Codeforces Round #424 E. Cards Sorting

题意 ： 给你N张牌，操作为拿起牌顶的牌，若当前牌的数值是最小的，就扔掉它，否则将牌放到牌堆底，问扔完所有牌需要多少次操作。

这个题肯定是从小到大删除，可以用一个 vector 保存每一个数的下标然后给每一个数的下标从小到大排个序（这样可以用O(1)时间内找到要删掉下一个的位置 ），用线段树或者树状数组维护每个区间内有多少个数没有背删去 （单点修改，区间查询）。

#include <iostream>#include <algorithm>#include <cstring>#include <cstdio>#include <vector>#define ll long longusing namespacestd;const int maxn =1e5 + 10;struct node {    int l,r;    int sum;}tr[maxn <<2];struct point {    int pos;    int cnt;};ll ans = 0;vector <int> a[maxn];void build (int l,int r,int root) {    tr[root].l = l;    tr[root].r = r;    if (l == r) {        tr[root].sum =1;        return;    }    int mid = (l + r) >>1;    build(l, mid, root <<1);    build(mid +1, r, root << 1 |1);    tr[root].sum =tr[root << 1].sum +tr[root << 1 |1].sum;//    return ;}void update (int pos,int root) {    if (tr[root].l ==tr[root].r) {        tr[root].sum =0;        return;    }    int mid = (tr[root].l +tr[root].r) >>1;    if (pos > mid) update(pos, root << 1 | 1);    else update(pos, root << 1);    tr[root].sum =tr[root << 1].sum +tr[root << 1 |1].sum;}int query (int l,int r,int root) {    if (l > r)return 0;    if (tr[root].l >= l &&tr[root].r <= r) {        returntr[root].sum;    }    int mid = (tr[root].l +tr[root].r) >>1;    if (l > mid)return query(l, r, root <<1 | 1);    else if (r <= mid) returnquery(l, r, root << 1);    else return query(l, r, root <<1) + query(l, r, root <<1 | 1);}int main () {    ios_base ::sync_with_stdio(false);    int n;    cin >> n;    build(1, n,1);    int mi =maxn;    for (int i =1;i <= n; ++ i) {        int x;        cin >> x;        mi = min (mi,x);        a[x].push_back (i);    }    for (int i = mi;i <maxn; ++ i) {        if (a[i].size()){//            cout << i << ' ' << a[i][0] << endl;            sort (a[i].begin(),a[i].end());        }    }    int pos =0;    for (int i = mi;i <maxn; ++ i) {        if (a[i].size ()) {            int u =a[i].size();            int cnt =0;            int j =lower_bound (a[i].begin(),a[i].end(),pos) - a[i].begin();//            cout << j << ' ' << a[i][j] << ' ';            if (j != u){            if (pos !=a[i][j])                ans +=query(pos + 1,a[i][j],1);            }            else {                j = 0;                ans +=query(pos + 1, n,1);                ans +=query(1,a[i][j],1);            }//            cout << ans << endl;            update(a[i][j],1);            pos = a[i][j];            for (int k =0;k < u - 1;++k) {                if (j != u -1) {                    j ++;                    ans +=query(pos + 1,a[i][j],1);                    update(a[i][j],1);                    pos = a[i][j];                }                else {                    j = 0;                    ans +=query(1,a[i][j],1);                    ans +=query(pos + 1, n,1);                    pos = a[i][j];                    update(a[i][j],1);                }            }        }    }    cout <<ans << endl;    return0;}