8.12 NOI 模拟赛 UOJ UNR#4 T1 序列妙妙值 dp 分块优化dp 根号平衡

LINK：序列妙妙值

怎么人均会T1啊

感觉上午想的时候就差一点了 因为已经注意到了 这道题可能是要折半什么的.

后来 去刚T2了. 所以就没了.

容易得到一个(n^2K)的做法。期望得分40.

状态转移：(f_{i,j}=f_{k,j-1}+a_iigotimes a_k)

其中(a_i)是异或前缀和.

设值域为V.

那么保存(g_i)表示 (a_k==i)是的最小(f_{k,j-1}) 那么就得到了一个(ncdot vcdot K)的做法。

期望得分 60.

其实可以乱搞 比如取 前255小的(a_iigotimes a_k)这样的(k)来进行更新什么的.

上午我考虑到了另外一种想法 发现(a_i)的出现最多为V的大小.

考虑优化上述第一个暴力，对于一个(a_i)我们肯定是要从(1-i-1)来扫获得答案.

可以把扫到的端点及值存起来 那么对于每个本质不同的(a_i)指针最多移动n次.

最坏复杂度(ncdot Vcdot K)

不过很大程度上跑不满？数据不太行 所以得分80.

code

//#include<bits/stdc++.h>
#include<iostream>
#include<cstdio>
#include<ctime>
#include<cctype>
#include<queue>
#include<deque>
#include<stack>
#include<iostream>
#include<iomanip>
#include<cstdio>
#include<cstring>
#include<string>
#include<ctime>
#include<cmath>
#include<cctype>
#include<cstdlib>
#include<queue>
#include<deque>
#include<stack>
#include<vector>
#include<algorithm>
#include<utility>
#include<bitset>
#include<set>
#include<map>
#define ll long long
#define db double
#define INF 1000000000
#define inf 1000000000000000ll
#define ldb long double
#define pb push_back
#define put_(x) printf("%d ",x);
#define get(x) x=read()
#define gt(x) scanf("%d",&x)
#define gi(x) scanf("%lf",&x)
#define put(x) printf("%d
",x)
#define putl(x) printf("%lld
",x)
#define rep(p,n,i) for(RE int i=p;i<=n;++i)
#define go(x) for(int i=lin[x],tn=ver[i];i;tn=ver[i=nex[i]])
#define fep(n,p,i) for(RE int i=n;i>=p;--i)
#define vep(p,n,i) for(RE int i=p;i<n;++i)
#define pii pair<int,int>
#define mk make_pair
#define RE register
#define P 13331ll
#define gf(x) scanf("%lf",&x)
#define pf(x) ((x)*(x))
#define uint unsigned long long
#define ui unsigned
#define EPS 1e-5
#define sq sqrt
#define S second
#define F first
#define mod 1000000007
#define md 998244353
#define max(x,y) ((x)<(y)?y:x)
#define l(i) t[i].l
#define r(i) t[i].r
#define mx(i) t[i].mx
#define w(i) t[i].w
#define zz p<<1
#define yy p<<1|1
using namespace std;
char buf[1<<15],*fs,*ft;
inline char getc()
{
	return (fs==ft&&(ft=(fs=buf)+fread(buf,1,1<<15,stdin),fs==ft))?0:*fs++;
}
inline int read()
{
	RE int x=0,f=1;RE char ch=getc();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getc();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getc();}
	return x*f;
}
inline ll Read()
{
	RE ll x=0,f=1;RE char ch=getc();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getc();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getc();}
	return x*f;
}
const int MAXN=60010,maxn=4100;
int n,K,maxx;
int a[MAXN];
int f[MAXN][9];
int vis[maxn],g[maxn];
int main()
{
	//freopen("1.in","r",stdin);
	//freopen("1.out","w",stdout);
	get(n);get(K);
	rep(1,n,i)get(a[i]),maxx=max(maxx,a[i]),a[i]^=a[i-1],f[i][1]=a[i];
	if(maxx<=4096)
	{
		maxx=255;
		rep(2,K,j)
		{
			memset(vis,0x3f,sizeof(vis));
			memset(g,0,sizeof(g));
			rep(1,n,i)
			{
				while(g[a[i]]+1<i)
				{
					++g[a[i]];
					vis[a[i]]=min(vis[a[i]],f[g[a[i]]][j-1]+(a[g[a[i]]]^a[i]));
				}
				f[i][j]=vis[a[i]];
			}
		}
		rep(K,n,i)printf("%d ",f[i][K]);
		return 0;
	}
	rep(1,n,i)rep(2,K,j)
	{
		f[i][j]=INF;
		rep(1,i-1,k)f[i][j]=min(f[i][j],f[k][j-1]+(a[i]^a[k]));
	}
	rep(K,n,i)printf("%d ",f[i][K]);
	return 0;
}

