hdu 4747 (线段树)

题意：有一个序列a[]，mex(L, R)表示区间a在区间[L, R]上第一个没出现的最小非负整数，对于序列a[]，求所有的mex(L, R)的和（1 <= L <= R <= n，1 <= n <= 200000，0 <= ai <= 10^9）。

求出所有的mex(1, i)；接着删去第1个结点，就是所有的mex(2, i)；接着再删去第1个结点，就是所有的mex(3, i)；……最后就是mex(n, n)，求和即是答案。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>

using namespace std;
#define ls rt<<1
#define rs rt<<1|1
#define lson l, m, rt<<1
#define rson m + 1, r, rt<<1|1
typedef long long ll;
const int maxn = 2e5 + 5;
int n, a[maxn], vis[maxn], next[maxn], mex1[maxn << 2];
struct SegTree{
    int Max, lazy;
    ll sum;
}seg[maxn << 2];

void pushUp(int rt){
    seg[rt].sum = seg[ls].sum + seg[rs].sum;
    seg[rt].Max = max(seg[ls].Max, seg[rs].Max);
}
void pushDown(int rt, int len){
    if (seg[rt].lazy != -1){
        seg[ls].lazy = seg[rs].lazy = seg[rt].lazy;
        seg[ls].Max = seg[rt].lazy;
        seg[rs].Max = seg[rt].lazy;
        seg[ls].sum = (ll)((len + 1) / 2) * ((ll)seg[rt].lazy);
        seg[rs].sum = (ll)(len / 2) * ((ll)seg[rt].lazy);
        seg[rt].lazy = -1;
    }
}
void build(int l, int r, int rt){
    seg[rt].lazy = -1;
    if (l == r){
        seg[rt].sum = seg[rt].Max = mex1[l];
        return ;
    }
    int m = (l + r) >> 1;
    build(lson);
    build(rson);
    pushUp(rt);
}
int find(int key, int l, int r, int rt){//找到第一个mex大于a[i]的下标
    if (l == r) return l;
    pushDown(rt, r - l + 1);
    int m = (l + r) >> 1;
    if (seg[ls].Max > key) return find(key, lson);
    else return find(key, rson);
}
void update(int val, int L, int R, int l, int r, int rt){
    if (L <= l && r <= R){
        //pushDown(rt, r - l + 1);
        seg[rt].Max = val;
        seg[rt].sum = (ll) val * (ll) (r - l + 1);
        seg[rt].lazy = val;
        return ;
    }
    int m = (l + r) >> 1;
    //cout << seg[rt].lazy << " l = " << l << " r= " << r << endl;
    pushDown(rt, r - l + 1);
    if (L <= m ) update(val, L, R, lson);
    if (R > m) update(val, L, R, rson);
    pushUp(rt);
    /*if (L == 2 && R == 4){
        cout << " ll = " << l << " rr = " << r << endl;
        cout << " sum = " << seg[rt].sum << " rt = " << rt << endl;
    }*/
}
int main(){
    while (~scanf("%d", &n) && n){
        for (int i = 1; i <= n; ++i){
            scanf("%d", &a[i]);
            if (a[i] > n) a[i] = n;
        }
        //得到mex[1, i]
        int tmp = 0;
        memset(vis, 0, sizeof(vis));
        for (int i = 1; i <= n; ++i){
            vis[a[i]] = 1;
            while(vis[tmp]) tmp++;
            mex1[i] = tmp;
        }
        //得到next值
        for (int i = 0; i <= n; ++i) vis[i] = n + 1;
        for (int i = n; i >= 1; --i){
            next[i] = vis[a[i]];
            vis[a[i]] = i;
        }
        build(1, n, 1);
        ll ans = 0;
        for (int i = 1; i <= n; ++i){
        //    cout << seg[1].sum << endl;
            ans += seg[1].sum;
            update(0, i, i, 1, n, 1);
            if (a[i] < seg[1].Max){
                int l = find(a[i], 1, n, 1), r = next[i] - 1;
                //cout << " l = " << l << " r = " << r << endl;
                if (l <= r) update(a[i], l, r, 1, n, 1);
            }
        }
        printf("%I64d
", ans);
    }
    return 0;
}