POJ-2104 K-th Number CDQ分治

题目传送门

题意：给你一个序列，长度为n，m次询问，询问一段区间的第k大。

题解：CDQ分治，对整个值域进行分治。每次取一个mid， 计算出整个区间内mid <= 的数目，如果 num >= k， 那么就可以第k大的数一定落在[ l, mid]之间， 否者就会落在 [mid+1, r]之间， 然后我们继续递归二分下去，直到 l == r 那么就找到第k大了。

代码：

#include<cstdio>
#include<algorithm>
#include<vector>
#include<queue>
#include<iostream>
#include<cstring>
using namespace std;
#define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
#define LL long long
#define ULL unsigned LL
#define fi first
#define se second
#define pb push_back
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lch(x) tr[x].son[0]
#define rch(x) tr[x].son[1]
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
typedef pair<int,int> pll;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const LL mod =  (int)1e9+7;
const int N = 2e5 + 100;
int n;
struct Node{
    int op;
    int L, R;
    int x, id;
}A[N], lf[N], ri[N];
int tot = 0;
void NowNode(int op, int L, int R, int x, int id){
    ++tot;
    A[tot].op = op; A[tot].L = L; A[tot].R = R;
    A[tot].x = x; A[tot].id = id;
}
int ans[N];
int tree[N];
int Mn = inf, Mx = 0;
void Add(int x, int v){
    for(int i = x; i <= n; i += i & (-i)){
        tree[i] += v;
    }
}
int Query(int x){
    int ret = 0;
    for(int i = x; i; i -= i&(-i)){
        ret += tree[i];
    }
    return ret;
}
void cdq(int Lval, int Rval, int st, int ed){
    if(Lval > Rval || st > ed) return ;
    if(Lval == Rval){
        for(int i = st; i <= ed; i++)
            if(A[i].op == 1) ans[A[i].id] = Lval;
        return ;
    }
    int mid = Lval + Rval >> 1;

    int lsz = 0, rsz = 0;
    for(int i = st; i <= ed; i++){
        int op = A[i].op, x = A[i].x, L = A[i].L, R = A[i].R, id = A[i].id;
        if(op == 0){
            if(x <= mid){
                Add(A[i].id, 1);
                lf[++lsz] = A[i];
            }
            else ri[++rsz] = A[i];
        }
        else if(A[i].op == 1){
            int num = Query(R) - Query(L-1);
            if(num >= x) lf[++lsz] = A[i];
            else {
                A[i].x -= num;
                ri[++rsz] = A[i];
            }
        }
    }
    for(int i = 1; i <= lsz; i++) {

        if(lf[i].op == 0) {
            Add(lf[i].id, -1);
        }
        A[st + i - 1] = lf[i];
    }
    for(int i = 1; i <= rsz; i++) A[st + lsz + i - 1] = ri[i];
    cdq(Lval, mid, st, st+lsz-1);
    cdq(mid+1, Rval, st+lsz, ed);
}
int main(){
    int m, val, l, r;
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; i++){
        scanf("%d", &val);
        NowNode(0,0,0,val,i);
        Mn = min(Mn, val);
        Mx = max(Mx, val);
    }
    for(int i = 1; i <= m; i++){
        scanf("%d%d%d", &l, &r, &val);
        NowNode(1,l,r,val,i);
    }
    cdq(Mn,Mx,1,n+m);
    for(int i = 1; i <= m; i++){
        printf("%d
", ans[i]);
    }
    return 0;
}