P1972 [SDOI2009] HH的项链

P1972 [SDOI2009] HH的项链

树状数组题解

将全部输入排序，进行离散化

#include<bits/stdc++.h>
using namespace std;

#define rint register int
const int N = 1e6 + 7;

int a[N], tree[N], n, ans[N], st[N];
struct Q{
	int l, r, id;
}q[N];

bool cmp(Q x, Q y){
	return x.r < y.r;
}

int lowbit(int x) { return x & (- x); }

void modify(int x, int k){
	for(rint i = x; i <= n;i += lowbit(i))
		tree[i] += k;
}

int query(int x){
	int ans = 0;
	for(rint i = x; i ;i -= lowbit(i)) ans += tree[i];
	return ans;
}

int main(){
	int m;
	scanf("%d", &n);
	for(rint i = 1;i <= n;i ++) scanf("%d", &a[i]);
	scanf("%d", &m);
	for(rint i = 1;i <= m;i ++){
		scanf("%d%d", &q[i].l, &q[i].r);
		q[i].id = i;
	}
	int p = 1;
	sort(q + 1, q + m + 1, cmp);
	for(rint i = 1;i <= m;i ++){
		for(rint j = p;j <= q[i].r;j ++){
			if(st[a[j]]) modify(st[a[j]], -1);
			modify(j, 1);
			st[a[j]] = j;
		}
		p = q[i].r + 1;
		ans[q[i].id] = query(q[i].r) - query(q[i].l - 1);
	}
	for(rint i = 1;i <= m;i ++){
		printf("%d\n", ans[i]);
	}
	return 0;
}

主席树题解（有两个样例卡时间过掉的）

这题的主席树代码出奇的短！！！

#include<bits/stdc++.h>
using namespace std;

const int N = 1e6 + 7, M = N * 40;

int a[N], cnt, la[N], rt[N];

struct Tree{
	int size, l, r;
}tr[M];

int change(int pre, int x, int v, int l, int r){
	int mid = (l + r) >> 1;
	int p = ++ cnt;
	tr[p] = tr[pre];
	tr[p].size += v;
	if(l == r) return p;
	if(mid >= x)tr[p].l = change(tr[pre].l, x, v, l, mid);
	else tr[p].r = change(tr[pre].r, x, v, mid + 1, r);
	return p;
}

int query(int p, int k, int l, int r){
	if(l == r) return tr[p].size;
	int mid = (l + r) >> 1;
	if(k <= mid) return query(tr[p].l, k, l, mid) + tr[tr[p].r].size;
	else return query(tr[p].r, k, mid + 1, r);
}

int main(){
	int n, m;
	scanf("%d", &n);
	for(int i = 1;i <= n;i ++){
		scanf("%d", &a[i]);
		if(la[a[i]]){
			int t = change(rt[i - 1], la[a[i]], -1, 1, n);
			rt[i] = change(t, i, 1, 1, n);
		}
		else
			rt[i] = change(rt[i - 1], i, 1, 1, n);
		la[a[i]] = i;
	}
	scanf("%d", &m);
	while(m --){
		int x, y;
		scanf("%d%d", &x, &y);
		printf("%d\n", query(rt[y], x, 1, n));
	}
	return 0;
}