid, cnt = 0;
    while(t1 < mid && t2 < r){
        if(q[t1].y <= q[t2].y){
            if(q[t1].op == 1)  num[q[t1].z]++;
            else if(q[t1].op == 2) num[q[t1].z]--;
            tmp[cnt++] = q[t1++];
        }
        else {
            if(q[t2].op == 3) ans[q[t2].id] += num[q[t2].z];
            tmp[cnt++] = q[t2++];
        }
    }
    while(t1 < mid)  tmp[cnt++] = q[t1++];
    while(t2 < r){
        if(q[t2].op == 3) ans[q[t2].id] += num[q[t2].z];
        tmp[cnt++] = q[t2++];
    }
    for(int i = 0; i < cnt; i++) {
        q[i + l] = tmp[i];
        num[tmp[i].z] = 0;
    }
}
int b[maxn];
int main()
{
    int n, m, qq;
    while(~scanf("%d", &n)){
        for(int i = 0; i < n; i++) {
            int x, y;
            scanf("%d%d%d", &q[i].op, &x, &y);
            q[i].val = 1;
            q[i].id = i + 1;
            q[i].x = i;
            q[i].y = x;
            q[i].z = y;
            b[i] = y;
        }
        sort(b, b + n);
        int len = unique(b, b + n) - b;
        for(int i = 0; i < n; i++) {
            q[i].z = lower_bound(b, b + len, q[i].z) - b + 1;
        }
        for(int i = 0; i <= n; i++) ans[i] = 0;
        CDQ(0, n);
        sort(q, q + n, cmp1);
        for(int i = 0; i < n; i++) {
            if(q[i].op == 3)
                printf("%d
", ans[q[i].id]);
        }
    }
    return 0;
}