牛客网多校赛第一场J题Different Integer

链接：https://www.nowcoder.com/acm/contest/139/J
来源：牛客网

Given a sequence of integers a1, a2, …, an and q pairs of integers (l1, r1), (l2, r2), …, (lq, rq), find count(l1, r1), count(l2, r2), …, count(lq, rq) where count(i, j) is the number of different integers among a1, a2, …, ai, aj, aj + 1, …, an.
输入描述:
The input consists of several test cases and is terminated by end-of-file.
The first line of each test cases contains two integers n and q.
The second line contains n integers a1, a2, …, an.
The i-th of the following q lines contains two integers li and ri.
输出描述:
For each test case, print q integers which denote the result.
示例1
输入
复制
3 2
1 2 1
1 2
1 3
4 1
1 2 3 4
1 3
输出
复制
2
1
3
备注:
* 1 ≤ n, q ≤ 105
* 1 ≤ ai ≤ n
* 1 ≤ li, ri ≤ n
* The number of test cases does not exceed 10.

算法：莫队算法

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MOD = 1e9+7;
const int maxn = 1e6 + 10;
const int inf = 0x3f3f3f3f;
#define met(a, b) memset(a, b, sizeof(a))
struct MATION {
    int l, r, id;
} q[2 * maxn];
ll cnt = 0;
int pos[maxn], block, val[maxn], ANS[maxn], a[maxn];
bool vis[maxn];
int d, n, m, top = 0;
int cmp(MATION a, MATION b) {
    if(pos[a.l] != pos[b.l]) return pos[a.l] < pos[b.l];
    else return a.r < b.r;
}
void solve() {
    met(val, 0);
    int ans = 0;
    int l = q[0].l, r = q[0].l - 1;
    for(int i = 0; i < top; i++) {
        while(l < q[i].l) {
            val[a[l]]--;
            if(val[a[l]] == 0) ans--;
            l++;
        }
        while(l > q[i].l) {
            l--;
            val[a[l]]++;
            if(val[a[l]] == 1) ans++;
        }
        while(r < q[i].r) {
            r++;
            val[a[r]]++;
            if(val[a[r]] == 1) ans++;
        }
        while(r > q[i].r) {
            val[a[r]]--;
            if(val[a[r]] == 0) ans--;
            r--;
        }
        ANS[q[i].id] = ans;
    }
}
int main() {
    while(~scanf("%d%d", &n, &m)) {
        met(vis, false);
        cnt = 0;
        block = sqrt(n);
        for(int i = 1; i <= n; i++) {
            scanf("%d", &a[i]);
            pos[i] = (i - 1) / block + 1;
            pos[i + n] = (i + n - 1) / block + 1;
            a[i + n] = a[i];
            if(!vis[a[i]]) {
                vis[a[i]] = true;
                cnt++;
            }
        }
        int l, r;
        top = 0;
        for(int i = 0; i < m; i++) {
            scanf("%d%d", &l, &r);
            if(r <= l) {
                ANS[i] = cnt;
                continue;
            }
            q[top].l = r;
            q[top].r = n + l;
            q[top].id = i;
            top++;
        }
        sort(q, q + top, cmp);

        solve();
        for(int i = 0; i < m; i++) {
            //cout << ANS[i * 2] << " * " << ANS[i * 2 + 1] << endl;
            printf("%d\n",ANS[i]);
        }
    }
    return 0;
}