codeforces 689D ST表+二分 模板
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Mike and !Mike are old childhood rivals, they are opposite in everything they do, except programming. Today they have a problem they cannot solve on their own, but together (with you) — who knows?
Every one of them has an integer sequences a and b of length n. Being given a query of the form of pair of integers (l, r), Mike can instantly tell the value of while !Mike can instantly tell the value of .
Now suppose a robot (you!) asks them all possible different queries of pairs of integers (l, r) (1 ≤ l ≤ r ≤ n) (so he will make exactly n(n + 1) / 2 queries) and counts how many times their answers coincide, thus for how many pairs is satisfied.
How many occasions will the robot count?
Input
The first line contains only integer n (1 ≤ n ≤ 200 000).
The second line contains n integer numbers a1, a2, …, an ( - 109 ≤ ai ≤ 109) — the sequence a.
The third line contains n integer numbers b1, b2, …, bn ( - 109 ≤ bi ≤ 109) — the sequence b.
Output
Print the only integer number — the number of occasions the robot will count, thus for how many pairs is satisfied.
Example
Input
6
1 2 3 2 1 4
6 7 1 2 3 2
Output
2
Input
3
3 3 3
1 1 1
Output
0
Note
The occasions in the first sample case are:
1.l = 4,r = 4 since max{2} = min{2}.
2.l = 4,r = 5 since max{2, 1} = min{2, 3}.
There are no occasions in the second sample case since Mike will answer 3 to any query pair, but !Mike will always answer 1.
题意:
给定两个等长区间,求相同l,r情况下,有多少个区间,第一个序列的最大值和第二个序列的最小值相同。
因为最大值是一个单调递增的函数,最小值是一个单调递减的函数,所以 i区间到mid区间的 (最大值-最小值) 是一个单调递增的函数。所以枚举左端点,不断的进行二分右端点,分别进行两个二分可以得到两个右端点,第一个右端点,代表的是满足,max(i,mid)==min(i,mid) 的右边界最大值,然后第二个右端点表示的是max(i,mid)==min(i,mid) 的右边界最小值。 两者一减,那么就是左边界为i,右边界在(t1~t2)满足,这些是都满足的区间数。
献上代码很好理解。。
无敌快的代码
#include <bits/stdc++.h>using namespace std;int n;int a[201000],b[201000];int mn[201000][30];int mx[201000][30];int num;void init(){ for(int i=1;i<=n;i++) { mx[i][0]=a[i],mn[i][0]=b[i]; } for(int j = 1; (1<<j) <= n; ++j) for(int i = 1; i + (1<<j) - 1 <= n; ++i) { mx[i][j] = max(mx[i][j-1],mx[i+(1<<(j-1))][j-1]); mn[i][j] = min(mn[i][j-1],mn[i+(1<<(j-1))][j-1]); } }int rmq(int i,int j,int f){ int k=0; while((1 << (k + 1)) <= j - i + 1) k++; int res=0; if(f==0) res=min(mn[i][k],mn[j-(1<<k)+1][k]); if(f==1) res=max(mx[i][k],mx[j-(1<<k)+1][k]); return res;}int main(){ scanf("%d",&n); for(int i=1;i<=n;i++) scanf("%d",&a[i]); for(int i=1;i<=n;i++) scanf("%d",&b[i]); init(); long long ans=0; for(int i=1;i<=n;i++) { int l=i,r=n; int t1=-1,t2=-1; while(l<=r) { int mid=(l+r)>>1; int re1=rmq(i,mid,1),re2=rmq(i,mid,0); if(re1>re2) { r=mid-1; } else if(re1<re2) { l=mid+1; } else if(re1==re2){ t1=mid; r=mid-1; } } if(t1==-1) continue; l=i,r=n; while(l<=r) { int mid=(l+r)>>1; int re1=rmq(i,mid,1),re2=rmq(i,mid,0); if(re1>re2) r=mid-1; else if(re1<re2) l=mid+1; else { t2=mid; l=mid+1; } } ans+=t2-t1+1; } printf("%lld\n",ans );}
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