CF 1000F - One Occurrence 线段树 离线处理 区间出现一次的数

http://codeforces.com/contest/1000/problem/F

题意：

给你一个数组，有Q次询问，每次询问需要回答一个[L, R]中只出现一次的数，没有则输出0。

题解：

线段树，离线处理。

把询问按照r从小到大排序。

然后扫整个数组，对于每个i，我们把pre[a[i]]插到线段树的i中，同时把线段树中pre[a[i]]设为INF。

然后区间查询[q[i].l, q[i].r]中小于 q[i].l的数即可。

线段树里面是pair，第一维是值，第二维是下标，即可查找最小值的同时找到最小值的下标。

代码：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <queue>
#include <bitset>
#include <map>
#include <vector>
#include <stack>
#include <set>
#include <cmath>
#ifdef LOCAL
#define debug(x) cout<<#x<<" = "<<(x)<<endl;
#else
#define debug(x) 1;
#endif

#define chmax(x,y) x=max(x,y)
#define chmin(x,y) x=min(x,y)
#define lson id<<1,l,mid
#define rson id<<1|1,mid+1,r
#define lowbit(x) x&-x
#define mp make_pair
#define pb push_back
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, int> pii;

const ll MOD = 1e9 + 7;
const double eps = 1e-10;
const int INF = 0x3f3f3f3f;
const ll INFLL = 0x3f3f3f3f3f3f3f3fll;
const int MAXN = 5e5 + 5;

struct Query {
    int l, r;
    int id;
    bool operator < (const Query & x) const {
        return r < x.r;
    }
} ask[MAXN];

int a[MAXN];
pii tree[MAXN * 4];
int ans[MAXN];

void build(int id, int l, int r) {
    if(l == r) {
        tree[id].first = tree[id].second = INF;
        return ;
    }
    int mid = (l + r) >> 1;
    build(lson);
    build(rson);
    tree[id].first = tree[id].second = INF;
}


void update(int id, int l, int r, int x, int v) {
    if(l == r) {
        tree[id] = mp(v, l);
        return ;
    }
    int mid = (l + r) >> 1;
    if(x <= mid)  update(lson, x, v);
    else update(rson, x, v);
    tree[id] = min(tree[id<<1], tree[id<<1|1]);
}

pii query(int id, int l, int r, int L, int R) {
    if(L <= l && R >= r) {
        return tree[id];
    }
    int mid = (l + r) >> 1;
    pii ret = mp(INF, INF);
    if(L <= mid) ret = query(lson, L, R);
    if(R > mid) ret = min(ret, query(rson, L, R));
    return ret;
}

int pre[MAXN];

int main() {
#ifdef LOCAL
    freopen ("input.txt", "r", stdin);
#endif
    int n;
    scanf("%d", &n);
    for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
    int q;
    scanf("%d", &q);
    build(1, 1, n);
    for(int i = 1; i <= q; i++) {
        scanf("%d %d", &ask[i].l, &ask[i].r);
        ask[i].id = i;
    }
    sort(ask + 1, ask + 1 + q);
    int p = 0;
    for(int i = 1; i <= q; i++) {
        while(p < ask[i].r) {
            ++p;
            update(1, 1, n, p, pre[a[p]]);
            if(pre[a[p]]) update(1, 1, n, pre[a[p]], INF);
            pre[a[p]] = p;
        }
        pii x = query(1, 1, n, ask[i].l, ask[i].r);
        if(x.first < ask[i].l) ans[ask[i].id] = a[x.second];
    }
    for(int i = 1; i <= q; i++) printf("%d\n", ans[i]);
    return 0;
}