GTY‘s gay friends

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1379    Accepted Submission(s): 355

Problem Description

GTY has n
gay friends. To manage them conveniently, every morning he ordered all
his gay friends to stand in a line. Every gay friend has a
characteristic value ai
, to express how manly or how girlish he is. You, as GTY‘s assistant,
have to answer GTY‘s queries. In each of GTY‘s queries, GTY will give
you a range [l,r] . Because of GTY‘s strange hobbies, he wants there is a permutation [1..r−l+1] in [l,r]. You need to let him know if there is such a permutation or not.

Input

Multi test cases (about 3) . The first line contains two integers n and m ( 1≤n,m≤1000000
), indicating the number of GTY‘s gay friends and the number of GTY‘s
queries. the second line contains n numbers seperated by spaces. The ith number ai ( 1≤ai≤n ) indicates GTY‘s ith
gay friend‘s characteristic value. The next m lines describe GTY‘s
queries. In each line there are two numbers l and r seperated by spaces (
1≤l≤r≤n ), indicating the query range.

Output

For each query, if there is a permutation [1..r−l+1] in [l,r], print ‘YES‘, else print ‘NO‘.

Sample Input

8 5
2 1 3 4 5 2 3 1
1 3
1 1
2 2
4 8
1 5
3 2
1 1 1
1 1
1 2

Sample Output

YES NO YES YES YES YES NO

题意：

n个数m个询问，询问（l,r）中的数是否为1 ~ r-l+1的一个排列。

分析：

若（l,r）中的数为1 ~ r-l+1中的一个排列，则必须满足：

1、（l,r）中的数之和为len*(len+1)/2，其中len = r-l+1。

2、区间内的数字各不相同，即用线段树维护位置i上的数上次出现的位置的最大值。

只要区间内所有的数上次出现的位置last[i] < l，则区间内的数各不相同。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
#include <math.h>
using namespace std;
typedef long long LL;
const int N = 1000005;
int pre[N],now[N];
LL sum[N];
struct Tree{
    int MAX_ID;
}tree[N<<2];
int MAX_ID;
void PushUp(int idx){
    tree[idx].MAX_ID = max(tree[idx<<1].MAX_ID,tree[idx<<1|1].MAX_ID);
}
void build(int l,int r,int idx){
    if(l==r){
        tree[idx].MAX_ID = pre[l];
        return;
    }
    int mid = (l+r)>>1;
    build(l,mid,idx<<1);
    build(mid+1,r,idx<<1|1);
    PushUp(idx);
}
void query(int l,int r,int L,int R,int idx){
    if(l >= L&& r <= R){
        MAX_ID = max(MAX_ID,tree[idx].MAX_ID);
        return;
    }
    int mid = (l+r)>>1;
    if(mid>=L) query(l,mid,L,R,idx<<1);
    if(mid<R)  query(mid+1,r,L,R,idx<<1|1);
}
int main()
{
    int n,k;
    while(scanf("%d%d",&n,&k)!=EOF){
        memset(now,0,sizeof(now));
        sum[0] = 0;
        for(int i=1;i<=n;i++){
            int x;
            scanf("%d",&x);
            sum[i] = sum[i-1]+x;
            pre[i] = now[x];
            now[x] = i;
        }
        build(1,n,1);
        while(k--){
            int l,r;
            scanf("%d%d",&l,&r);
            LL len = r-l+1;
            LL S = (len+1)*(len)/2;
            if(sum[r]-sum[l-1]!=S){
                printf("NO\n");
                continue;
            }
            MAX_ID = -1;
            query(1,n,l,r,1);
            if(MAX_ID<l) printf("YES\n");
            else printf("NO\n");
        }
    }
    return 0;
}

不过我们还有更简单的hash做法,对于[1..n][1..n][1..n]中的每一个数随机一个64位无符号整型作为它的hash值,一个集合的hash值为元素的异或和,预处理[1..n]的排列的hash和原序列的前缀hash异或和,就可以做到线性预处理,O(1)O(1)O(1)回答询问.

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cmath>
#include<vector>
#include<queue>
#include<map>
#include<ctime>
#define eps (1e-8)

using namespace std;

typedef long long ll;
unsigned long long ra[1000005],a[1000005],ha[1000005];
int n,m,t,p;

//int main(int argc,char const *argv[])
int main()
{
    srand(time(NULL));
    a[0] = ha[0] = 0;
    for(int i=1; i<=1000000; i++)
    {
        ra[i] = rand()*rand();
        ha[i] = ha[i-1]^ra[i];
    }
    for(int i=1; i<=20; i++)
    {
        printf("%lld\n",ra[i]);
    }
    while(~scanf("%d%d",&n,&m))
    {
        for(int i=1; i<=n; i++)
        {
            scanf("%d",&t);
            a[i] = ra[t];
            a[i]^=a[i-1];
        }
        for(int i=1; i<=m; i++)
        {
            scanf("%d%d",&t,&p);
            puts((ha[p-t+1]==(a[p]^a[t-1])?"YES":"NO"));
        }
    }
    return 0;
}