CF19D Points

Posted 2022-12-22 resftlmuttmotw

\(Firstly\),离散

坐标范围太大

考虑离散化

\(Secondly\),线段树

在一个笛卡尔坐标系中，定义三种操作:

由题意的这句话非常容易想到这是一道数据结构题

\(1<=n<=2?10^5\)\(\Rightarrow\)最多有\(2?10^5\)个横坐标

针对每一个\(x_i\)对应的\(y_i_max\)我们用线段树来维护

3.find x y :找到所有已标记并在（x，y）右上方的点中,最左边的点，若点不唯一,选择最下面的一个点; 如果没有符合要求的点,给出"-1",否则给出x y.

酱紫,我们可以再二分线段树\(\Rightarrow\)在\(logn\)的时间复杂度内完成操作\(3\)

操作\(1\),\(2\)就只是更改某个\(x\)的\(y_i_max\)\(\Rightarrow\)单点修改

\(Additionally\),\(set\)

也许你认为到这就可以了

但是你忽略了一点 : 操作\(1\),\(2\)是否该更改

一些数据结构大神跳出来话说 针对每一个\(x\)上的\(y\)我们构建一个平衡树

支持插入,删除,维护最大值

其实 \(set\)就可以完全实现这些操作

\(Finally\)

这道题有卡常倾向

线段树\(+\)平衡树貌似是过不了的

线段树\(+\) \(set\) \(+\)直接二分只能\(A51\)个点

线段树\(+\) \(set\) \(+\)线段树二分才能\(AC\)

\(Code\)

\(AC Code\)

#include <map>
#include <set>
#include <cstdio>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;
#define reg register
const int MAXN = 2e5 + 10;
struct node 
    int sit,x,y;
    void assignedment(int SIT) 
        sit = SIT;
    
;
int T,Right,seg[MAXN];
set<int> st[MAXN];
node option[MAXN];
map<string,int> mp;
map<int,int> past_x,past_y;
//记录以前的值  
namespace pre 
    pair<int,int> a[MAXN],b[MAXN];
    inline void init() 
        mp["add"] = 1,mp["find"] = 2,mp["remove"] = 3;
        scanf("%d",&T);
        for(reg int i = 1; i <= T; i++) 
            string sit;
            cin >> sit;
            int x,y;
            scanf("%d%d",&x,&y);
            a[i] = make_pair(x,i);
            b[i] = make_pair(y,i);
            option[i].assignedment(mp[sit]);
        
    
    inline void hash() 
        sort(a + 1,a + 1 + T);
        sort(b + 1,b + 1 + T);
        int va,vb;
        va = vb = 0;
        for(reg int i = 1; i <= T; i++) 
            if(i == 1||a[i].first > a[i - 1].first) va++;
            if(i == 1||b[i].first > b[i - 1].first) vb++;
            option[a[i].second].x = va;
            past_x[va] = a[i].first;
            past_y[vb] = b[i].first;
            option[b[i].second].y = vb;
        
        Right = va;
    
    //输入及离散化 

namespace segment 
    int tree[MAXN << 2];
    inline void change(int l,int r,int k,int pos,int x) 
        if(l == r&&l == pos) 
            tree[k] = x;
            return;
        
        int mid = l + r >> 1;
        if(pos <= mid) change(l,mid,k << 1,pos,x);
        else change(mid + 1,r,k << 1 | 1,pos,x);
        tree[k] = max(tree[k << 1],tree[k << 1 | 1]);
    
    inline int query(int l,int r,int k,int ql,int qr) 
        if(ql <= l&&r <= qr)
            return tree[k];
        int mid = l + r >> 1,k1,k2;
        k1 = k2 = -1;
        if(ql <= mid) k1 = query(l,mid,k << 1,ql,qr);
        if(qr > mid) k2 = query(mid + 1,r,k << 1 | 1,ql,qr);
        return max(k1,k2);
    
    //线段树 
    inline int findans(int l,int r,int k,int pos,int x)
    
        if(l == r) return l;
        int mid = l + r >> 1,k1,k2;
        k1 = k2 = Right + 1;
        if(pos < mid&&tree[k << 1] > x) k1 = findans(l,mid,k << 1,pos,x);
        if(k1 < Right + 1) return k1;
        //这一句必须加 不然T(左边已经找到了 没必要找右边的) 
        if(tree[k << 1 | 1] > x) k2 = findans(mid + 1,r,k << 1 | 1,pos,x);
        return min(k1,k2);
    
    //线段树上的二分 

inline void solve() 
    for(reg int i = 1; i <= T; i++) 
        switch(option[i].sit) 
            case 1: 
                if(st[option[i].x].size() == 0)
                    segment::change(1,Right,1,option[i].x,option[i].y);
                else 
                    auto it = st[option[i].x].end();
                    if(option[i].y > *(--it)) segment::change(1,Right,1,option[i].x,option[i].y);
                
                st[option[i].x].insert(option[i].y);
                //加点 用set 
                break;
            
            case 2: 
                int l = option[i].x + 1,r = Right;
                int res = segment::query(1,Right,1,l,r);
                if(res <= option[i].y) printf("-1\n");
                else 
                    int j = segment::findans(1,Right,1,option[i].x,option[i].y);
                    printf("%d %d\n",past_x[j],past_y[*(st[j].upper_bound(option[i].y))]);
                
