CodeForces 689 D Friends and Subsequences

Friends and Subsequences

题解：

如果左端点来说， 那么对于a[i]来说是向上的一条折线， b[i]来说是向下的一条折线， 那么如果这2个折线求交点个数的话， 我们可以二分去求第一个 a[i] == b[i] 的地方， 求最后一个a[i] == b[i]的地方。

代码：

#include<bits/stdc++.h>
using namespace std;
#define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
#define LL long long
#define ULL unsigned LL
#define fi first
#define se second
#define pb push_back
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lch(x) tr[x].son[0]
#define rch(x) tr[x].son[1]
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
typedef pair<int,int> pll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const LL _INF = 0xc0c0c0c0c0c0c0c0;
const LL mod =  (int)1e9+;
const int N = 2e5 + ;
int a[N], b[N];
int Log[N];
int Max[N][];
int Min[N][];
void init(int n) {
    for(int i = -(Log[]=-); i < N; i++)
        Log[i] = Log[i - ] + ((i & (i - )) == );
    for(int i = ; i <= n; ++i) Max[i][] = a[i], Min[i][] = b[i];
    for(int j = ; j <= Log[n]; j++)
        for(int i = ; i+(<<j)- <= n; i++)
            Max[i][j] = max(Max[i][j-], Max[i+(<<(j-))][j-]),
            Min[i][j] = min(Min[i][j-], Min[i+(<<(j-))][j-]);
}
int QMin(int l, int r) {
    int k = Log[r - l + ];
    return min(Min[l][k], Min[r-(<<k)+][k]);
}
int QMax(int l, int r){
    int k = Log[r - l + ];
    return max(Max[l][k], Max[r-(<<k)+][k]);
}
int main(){
    int n;
    scanf("%d", &n);
    for(int i = ; i <= n; ++i) scanf("%d", &a[i]);
    for(int i = ; i <= n; ++i) scanf("%d", &b[i]);
    init(n);
    LL ans = ;
    for(int i = ; i <= n; ++i){
        int l = i, r = n;
        while(l <= r){
            int m = l+r >> ;
            if(QMin(i, m) > QMax(i, m)) l = m + ;
            else r = m - ;
        }
        if(QMin(i, l) != QMax(i, l)) continue;
        int tl = l;
        l = i, r = n;
        while(l <= r){
            int m = l+r >> ;
            if(QMin(i, m) >= QMax(i, m)) l = m + ;
            else  r = m - ;
        }
        ans += l - tl;
    }
    cout << ans << endl;
    return ;
}