BZOJ-3207 花神的嘲讽计划Ⅰ

貌似是先把各个子串处理成各个Hash值，然后离散化，然后这道题就变成询问区间[x,y]中有没有数字k。

主席树直接上。。。

#include <cstdlib>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
#define rep(i, l, r) for(int i=l; i<=r; i++)
#define clr(x, c) memset(x, c, sizeof(x))
#define lowbit(x) (x&-x)
#define maxn 600009
#define maxm 5000009
#define hash 2333
#define inf 0x7fffffff
#define s(x) Sum[x]
#define t(x) Tree[x]
#define l(x) Left[x]
#define r(x) Right[x]
#define k(x) Key[x]
#define ull unsigned long long
using namespace std;
inline int read()
{
	int x=0, f=1; char ch=getchar();
	while (!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
	while (isdigit(ch)) x=x*10+ch-'0', ch=getchar();
	return x*f;
}
struct node{ull v; int t;} q[maxn];
bool cmp(node a, node b){return a.v<b.v;}
int n, m, k, z, ln, qx[maxn], qy[maxn], a[maxn];
int Tree[maxn], Left[maxm], Right[maxm], Sum[maxm], Key[maxm];
inline ull Hash(int x){ull now=1; rep(i, 0, k-1) now=now*hash+a[i+x]; return now;}
void Add(int k, int l, int r, int u, int &t)
{
	if (!t) t=++z;
	s(t)=s(u)+1;
	if (l==r) return; int mid=(l+r)>>1;
	if (k<=mid) r(t)=r(u), Add(k, l, mid, l(u), l(t));
	else l(t)=l(u), Add(k, mid+1, r, r(u), r(t));
}
inline bool Query(int l, int r, int k)
{
	int L=1, R=ln, x=t(l-1), y=t(r);
	while (L<R)
	{
		if (s(y)-s(x)==0) return false;
		int mid=(L+R)>>1;
		if (k<=mid) R=mid, x=l(x), y=l(y);
		else L=mid+1, x=r(x), y=r(y);
	}
	if (s(y)-s(x)) return true; else return false;
}
int main()
{
	n=read(), m=read(), k=read(); t(0)=++z, l(z)=r(z)=z;
	rep(i, 1, n) a[i]=read();
	rep(i, 1, n-k+1) q[i].v=Hash(i), q[i].t=i; n=n-k+1;
	rep(i, 1, m)
	{
		qx[i]=read(), qy[i]=read()-k+1;
		rep(j, 1, k) a[j]=read(); q[i+n].v=Hash(1), q[i+n].t=i+n;
	}
	sort(q+1, q+m+n+1, cmp); k(q[1].t)=++ln;
	rep(i, 2, n+m) 
		k(q[i].t) = q[i].v!=q[i-1].v ? ++ln : ln;
	rep(i, 1, n) Add(k(i), 1, ln, t(i-1), t(i));
	rep(i, 1, m) if (Query(qx[i], qy[i], k(i+n))) printf("No\n"); else printf("Yes\n");
	return 0;
}