省队集训 Day4 a

【题目大意】

求有多少区间只包含1个出现次数为1的数。

$1leq n leq 5*10^5, 0 leq a_i leq 10^9$

【题解】

考虑枚举右端点，设这个数上一次出现位置为pre[i]，那么就是$[pre[i]+1,i]$区间加1，$[pre[pre[i]]+1, pre[i]]$区间减1，和统计区间中1的个数。

注意到数不会减到负的，那么1只可能是最小值或次小值，那么线段树直接维护即可。

复杂度$O(nlogn)$。

考场交错代码导致爆0...QAQ

# include <vector>
# include <stdio.h>
# include <string.h>
# include <iostream>
# include <algorithm>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;

const int M = 5e5 + 10, N = 1e5 + 10, inf = 1e9;

inline int getint() {
    int x = 0; char ch = getchar();
    while(!isdigit(ch)) ch = getchar();
    while(isdigit(ch)) x = (x<<3) + (x<<1) + ch - '0', ch = getchar();
    return x;
}

int n, m, a[M], pre[M], lst[M];
vector<int> ps;

struct pa {
    int x, tx, y, ty;
    pa() {}
    pa(int x, int tx, int y, int ty) : x(x), tx(tx), y(y), ty(ty) {}
    inline friend pa operator + (pa a, pa b) {
        pa c;
        if(a.x < b.x) {
            c.x = a.x, c.tx = a.tx;
            c.y = min(a.y, b.x); c.ty = 0;
            if(a.y == c.y) c.ty += a.ty;
            if(b.x == c.y) c.ty += b.tx;
        } else if(a.x > b.x) {
            c.x = b.x, c.tx = b.tx;
            c.y = min(a.x, b.y); c.ty = 0;
            if(a.x == c.y) c.ty += a.tx;
            if(b.y == c.y) c.ty += b.ty;
        } else {
            c.x = a.x, c.tx = a.tx + b.tx;
            c.y = min(a.y, b.y); c.ty = 0;
            if(a.y == c.y) c.ty += a.ty;
            if(c.y == b.y) c.ty += b.ty;
        }
        return c;
    }
};

struct SMT {
    pa w[M << 2]; int tag[M << 2];    
    # define ls (x<<1)
    # define rs (x<<1|1) 
    inline void up(int x) {
        w[x] = w[ls] + w[rs];
    }
    
    inline void pushtag(int x, int t) {
        w[x].x += t;
        if(w[x].y != inf) w[x].y += t;
        tag[x] += t;
    }
    
    inline void down(int x) {
        if(!tag[x]) return ;
        pushtag(ls, tag[x]);
        pushtag(rs, tag[x]);
        tag[x] = 0;
    }
    
    inline void build(int x, int l, int r) {
        if(l == r) {
            w[x].x = 0, w[x].tx = 1;
            w[x].y = inf, w[x].ty = 0;
            return ;
        }
        int mid = l+r >> 1;
        build(ls, l, mid);
        build(rs, mid+1, r);
        up(x);
    }    
    
    inline void edt(int x, int l, int r, int L, int R, int d) {
        if(L <= l && r <= R) {
            pushtag(x, d);
            return ;
        }
        down(x);
        int mid = l+r>>1;
        if(L <= mid) edt(ls, l, mid, L, R, d);
        if(R > mid) edt(rs, mid+1, r, L, R, d);
        up(x);
    }
    
    inline pa sum(int x, int l, int r, int L, int R) {
        if(L <= l && r <= R) return w[x];
        down(x);
        int mid = l+r>>1;
        if(R <= mid) return sum(ls, l, mid, L, R);
        else if(L > mid) return sum(rs, mid+1, r, L, R);
        else return sum(ls, l, mid, L, mid) + sum(rs, mid+1, r, mid+1, R);
    }    
    
    inline void debug(int x, int l, int r) {
        printf("x = %d,l = %d,r = %d,mi = %d,semi = %d,minum = %d,seminum = %d,tag = %d
", x, l, r, w[x].x, w[x].y, w[x].tx, w[x].ty, tag[x]);
        if(l == r) return ;
        int mid = l+r>>1;
        debug(ls, l, mid); debug(rs, mid+1, r);
    }
}T;
     

int main() {
    freopen("a1.in", "r", stdin);
//    freopen("a.out", "w", stdout);
    n = getint();
    for (int i=1; i<=n; ++i) {
        a[i] = getint();
        ps.push_back(a[i]);
    }
    sort(ps.begin(), ps.end());
    ps.erase(unique(ps.begin(), ps.end()), ps.end()); 
    m = ps.size();
    for (int i=1; i<=n; ++i) a[i] = lower_bound(ps.begin(), ps.end(), a[i]) - ps.begin() + 1;
    for (int i=1; i<=n; ++i) pre[i] = lst[a[i]], lst[a[i]] = i;
    ll ans = 0; pa t; T.build(1, 1, n);
    for (int i=1; i<=n; ++i) {
//        T.debug(1, 1, n); cout << endl;
        T.edt(1, 1, n, pre[i] + 1, i, 1);
        if(pre[i]) T.edt(1, 1, n, pre[pre[i]] + 1, pre[i], -1);
        t = T.sum(1, 1, n, 1, i);
        if(t.x == 1) ans += t.tx;
        if(t.y == 1) ans += t.ty;
        // 区间加1，区间减1，查询区间内1的出现次数
    }
    cout << ans;    
    return 0;
}