Codeforces 689D Friends and Subsequences(RMQ+二分)

题意：大小为n的序列a和b中，求(l,r)的个数，使得max(a[l..r]) = min(b[l...r])。

解析：在固定l后，随着r的增加，max(a[l..r]) - min(b[l...r])是不减的。所以可以通过二分求得max(a[l..r]) = min(b[l...r])的两个边界。

[code]:

#include<cstdio>#include<cstring>#include<algorithm>#include<vector>using namespace std;typedef long long LL;const int maxn = 2e5+5;int n,a[maxn],b[maxn];int mi[21][maxn],mx[21][maxn];int tlog[maxn],p[21];void init(){    int i,j;    p[0] = 1;tlog[0] = -1;    for(i = 1;i <= 20;i++) p[i] = 2*p[i-1];    for(i = 1;i <= n;i++){        tlog[i] = (i&(i-1))?tlog[i-1]:(tlog[i-1]+1);    }    for(i = 1;i <= n;i++) mi[0][i] = b[i];    for(j = 1;p[j] <= n;j++){        for(i = 1;i+p[j]-1<=n;i++)            mi[j][i] = min(mi[j-1][i],mi[j-1][i+p[j-1]]);    }    for(i = 1;i <= n;i++) mx[0][i] = a[i];    for(j = 1;p[j] <= n;j++){        for(i = 1;i+p[j]-1<=n;i++)            mx[j][i] = max(mx[j-1][i],mx[j-1][i+p[j-1]]);    }}int get_mx(int l,int r){    int k = tlog[r-l+1];    return max(mx[k][l],mx[k][r-p[k]+1]);}int get_mi(int l,int r){    int k = tlog[r-l+1];    return min(mi[k][l],mi[k][r-p[k]+1]);}int lower(int p){    int i,j,lb,rb,mid;    lb = p-1,rb = n+1;    while(rb-lb>1){        mid = (lb+rb)>>1;        if(get_mx(p,mid)>=get_mi(p,mid)) rb = mid;        else lb = mid;    }    return rb;}int upper(int p){    int i,j,lb,rb,mid;    lb = p-1,rb = n+1;    while(rb-lb>1){        mid = (lb+rb)>>1;        if(get_mx(p,mid)>get_mi(p,mid)) rb = mid;        else lb = mid;    }    return rb;}int main(){    int i,j,cas;    scanf("%d",&n);    for(i = 1;i <= n;i++){        scanf("%d",&a[i]);    }    for(i = 1;i <= n;i++){        scanf("%d",&b[i]);    }    init();    LL ans = 0;    int l,r;    for(i = 1;i <= n;i++){        l = lower(i),r = upper(i);        ans += r-l;    }    printf("%I64d\n",ans);    return 0;}