                //求答案 
                break;
            
            case 3: 
                auto it = st[option[i].x].upper_bound(option[i].y);
                bool f = 0;
                if(it == st[option[i].x].end()) f = 1;
                st[option[i].x].erase((--it));
                if(f) 
                    int pas;
                    if(st[option[i].x].size() == 0)
                        pas = 0;
                    else 
                        auto it = st[option[i].x].end();
                        pas = *--it;
                    
                    segment::change(1,Right,1,option[i].x,pas);
                
                //删点 用set 
                break;
            
        
    

int main() 
    pre::init();
    pre::hash();
    solve();
    return 0;

\(Tle code\)

#include <map>
#include <set>
#include <cstdio>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;
#define reg register
const int MAXN = 2e5 + 10;
struct node 
    int sit,x,y;
    void assignedment(int SIT) 
        sit = SIT;
    
;
int T,Right,seg[MAXN];
set<int> st[MAXN];
node option[MAXN];
map<string,int> mp;
map<int,int> past_x,past_y;
namespace pre 
    pair<int,int> a[MAXN],b[MAXN];
    inline void init() 
        mp["add"] = 1,mp["find"] = 2,mp["remove"] = 3;
        scanf("%d",&T);
        for(reg int i = 1; i <= T; i++) 
            string sit;
            cin >> sit;
            int x,y;
            scanf("%d%d",&x,&y);
            a[i] = make_pair(x,i);
            b[i] = make_pair(y,i);
            option[i].assignedment(mp[sit]);
        
    
    inline void hash() 
        sort(a + 1,a + 1 + T);
        sort(b + 1,b + 1 + T);
        int va,vb;
        va = vb = 0;
        for(reg int i = 1; i <= T; i++) 
            if(i == 1||a[i].first > a[i - 1].first) va++;
            if(i == 1||b[i].first > b[i - 1].first) vb++;
            option[a[i].second].x = va;
            past_x[va] = a[i].first;
            past_y[vb] = b[i].first;
            option[b[i].second].y = vb;
        
        Right = va;
    

namespace segment 
    int tree[MAXN << 2];
    inline void change(int l,int r,int k,int pos,int x) 
        if(l == r&&l == pos) 
            tree[k] = x;
            return;
        
        int mid = l + r >> 1;
        if(pos <= mid) change(l,mid,k << 1,pos,x);
        else change(mid + 1,r,k << 1 | 1,pos,x);
        tree[k] = max(tree[k << 1],tree[k << 1 | 1]);
    
    inline int query(int l,int r,int k,int ql,int qr) 
        if(ql <= l&&r <= qr)
            return tree[k];
        int mid = l + r >> 1,k1,k2;
        k1 = k2 = -1;
        if(ql <= mid) k1 = query(l,mid,k << 1,ql,qr);
        if(qr > mid) k2 = query(mid + 1,r,k << 1 | 1,ql,qr);
        return max(k1,k2);
    

inline void solve() 
    for(reg int i = 1; i <= T; i++) 
        switch(option[i].sit) 
            case 1: 
                if(st[option[i].x].size() == 0)
                    segment::change(1,Right,1,option[i].x,option[i].y);
                else 
                    auto it = st[option[i].x].end();
                    if(option[i].y > *(--it)) segment::change(1,Right,1,option[i].x,option[i].y);
                
                st[option[i].x].insert(option[i].y);
                break;
            
            case 2: 
                int l = option[i].x + 1,r = Right;
                int res = segment::query(1,Right,1,l,r);
                if(res <= option[i].y) printf("-1\n");
                else 
                    while(l < r) 
                        int mid = l + r >> 1;
                        int res = segment::query(1,Right,1,l,mid);
                        if(res > option[i].y) r = mid;
                        else l = mid + 1;
                    
                    int j = l;
                    printf("%d %d\n",past_x[j],past_y[*(st[j].upper_bound(option[i].y))]);
                
                
                //与上面AC的代码不一样的只有这里 上面是在树上二分的 时间复杂度O(logn) 而这里直接二分+线段树求区间最大值O(log^2n)
                break;
            
            case 3: 
                auto it = st[option[i].x].upper_bound(option[i].y);
                bool f = 0;
                if(it == st[option[i].x].end()) f = 1;
                st[option[i].x].erase((--it));
                if(f) 
                    int pas;
                    if(st[option[i].x].size() == 0)
                        pas = 0;
                    else 
                        auto it = st[option[i].x].end();
                        pas = *--it;
                    
                    segment::change(1,Right,1,option[i].x,pas);
                
                break;
            
        
    

int main() 
    pre::init();
    pre::hash();
    solve();
    return 0;