CodeForces-1262D Optimal Subsequences

传送门：CodeForces-1262D

思路分析

将数组按权值和ID排序
题意就可以转化为取前$k$个数，在原数组上第$pos$个数是哪个
将询问按k从小到大排序，前k个一定是一样的
将前k个数的ID放进树状数组里，二分查找到第一个前缀和等于$pos$的位置就是答案
还有一个黑科技，用红黑树维护Kth值

样例输入

3
10 20 10
6
1 1
2 1
2 2
3 1
3 2
3 3

样例输出

20
10
20
10
20
10

AC代码

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <bits/stdc++.h>
#define ls x<<1
#define rs x<<1|1
#define fi first
#define se second
#define ll long long
#define pb push_back
#define mp make_pair
#define fun function
#define sz(x) (x).size()
#define lowbit(x) (x)&(-x)
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))

namespace FastIO {
#define BUF_SIZE 100000
#define OUT_SIZE 100000
    bool IOerror=0;
    inline char nc() {
        static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE;
        if(p1==pend) {
            p1=buf;
            pend=buf+fread(buf,1,BUF_SIZE,stdin);
            if(pend==p1) {
                IOerror=1;
                return -1;
            }
        }
        return *p1++;
    }
    inline bool blank(char ch) {
        return ch==' '||ch=='\n'||ch=='\r'||ch=='\t';
    }
    template<class T> inline bool read(T &x) {
        bool sign=0;
        char ch=nc();
        x=0;
        for(; blank(ch); ch=nc());
        if(IOerror)return false;
        if(ch=='-')sign=1,ch=nc();
        for(; ch>='0'&&ch<='9'; ch=nc())x=x*10+ch-'0';
        if(sign)x=-x;
        return true;
    }
    template<class T,class... U>bool read(T& h,U&... t) {
        return read(h)&&read(t...);
    }
#undef OUT_SIZE
#undef BUF_SIZE
};
using namespace std;
using namespace FastIO;
using namespace __gnu_pbds;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> rbtree;

const int INF = 0x3f3f3f3f;
const int N = 2e5+10;

int a[N],bit[N],ans[N];
rbtree Tree;
 
struct date {
    int x,id;
} t[N];

struct node {
    int k,pos,id;
} q[N];

void add(int x) {
    for(int i=x; i<N; i+=lowbit(i)) bit[i]+=1;
}

int query(int x) {
    int sum=0;
    for(int i=x; i; i-=lowbit(i)) sum+=bit[i];
    return sum;
}

int cmp1(date x,date y) {
    if(x.x==y.x) return x.id<y.id;
    else return x.x>y.x;
}

int cmp2(node x,node y) {
    return x.k<y.k;
}

int main() {

#ifdef xiaofan
    freopen("1.in","r",stdin);
    freopen("1.out","w",stdout);
#endif

    int n;
    read(n);
    for(int i=1; i<=n; i++) {
        int x;
        read(x);
        a[i]=x;
        t[i].x=x;
        t[i].id=i;
    }
    int m;
    read(m);
    for(int i=1; i<=m; i++) {
        read(q[i].k,q[i].pos);
        q[i].id=i;
    }
    sort(t+1,t+1+n,cmp1);
    sort(q+1,q+1+m,cmp2);
    int now=1;
    for(int i=1; i<=m; i++) {
        //名次树 
        while(now<=q[i].k) Tree.insert(t[now++].id);
        int res=*Tree.find_by_order(q[i].pos-1);
        ans[q[i].id]=a[res];
        //二分树状数组 
        while(now<=q[i].k) add(t[now++].id);
        int l=1,r=n,res=0;
        while(r>=l){
            int mid=l+r>>1;
            if(query(mid)>=q[i].pos) res=mid,r=mid-1;
            else l=mid+1;
        }
        ans[q[i].id]=a[res];
        
    }
    for(int i=1; i<=m; i++) cout<<ans[i]<<endl;




    return 0;
}