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);
}
}