HDU 4747 Mex

4747

思路：

线段树

先求出mex(1,1)， mex(1, 2) ， mex(1,3)，...，mex(1，n)（单调上升），先将这些mex放进线段树里求和

然后再求出next[i]表示下一次出现a[i] 的位置

然后从前往后不停的删数，对于一个数a[i]，我们删掉他的影响是：l为mex大于a[i]的位置，r 为next[i]，l 到 r-1 之间的 mex都变为 a[i]

然后这个线段树只需要维护区间最大值（方便查找第一个大于a[i]的位置）和区间和就可以了

代码：

#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
//#define mp make_pair
#define pb push_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pii pair<int, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);
//head

const int N = 2e5 + 5;
int a[N], nxt[N], mx[N<<2], lazy[N<<2], mex[N];
LL sum[N<<2];
map<int, int>mp;
void push_up(int rt) {
    sum[rt] = sum[rt<<1] + sum[rt<<1|1];
    mx[rt] = max(mx[rt<<1], mx[rt<<1|1]);
}
void push_down(int rt, int len) {
    sum[rt<<1] = 1LL * lazy[rt] * (len - (len >> 1));
    mx[rt<<1] = lazy[rt];
    lazy[rt<<1] = lazy[rt];
    sum[rt<<1|1] = 1LL * lazy[rt] * (len >> 1);
    mx[rt<<1|1] = lazy[rt];
    lazy[rt<<1|1] = lazy[rt];
    lazy[rt] = 0;
}
void build(int rt, int l, int r) {
    if(l == r) {
        mx[rt] = sum[rt] = mex[l];
        return ;
    }
    int m = (l+r) >> 1;
    build(ls);
    build(rs);
    push_up(rt);
}
void update(int x, int L, int R, int rt, int l, int r) {
    if(L <= l && r <= R) {
        mx[rt] = x;
        sum[rt] = 1LL * (r-l+1) * x;
        lazy[rt] = x;
        return ;
    }
    if(lazy[rt]) push_down(rt, r-l+1);
    int m = (l+r) >> 1;
    if(L <= m) update(x, L, R, ls);
    if(R > m) update(x, L, R, rs);
    push_up(rt);
}
LL query(int L, int R, int rt, int l, int r) {
    if(L <= l && r <= R) return sum[rt];
    if(lazy[rt]) push_down(rt, r-l+1);
    int m = (l+r) >> 1;
    LL ans = 0;
    if(L <= m) ans += query(L, R, ls);
    if(R > m) ans += query(L, R, rs);
    push_up(rt);
    return ans;
}
int Find(int x, int rt, int l, int r) {
    if(l == r) return l;
    int m = (l+r) >> 1;
    if(lazy[rt]) push_down(rt, r-l+1);
    if(mx[rt<<1] > x) return Find(x, ls);
    else return Find(x, rs);
}
int main() {
    int n;
    while(~scanf("%d", &n) && n) {
        mem(lazy, 0);
        build(1, 1, n);
        for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
        mp.clear();
        int tmp = 0;
        for (int i = 1; i <= n; i++) {
            mp[a[i]]++;
            while(mp.find(tmp) != mp.end()) tmp++;
            mex[i] = tmp;
        }
        build(1, 1, n);
        mp.clear();
        for (int i = n; i >= 1; i--) {
            if(mp.find(a[i]) == mp.end()) nxt[i] = n+1;
            else nxt[i] = mp[a[i]];
            mp[a[i]] = i;
        }
        LL ans = 0;
        for (int i = 1; i <= n; i++) {
            ans += query(1, n, 1, 1, n);
            if(mx[1] <= a[i]) continue;
            int l = Find(a[i], 1, 1, n);
            int r = nxt[i];
            if(l < r) update(a[i], l, r-1, 1, 1, n);
        }
        printf("%lld\n", ans);
    }
    return 0;
}