莫队算法 sqrt(n)分块思想

在此说一下本渣对莫队算法思想的一些浅薄理解

莫队算法的思想就是对真个区间的分块,然后按照每块来分别进行计算,这样最终的复杂度可以达到n*sqrt(n)

小Z的袜子是一道非常经典的题目.:题目链接http://acm.hust.edu.cn/vjudge/contest/view.action?cid=29469#problem/A

我们先对整个区间分块,然后按照左区间所在的块进行排序,如果左区间在相同的块内,则按照右区间进行排序.

如此相当于对每个左区间在相同块内的询问,我们只要右区间从最小迭代到最大即可.也就是说我们把算法的复杂度的锅让区间的左端点来背 = =,恩......

当然了,既然使用如上算法,那么显然我们要进行离线处理.

具体代码如下:

#include <cstdio>#include <iostream>#include <algorithm>#include <queue>#include <cstring>#include <cmath>#include <set>using namespace std;int n, m, a[100010], num[100010], unit;struct N {//存询问的信息int l, r, id;bool operator < (const N &rhs) const {//先按照左端点是否所在块进行排序,然后按照右端点排序if (l / unit == rhs.l / unit) return r < rhs.r;else return l / unit < rhs.l / unit;}} q[100010];pair<long long, long long>ans[100010];//储存答案long long gcd(long long x, long long y) {if (y == 0) return x;return gcd(y, x % y);}int main() {//freopen("in.in", "r", stdin);//freopen("out.out", "w", stdout);scanf("%d %d", &n, &m);for (int i = 1; i <= n; i++)scanf("%d", &a[i]);for (int i = 0; i < m; i++) {scanf("%d %d", &q[i].l, &q[i].r);q[i].id = i;}unit = sqrt(n);sort(q, q + m);int l = 1,r = 0;long long tmp = 0;memset(num,0,sizeof(num));for(int i=0;i<m;i++){//迭代求答案过程while(r < q[i].r){r++;tmp -= (long long)num[a[r]] * (num[a[r]] - 1);num[a[r]]++;tmp += (long long)num[a[r]] * (num[a[r]] - 1);}while(r > q[i].r){tmp -= (long long)num[a[r]] * (num[a[r]] - 1);num[a[r]]--;tmp += (long long)num[a[r]] * (num[a[r]] - 1);r--;}while(l < q[i].l){tmp -= (long long)num[a[l]] * (num[a[l]] - 1);num[a[l]]--;tmp += (long long)num[a[l]] * (num[a[l]] - 1);l++;}while(l > q[i].l){l--;tmp -= (long long)num[a[l]] * (num[a[l]] - 1);num[a[l]]++;tmp += (long long)num[a[l]] * (num[a[l]] - 1);}ans[q[i].id].first  = tmp;ans[q[i].id].second = (long long)(q[i].r - q[i].l + 1) * (q[i].r - q[i].l);}for(int i=0;i<m;i++){if(ans[i].first + ans[i].second == 0) { printf("0/1\n"); continue; }long long tmp = gcd(ans[i].second,ans[i].first);printf("%lld/%lld\n",ans[i].first/tmp,ans[i].second/tmp);}return 0;}