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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 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?
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.
Print the only integer number — the number of occasions the robot will count, thus for how many pairs is satisfied.
6
1 2 3 2 1 4
6 7 1 2 3 2
2
3
3 3 3
1 1 1
0
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.
分析:RMQ+二分；
代码：
#include <iostream> #include <cstdio> #include <cstdlib> #include <cmath> #include <algorithm> #include <climits> #include <cstring> #include <string> #include <set> #include <map> #include <queue> #include <stack> #include <vector> #include <list> #include <ext/rope> #define rep(i,m,n) for(i=m;i<=n;i++) #define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++) #define vi vector<int> #define pii pair<int,int> #define mod 1000000007 #define inf 0x3f3f3f3f #define pb push_back #define mp make_pair #define fi first #define se second #define ll long long #define pi acos(-1.0) const int maxn=2e5+10; const int dis[][2]={0,1,-1,0,0,-1,1,0}; using namespace std; using namespace __gnu_cxx; ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);} ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;} int n,m,p[maxn],a[20][maxn],b[20][maxn]; ll ans; void init() { for(int i=2;i<n;i++)p[i]=1+p[i/2]; for(int i=1;i<20;i++) for(int j=0;j+(1<<i)-1<n;j++) a[i][j]=max(a[i-1][j],a[i-1][j+(1<<(i-1))]),b[i][j]=min(b[i-1][j],b[i-1][j+(1<<(i-1))]); return; } int getma(int l,int r) { int x=p[r-l+1]; return max(a[x][l],a[x][r-(1<<x)+1]); } int getmi(int l,int r) { int x=p[r-l+1]; return min(b[x][l],b[x][r-(1<<x)+1]); } int getl(int now) { int l=now-1,r=n; while(r-l>1) { int mid=(l+r)>>1; if(getma(now,mid)<getmi(now,mid))l=mid; else r=mid; } return r; } int getr(int now) { int l=now-1,r=n; while(r-l>1) { int mid=(l+r)>>1; if(getma(now,mid)<=getmi(now,mid))l=mid; else r=mid; } return r; } int main() { int i,j,k,t; scanf("%d",&n); rep(i,0,n-1)scanf("%d",&a[0][i]); rep(i,0,n-1)scanf("%d",&b[0][i]); init(); rep(i,0,n-1)ans+=getr(i)-getl(i); printf("%lld\n",ans); return 0; }
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原文地址：http://www.cnblogs.com/dyzll/p/5666140.html
