Codeforces 387E 树状数组

http://codeforces.com/problemset/problem/387/E

记录每个数的下标，对数从小到大考虑。如果该数需要保留，那么就往set中插入该数的下标。否则可以查询出当前数左边和右边第一个比它小的数的下标，可以统计这个区间中已经删掉的数的数量，就可以计算出当前数的贡献。

#include
using namespace std;
const int MAX_LEN = 1e6+10;
typedef long long ll;

int n, m;
ll tree[MAX_LEN];
int id[MAX_LEN], keep[MAX_LEN];

void add(int pos, int value){
    while(pos <= n){
        tree[pos] += value;
        pos += (pos&-pos);
    }
    return ;
}

ll query(int l, int r){
    int ans = 0;
    while(r){
        ans += tree[r];
        r -= (r&-r);
    }
    while(l){
        ans -= tree[l];
        l -= (l&-l);
    }
    return ans;
}

set se;
set reverseSe;

int main(){
    std::ios::sync_with_stdio(false);
    memset(keep, false, sizeof(keep));
    scanf("%d%d",&n, &m);
    int tmp;
    for (int i = 1; i <= n; i++)
        scanf("%d", &tmp), id[tmp] = i;
    for (int i = 1; i <= m; i++)
        scanf("%d", &tmp), keep[tmp] = true;
    se.insert(0); se.insert(n+1);
    reverseSe.insert(0); reverseSe.insert(-n-1);

    if (n == m){
        cout << 0 << endl;
        return 0;
    }

    ll ans = 0;
    for (int i = 1; i <= n; i++){
        if (keep[i])
            se.insert(id[i]), reverseSe.insert(-id[i]);
        else{
            int l = -*reverseSe.upper_bound(-id[i]);
            int r = *se.upper_bound(id[i]);
            //cout << "l: " << l << " r: " << r << endl;
            ans += r-l-1;
            ans -= query(l, r-1);
            add(id[i], 1);
        }
    }

    printf("%I64\n", ans);
    return 0;
}