George and Cards

我們找到每個要被刪的數字左邊和右邊第一個比它小的沒被刪的數字的位置。然後從小到大枚舉要被刪的數， 求答案。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define
 
 ull unsigned long long
using namespace std;

const int N = 1e6 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;

int n, k, top, p[N], b[N];
int L[N], R[N];
bool ban[N];
PII stk[N];
LL ans;
vector<int> vc;

struct
 
 Bit {
    int a[N];
    inline void modify(int x, int v) {
        for(int i = x; i < N; i += i & -i)
            a[i] += v;
    }
    inline int sum(int x) {
        int ans = 0;
        for(int i = x; i; i -= i & -i)
            ans += a[i];
        return ans;
    }
    inline 
 
int query(int L, int R) {
        if(L > R) return 0;
        return sum(R) - sum(L - 1);
    }
} bit;

bool cmp(const int& a, const int& b) {
    return p[a] < p[b];
}

int main() {
    scanf("%d%d", &n, &k);
    for(int i = 1; i <= n; i++) bit.modify(i, 1);
    for(int i = 1; i <= n; i++) scanf("%d", &p[i]);
    for(int i = 1; i <= k; i++) scanf("%d", &b[i]), ban[b[i]] = true;
    for(int i = 1; i <= n; i++) {
        if(ban[p[i]]) {
            while(top && stk[top].fi > p[i]) top--;
            stk[++top] = mk(p[i], i);
        } else {
            int pos = lower_bound(stk + 1, stk + top + 1, mk(p[i], 0)) - stk - 1;
            L[i] = pos ? stk[pos].se : 0;
        }
    }
    top = 0;
    for(int i = n; i >= 1; i--) {
        if(ban[p[i]]) {
            while(top && stk[top].fi > p[i]) top--;
            stk[++top] = mk(p[i], i);
        } else {
            int pos = lower_bound(stk + 1, stk + top + 1, mk(p[i], 0)) - stk - 1;
            R[i] = pos ? stk[pos].se : n + 1;
        }
    }
    for(int i = 1; i <= n; i++)
        if(!ban[p[i]]) vc.push_back(i);
    sort(vc.begin(), vc.end(), cmp);
    for(auto& x : vc) {
        ans += bit.query(L[x] + 1, R[x] - 1);
        bit.modify(x, -1);
    }
    printf("%lld\n", ans);
    return 0;
}

/*
*/

Codeforces 387E George and Cards