CF833B The Bakery

此题可用决策单调性来做

证明 略

引入

决策单调性((My blog))

思路

于是此题便可先用分治优化决策单调性

时间 (O(n^2*k)) (101 tests) 中 只能过前(9)个 (9pts) ?

#include <map>
#include <string>
#include <cstdio>
#include <string>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define reg register int
#define isdigit(x) ('0' <= x&&x <= '9')
template<typename T>
inline T Read(T Type)
{
	T x = 0,f = 1;
	char a = getchar();
	while(!isdigit(a)) {if(a == '-') f = -1;a = getchar();}
	while(isdigit(a)) {x = (x << 1) + (x << 3) + (a ^ '0');a = getchar();}
	return x * f;
}
const int MAXN = 35010,inf = MAXN;
int a[MAXN],dp[MAXN][55];
bool vis[MAXN];
inline int cost(int l,int r)
{
	int res = 0;
	for(reg i = l;i <= r;i++)
	{
		if(!vis[a[i]]) res++;
		vis[a[i]] = 1;
	}
	for(reg i = 1;i <= r;i++) vis[a[i]] = 0;	
	return res;
}
inline void dfs(int k,int l,int r,int opl,int opr)
{
	if(l > r) return;
	int mid = l + r >> 1,maxl = -inf,id;
	for(reg i = opl;i <= min(opr,mid);i++)
	{
		int cur = dp[i - 1][k - 1] + cost(i,mid);
		if(cur > maxl) maxl = cur,id = i;
	}
	dp[mid][k] = maxl;
	dfs(k,l,mid - 1,opl,id);
	dfs(k,mid + 1,r,id,opr);
}
int main()
{
	int n = Read(1),k = Read(1);
	for(reg i = 1;i <= n;i++)
		a[i] = Read(1);
	for(reg i = 1;i <= k;i++) dfs(i,1,n,1,n);
	printf("%d
",dp[n][k]);
	return 0;
}

考虑优化

观察代码 发现(cost)(查询区间不同数个数)就有一层(n)

考虑降成(O(1)) 但预处理是(O(n^2))

于是只能降成(O(log_2^n))

从我们学习的数据结构中寻找

发现主席树恰好可以胜任

用主席树(A)掉这道题

SP3267 DQUERY - D-query

再粘过来就行了

然而我这道题写的莫队 现在又写了一遍主席树 233

(Code)

时间复杂度 (O(n*log_2^n*k))

#include <map>
#include <string>
#include <cstdio>
#include <string>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define reg register int
#define isdigit(x) ('0' <= x&&x <= '9')
template<typename T>
inline T Read(T Type)
{
	T x = 0,f = 1;
	char a = getchar();
	while(!isdigit(a)) {if(a == '-') f = -1;a = getchar();}
	while(isdigit(a)) {x = (x << 1) + (x << 3) + (a ^ '0');a = getchar();}
	return x * f;
}
const int MAXN = 35010,inf = MAXN;
int a[MAXN],dp[MAXN][55],n,k;
namespace segment_tree
{
	struct node
	{
		int l,r,val;
	}tree[MAXN * 40];
	map<int,int> vis;
	int cnt,root[MAXN];
	inline int update(int k,int l,int r,int pos,int v)
	{
	    tree[++cnt] = tree[k],tree[cnt].val += v;
	    k = cnt;
	    if(l == r) return k;
	    int mid = l + r >> 1;
	    if(pos <= mid) tree[k].l = update(tree[k].l,l,mid,pos,v);
	    else tree[k].r = update(tree[k].r,mid + 1,r,pos,v);
	    return k;
	}
	inline int query(int k,int pos,int l,int r)
	{
	    if(l >= pos) return tree[k].val;
	    int mid = l + r >> 1;
	    if(pos <= mid) return query(tree[k].l,pos,l,mid) + tree[tree[k].r].val;
	    return query(tree[k].r,pos,mid + 1,r);
	}
	inline void init()
	{
		n = Read(1),k = Read(1);
		root[0] = tree[0].l = tree[0].r = tree[0].val = 0;
		for(reg i = 1;i <= n;i++)
		{
			int x = Read(1);
			if(!vis[x]) root[i] = update(root[i - 1],1,n,i,1);
			else {
				int k = update(root[i - 1],1,n,vis[x],-1);
				root[i] = update(k,1,n,i,1);
			}
			vis[x] = i;
		}
	}
}
using namespace segment_tree;
inline int cost(int l,int r) {return query(root[r],l,1,n);}
inline void dfs(int k,int l,int r,int opl,int opr)
{
	if(l > r) return;
	int mid = l + r >> 1,maxl = -inf,id;
	for(reg i = opl;i <= min(opr,mid);i++)
	{
		int cur = dp[i - 1][k - 1] + cost(i,mid);
		if(cur > maxl) maxl = cur,id = i;
	}
	dp[mid][k] = maxl;
	dfs(k,l,mid - 1,opl,id);
	dfs(k,mid + 1,r,id,opr);
}
int main()
{
	init();
	for(reg i = 1;i <= k;i++) dfs(i,1,n,1,n);
	printf("%d
",dp[n][k]);
	return 0;
}