Codeforces Round #424 Div2 E. Cards Sorting

我只能说真的看不懂题解的做法
我的做法就是线段树维护，毕竟每个数的顺序不变嘛
那么单点维护 区间剩余卡片和最小值

每次知道最小值之后，怎么知道需要修改的位置呢
直接从每种数维护的set找到现在需要修改的数的在初始卡片的位置

#include <cstdio>#include <cstdlib>#include <cstring>#include <ctime>#include <algorithm>#include <iostream>#include <map>#include <set>#include <queue>#include <cmath>using namespace std;typedef long long ll;#define MP(x, y) make_pair(x, y)#define lson l,m, rt<<1#define rson m+1, r, rt<<1|1const int N = 1e5+5;const int INF = 0x3f3f3f3f;int MOD;int A[N];int Min[N << 2];int Sum[N << 2];void Build(int l, int r, int rt) {    if(l == r) {        Min[rt] = A[l]; Sum[rt] = 1;        return;    }    int m = (l + r) >> 1;    Build(lson); Build(rson);    Sum[rt] = Sum[rt << 1] + Sum[rt << 1|1];    Min[rt] = min(Min[rt <<1] , Min[rt << 1|1]);}void Update(int pos, int l, int r, int rt) {    if(l == r) {        Min[rt] = INF; Sum[rt] = 0;        return;    }    int m = (l + r) >> 1;    if(pos <= m) Update(pos, lson);    else Update(pos, rson);    Sum[rt] = Sum[rt<<1] + Sum[rt<<1|1];    Min[rt] = min(Min[rt<<1], Min[rt<<1|1]);}int Total(int pos, int l, int r, int rt) {    if(pos == 0) return 0;    if(pos == r) {        return Sum[rt];    }    int m = (l + r) >> 1;    if(pos <= m) return Total(pos, lson);    else return Sum[rt<<1] + Total(pos, rson);}map<int, set<int> > mp;set<int> ::iterator it;int main() {    int n;    while(~scanf("%d", &n)) {        ll ans = 0;        mp.clear();        for(int i = 1; i <= n; ++i) {            scanf("%d",&A[i]);            mp[A[i]].insert(i);        }        Build(1, n, 1);        int pos = 1;        for(int i = 1; i <= n; ++i) {            int tar = Min[1]; int nwpos;            it = mp[tar].lower_bound(pos);            if(it == mp[tar].end()) {                it = mp[tar].begin();            }                       nwpos = *it;            mp[tar].erase(it);        //  printf("%d %d\n", tar, nwpos);            Update(nwpos, 1, n, 1);            int t1 = Total(nwpos, 1,n,1); int t2 = Total(pos - 1, 1,n,1);        //  printf("%d %d\n", t1, t2);            if(pos <= nwpos) {                ans += t1 - t2 + 1;            }else {                ans += Sum[1] - t2 + t1 + 1;            }            pos = nwpos;        }        printf("%d\n", ans);    }    return 0;}