11/1/2018模拟 Max

题面

也就是说, 随机序列RMQ.((n le 8388608), (m le 8*10^6))

解法

我写了笛卡尔树+tarjan

然而听神仙说, 因为数据随机, 建完树暴力找lca就行, 跑的飞快...吊打std...

还有题解, 真是神仙做法...

设 (p_i) 表示比 (a_i) 大的前一个数所在的位置,那么 p 构成了一棵树。
若我们需要查询 [l, r] 的答案,只需找到 r 在这棵树上不小于 l 的祖先。于是我们可以按照 l
从大到小排序,一边向上查询祖先一边路径压缩(类似并查集)。
由于树上的每条边至多被压缩一次,复杂度 O(n) 。

我的代码:

#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<set>
#include<map>
using namespace std;
#define rep(i,l,r) for(register int i=(l);i<=(r);++i)
#define repdo(i,l,r) for(register int i=(l);i>=(r);--i)
#define il inline
typedef double db;
typedef long long ll;

//---------------------------------------
int n,m;
int gen,p1,p2;
int number(){
    gen=(1LL*gen*p1)^p2;
    return (gen&(n-1))+1;
}
const int nsz=8388700;
int a[nsz],ans[nsz];

struct tnd{int ch[2];}car[nsz];
int rt,pc=0;
int stk[nsz],top=0;
void build(){
	rep(i,1,n){
		while(top&&a[stk[top]]<a[i])car[i].ch[0]=stk[top--];
		car[stk[top]].ch[1]=i;
		stk[++top]=i;
	}
	rt=stk[1],pc=n;
}

struct tq{int t,pr;}qu[nsz*2];
int hd[nsz],pq=1;
void adde(int f,int t){qu[++pq]=(tq){t,hd[f]};hd[f]=pq;}
void adddb(int f,int t){adde(f,t);adde(t,f);}

int fa[nsz];
void init(){rep(i,1,n)fa[i]=i;}
void merge(int a,int b){fa[b]=a;}
int find(int p){return p==fa[p]?p:fa[p]=find(fa[p]);}

int vi[nsz];
void tar(int p){
	vi[p]=1;
	int v;
	rep(i,0,1){
		v=car[p].ch[i];
		if(v==0)continue;
		tar(v);
		merge(p,v);
	}
	for(int i=hd[p];i;i=qu[i].pr){
		if(vi[qu[i].t])
			ans[i/2]=find(qu[i].t);
	}
}
int main() {
//	freopen("max.in", "r", stdin);
//	freopen("max.out", "w", stdout);

	scanf("%d%d", &n, &m);
	scanf("%d%d%d", &gen, &p1, &p2);
	for (int i = 1; i <= n; ++i)
		a[i] = number();
	int l,r;
	for (int i = 1; i <= m; ++i) {
		l = number(), r = number();
		if (l > r) swap(l,r);
		adddb(l,r);
	}

	build();
	init();
	tar(rt);

	ll sum = 0;
	for (int i = 1; i <= m; ++i) {
		sum=(sum+a[ans[i]])%p2;
	}
	sum=sum*p1%p2;
	printf("%lld
", sum);
}

std:

#include <cstdio>
#include <ctime>
#include <cstdlib>
#include <algorithm>
using namespace std;

const int N = 1e7 + 5;

int n, m;
int gen, cute1, cute2;
int number() {
    gen = (1LL * gen * cute1) ^ cute2;
    return (gen & (n - 1)) + 1;
}


int hd[N], nxt[N], id[N], to[N], cnt;
int ans[N], a[N], p[N], q[N];

int add(int x, int y, int i) {
    ++cnt;
    nxt[cnt] = hd[x];
    to[cnt] = y;
    id[cnt] = i;
    hd[x] = cnt;
}


int getfa(int x, int y) {
    int fa = x;
    for (int i = x; i; i = p[i])
        if (p[i] < y || p[i] == i) {
            fa = i;
            break;
        }
    for (int j, i = x; i != fa; i = j) {
        j = p[i], p[i] = fa;
    }
    return fa;
}

int main() {
    freopen("max.in", "r", stdin);
    freopen("max.out", "w", stdout);

    scanf("%d%d", &n, &m);
    scanf("%d%d%d", &gen, &cute1, &cute2);

    for (int i = 1; i <= n; ++i)
        a[i] = number();
    for (int i = 1; i <= m; ++i) {
        int l = number(), r = number();
        if (l > r) swap(l, r);
        add(l, r, i);
    }
    double t1;
    fprintf(stderr, "%lf
", t1 = (double)clock()/CLOCKS_PER_SEC);

    int ind = 0;
    for (int i = 1; i <= n; ++i) {
        while (ind && a[q[ind]] <= a[i]) --ind;
        if (ind) p[i] = q[ind];
        else p[i] = i;
        q[++ind] = i;
    }

    for (int i = n; i; --i) {
        for (int j = hd[i]; j; j = nxt[j])
            ans[id[j]] = a[getfa(to[j], i)];
    }

    
    fprintf(stderr, "%lf
", (double)clock()/CLOCKS_PER_SEC - t1);

    int sum = 0;
    for (int i = 1; i <= m; ++i)
        (sum += 1LL * ans[i] * cute1 % cute2) %= cute2;
    printf("%d
", sum);
}