2019牛客暑期多校训练营(第一场) A Equivalent Prefixes ( st 表 + 二分+分治)

链接:https://ac.nowcoder.com/acm/contest/881/A
来源:牛客网

Equivalent Prefixes
时间限制:C/C++ 2秒,其他语言4秒
空间限制:C/C++ 524288K,其他语言1048576K
64bit IO Format: %lld
题目描述
Two arrays u and v each with m distinct elements are called equivalent if and only if
R
M
Q
(
u
,
l
,
r
)
=
R
M
Q
(
v
,
l
,
r
)
RMQ(u,l,r)=RMQ(v,l,r) for all
1

l

r

m
1≤l≤r≤m
where
R
M
Q
(
w
,
l
,
r
)
RMQ(w,l,r) denotes the index of the minimum element among
w
l
,
w
l

  • 1
    ,

    ,
    w
    r
    wl,wl+1,…,wr.
    Since the array contains distinct elements, the definition of minimum is unambiguous.

Bobo has two arrays a and b each with n distinct elements. Find the maximum number
p

n
p≤n where
{
a
1
,
a
2
,

,
a
p
}
{a1,a2,…,ap} and
{
b
1
,
b
2
,

,
b
p
}
{b1,b2,…,bp} are equivalent.
输入描述:
The input consists of several test cases and is terminated by end-of-file.

The first line of each test case contains an integer n.
The second line contains n integers
a
1
,
a
2
,

,
a
n
a1,a2,…,an.
The third line contains n integers
b
1
,
b
2
,

,
b
n
b1,b2,…,bn.

  • 1

    n

    10
    5
    1≤n≤105
  • 1

    a
    i
    ,
    b
    i

    n
    1≤ai,bi≤n
  • {
    a
    1
    ,
    a
    2
    ,

    ,
    a
    n
    }
    {a1,a2,…,an} are distinct.
  • {
    b
    1
    ,
    b
    2
    ,

    ,
    b
    n
    }
    {b1,b2,…,bn} are distinct.
  • The sum of n does not exceed
    5
    ×
    10
    5
    5×105.
    输出描述:
    For each test case, print an integer which denotes the result.
    示例1
    输入
    复制
    2
    1 2
    2 1
    3
    2 1 3
    3 1 2
    5
    3 1 5 2 4
    5 2 4 3 1
    输出
    复制
    1
    3
    4

题意:
给你两个数组a,b,大小为n,让你寻找一个数p (1<= p <= n) ,使之在 1~p 任意一个区间中a,b数组的最小值下标相同。
思路:

容易知道p的取值具有单调性,首先我们用st表在对数组a,b进行预处理,方便后续的RMQ,因为数组中的数相互不同,那么我们就可以直接RMQ获得区间最小值下标。

在二分p的过程中,我们这样来判断mid是否合法:

询问1~mid 区间两个数组中的最小值下标是否一致,如果不一致直接返回false,否则 以最小值下标minid为分界点递归处理1~minid,minid+1~mid。这个过程是O(n)的

所以总体时间复杂度是 O( n* log n )

细节见代码:

#include
using namespace std;
const int maxn=2e5+10;
int n;
int a[maxn],b[maxn];
int sa[maxn][20],sb[maxn][20],mn[maxn];
void init()
{
    mn[0]=-1;
    for (int i=1;i<=n;i++)
    {
        mn[i]=((i & (i-1))==0) ? mn[i-1]+1 : mn[i-1];
        sa[i][0]=a[i];
        sb[i][0]=b[i];
    }
    for (int j=1;j<=mn[n];j++)
        for (int i=1;i+(1<=r)
    {
        return 1;
    }
    int w=rqb(l,r);
    int q=rqa(l,r);
    if(w!=q)
    {
        return 0;
    }else{
        return pan(l,w-1)&&(pan(q+1,r));
    }
}
bool check(int mid)
{
//    bool res=1;
    return pan(1,mid);
}
int main(){
    while(~scanf("%d",&n)){
        for(int i=1;i<=n;i++){
            scanf("%d",&a[i]);
            ida[a[i]]=i;
        }
        for(int i=1;i<=n;i++){
            scanf("%d",&b[i]);
            idb[b[i]]=i;;
        }
        init();
        int l=1;
        int r=n;
        int mid;
        int ans=1;
        while(l<=r)
        {
            mid=(l+r)>>1;
            if(check(mid))
            {
                l=mid+1;
                ans=mid;
            }else{
                r=mid-1;
            }
 
        }
        printf("%d\n",ans);
    }
}

转载于:https://www.cnblogs.com/qieqiemin/p/11209227.html

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