2554. [福利]可持久化线段树 单点修改

https://syzoj.com/problem/247

Q k l r 查询数列在第k个版本时,区间[l, r]上的最大值

M k p v 把数列在第k个版本时的第p个数修改为v,并产生一个新的数列版本


///                 .-~~~~~~~~~-._       _.-~~~~~~~~~-.
///             __.'              ~.   .~              `.__
///           .'//                  \./                  \\`.
///        .'//                     |                     \\`.
///       .'// .-~"""""""~~~~-._     |     _,-~~~~"""""""~-. \\`.
///     .'//.-"                 `-.  |  .-'                 "-.\\`.
///   .'//______.============-..   \ | /   ..-============.______\\`.
/// .'______________________________\|/______________________________`.
#pragma GCC optimize(2)
#pragma comment(linker, "/STACK:102400000,102400000")
#include <vector>
#include <iostream>
#include <string>
#include <map>
#include <stack>
#include <cstring>
#include <queue>
#include <list>
#include <stdio.h>
#include <set>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <iomanip>
#include <cctype>
#include <sstream>
#include <functional>
#include <stdlib.h>
#include <time.h>
#include <bitset>
using namespace std;

#define pi acos(-1)
#define s_1(x) scanf("%d",&x)
#define s_2(x,y) scanf("%d%d",&x,&y)
#define s_3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define s_4(x,y,z,X) scanf("%d%d%d%d",&x,&y,&z,&X)
#define S_1(x) scan_d(x)
#define S_2(x,y) scan_d(x),scan_d(y)
#define S_3(x,y,z) scan_d(x),scan_d(y),scan_d(z)
#define PI acos(-1)
#define endl '\n'
#define srand() srand(time(0));
#define me(x,y) memset(x,y,sizeof(x));
#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)
#define close() ios::sync_with_stdio(0); cin.tie(0);
#define FOR(x,n,i) for(int i=x;i<=n;i++)
#define FOr(x,n,i) for(int i=x;i<n;i++)
#define fOR(n,x,i) for(int i=n;i>=x;i--)
#define fOr(n,x,i) for(int i=n;i>x;i--)
#define W while
#define sgn(x) ((x) < 0 ? -1 : (x) > 0)
#define bug printf("***********\n");
#define db double
#define ll long long
#define mp make_pair
#define pb push_back
#define sz size()
typedef long long LL;
typedef pair <int, int> ii;
const int INF=~0U>>1;
const LL LINF=0x3f3f3f3f3f3f3f3fLL;
const int dx[]={-1,0,1,0,1,-1,-1,1};
const int dy[]={0,1,0,-1,-1,1,-1,1};
const int maxn=1e5+10;
const int maxx=1e3+10;
const double EPS=1e-8;
const double eps=1e-8;
const int mod=19260817;
template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}
template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}
template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}
template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}
template <class T>
inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return 0;}
while (c != '-' && (c < '0' || c > '9')){c = getchar();}sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0');
while (c = getchar(), c >= '0' && c <= '9'){ret = ret * 10 + (c - '0');}ret *= sgn;return 1;}

inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false;
while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;}
else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}}
if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}}
if(IsN) num=-num;return true;}

void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');}
void print(LL a){ Out(a),puts("");}
//freopen( "in.txt" , "r" , stdin );
//freopen( "data.txt" , "w" , stdout );
//cerr << "run time is " << clock() << endl;

//void readString(string &s)
//{
//	static char str[maxn];
//	scanf("%s", str);
//	s = str;
//}

int n,q,Sz,cnt;
int a[maxn];
int T[maxn];
struct node 
{
	int mx,ls,rs;
}F[maxn*40];
void update(int k)
{
	F[k].mx=max(F[F[k].ls].mx,F[F[k].rs].mx);
}
void build(int &k,int l,int r)
{
	if(!k) k=++Sz;
	if(l==r) {F[k].mx=a[l];return ;}
	int mid=l+r>>1;
	build(F[k].ls,l,mid);
	build(F[k].rs,mid+1,r);
	update(k);
}
void insert(int &k,int last,int l,int r,int p,int v)
{
	k=++Sz;
	F[k]=F[last];
	if(l==r) {F[k].mx=v;return ;}
	int mid=l+r>>1;
	if(p<=mid)
		insert(F[k].ls,F[last].ls,l,mid,p,v);
	else 
		insert(F[k].rs,F[last].rs,mid+1,r,p,v);
	update(k);
}
int query(int &k,int l,int r,int x,int y)
{
	if(y<l||r<x) return -INF;
	if(x<=l&&r<=y) return F[k].mx;
	int mid=l+r>>1;
	return max(query(F[k].ls,l,mid,x,y),query(F[k].rs,mid+1,r,x,y));
}
void solve()
{
	W(s_2(n,q)!=EOF)
	{
		cnt=0;
		FOR(1,n,i) s_1(a[i]);
		build(T[++cnt],1,n);
		int op,k,x,y;
		FOR(1,q,i)
		{
			s_4(op,k,x,y);
			if(op&1)
				insert(T[++cnt],T[k],1,n,x,y);
			else print(query(T[k],1,n,x,y));
		}
	}
}
int main()
{
    //freopen( "in.txt" , "r" , stdin );
    //freopen( "data.txt" , "w" , stdout );
    int t=1;
    //init();
    //s_1(t);
    for(int cas=1;cas<=t;cas++)
    {
        //printf("Case #%d:\n",cas);
        solve();
    }
}


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转载自blog.csdn.net/qq_36553623/article/details/78772951