标签：
D - Friends and Subsequences
Description
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 exactlyn(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.
Sample Input
6
1 2 3 2 1 4
6 7 1 2 3 2
2
3
3 3 3
1 1 1
0
Hint
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：
给出两个长为n的序列a和b，问有多少个区间[L,R]满足max<a>[L,R] == min<b>[L,R]。
分析:
枚举左端点,二分找到右端点可行区间的左右边界;二分右端点需要用RMQ（RQM预处理和查询的相关知识点需要另外了解）。
左边界:要是a>=b,左移;否则右移,找到第一个a=b的点;右边界：要是a>b,左移,否则右移,找到最后一个a=b的点；最后累加右端点可
行区间长度;
AC的代码：
#include <iostream> #include<cstdio> #include<cmath> using namespace std; #define LL long long int rmq[2][20][200200]; void RMQ(int n) { for(int k = 1; (1<<k) <= n; ++k) for(int i = 0; i+(1<<k) <= n; ++i) { rmq[0][k][i] = max(rmq[0][k-1][i],rmq[0][k-1][i+(1<<(k-1))]); rmq[1][k][i] = min(rmq[1][k-1][i],rmq[1][k-1][i+(1<<(k-1))]); } } int Search(int pos,int l,int r) { int k = log((r-l+1)*1.0)/log(2.0); if(pos) return min(rmq[pos][k][l],rmq[pos][k][r-(1<<k)+1]); else return max(rmq[pos][k][l],rmq[pos][k][r-(1<<k)+1]); } int main() { int n; scanf("%d",&n); for(int i=0;i<n;i++) scanf("%d",&rmq[0][0][i]); for(int i=0;i<n;i++) scanf("%d",&rmq[1][0][i]); RMQ(n); LL ans=0; int l,r,a,b,s,e; for(int i=0;i<n;i++) { l=i; r=n-1; s=-1; while(l<=r) { int mid=(l+r)/2; a=Search(0,i,mid); b=Search(1,i,mid); if(a>=b) { if(a==b) s=mid; r=mid-1; } else l=mid+1; } if(s==-1) continue; l=i; r=n-1; e=-1; while(l<=r) { int mid=(l+r)/2; a=Search(0,i,mid); b=Search(1,i,mid); if(a>b) r=mid-1; else { e=mid; l = mid+1; } } ans+=(e-s+1); } printf("%lld\n",ans); return 0; }
另一种方法：
#include<bits/stdc++.h> using namespace std; int n,a[200005],b[200005]; long long ans; deque<int>x,y; 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]); for(int i=1,j=1; i<=n; i++) { while(!x.empty()&&a[x.back()]<=a[i]) x.pop_back(); while(!y.empty()&&b[y.back()]>=b[i]) y.pop_back(); x.push_back(i); y.push_back(i); while(j<=i&&a[x.front()]-b[y.front()]>0) { j++; while(!x.empty()&&x.front()<j) x.pop_front(); while(!y.empty()&&y.front()<j) y.pop_front(); } if(!x.empty()&&!y.empty()&&a[x.front()]==b[y.front()]) { ans+=min(x.front(),y.front())-j+1; } } printf("%lld",ans); }
Codeforces Round #361 (Div. 2) D
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原文地址：http://www.cnblogs.com/fenhong/p/5665025.html
