Codeforces 689D Friends and Subsequences(RMQ+二分)

題意：大小爲n的序列a和b中，求(l,r)的個數，使得max(a[l..r]) = min(b[l...r])。

解析：在固定l後，隨着r的增加，max(a[l..r]) - min(b[l...r])是不減的。所以可以通過二分求得max(a[l..r]) = min(b[l...r])的兩個邊界。

[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;
}