考虑正解 其实可以发现第二种方法每次更新值的时候 都是O(1)更新 O(V)查询.

或者观察到题目中的部分分 一档(2^8)最高档(2^{16})

其实我们折半. 也就是根号平衡.

也就是(sqrt V)更新 (sqrt V)查询.

那么可以设(g_{i,j})表示前8为为i 那么对于(a_i)的后8位为j的最优解.

这样就完成了.

code

//#include<bits/stdc++.h>
#include<iostream>
#include<cstdio>
#include<ctime>
#include<cctype>
#include<queue>
#include<deque>
#include<stack>
#include<iostream>
#include<iomanip>
#include<cstdio>
#include<cstring>
#include<string>
#include<ctime>
#include<cmath>
#include<cctype>
#include<cstdlib>
#include<queue>
#include<deque>
#include<stack>
#include<vector>
#include<algorithm>
#include<utility>
#include<bitset>
#include<set>
#include<map>
#define ll long long
#define db double
#define INF 1000000000
#define inf 1000000000000000ll
#define ldb long double
#define pb push_back
#define put_(x) printf("%d ",x);
#define get(x) x=read()
#define gt(x) scanf("%d",&x)
#define gi(x) scanf("%lf",&x)
#define put(x) printf("%d
",x)
#define putl(x) printf("%lld
",x)
#define rep(p,n,i) for(RE int i=p;i<=n;++i)
#define go(x) for(int i=lin[x],tn=ver[i];i;tn=ver[i=nex[i]])
#define fep(n,p,i) for(RE int i=n;i>=p;--i)
#define vep(p,n,i) for(RE int i=p;i<n;++i)
#define pii pair<int,int>
#define mk make_pair
#define RE register
#define P 13331ll
#define gf(x) scanf("%lf",&x)
#define pf(x) ((x)*(x))
#define uint unsigned long long
#define ui unsigned
#define EPS 1e-5
#define sq sqrt
#define S second
#define F first
#define mod 1000000007
#define md 998244353
#define max(x,y) ((x)<(y)?y:x)
#define l(i) t[i].l
#define r(i) t[i].r
#define mx(i) t[i].mx
#define w(i) t[i].w
#define zz p<<1
#define yy p<<1|1
using namespace std;
char buf[1<<15],*fs,*ft;
inline char getc()
{
	return (fs==ft&&(ft=(fs=buf)+fread(buf,1,1<<15,stdin),fs==ft))?0:*fs++;
}
inline int read()
{
	RE int x=0,f=1;RE char ch=getc();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getc();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getc();}
	return x*f;
}
inline ll Read()
{
	RE ll x=0,f=1;RE char ch=getc();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getc();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getc();}
	return x*f;
}
const int MAXN=60010,maxn=256;
int n,K,maxx;
int a[MAXN];
int f[MAXN][9];
int g[maxn][maxn];
int main()
{
	//freopen("1.in","r",stdin);
	//freopen("1.out","w",stdout);
	get(n);get(K);maxx=255;
	rep(1,n,i)get(a[i]),a[i]^=a[i-1],f[i][1]=a[i];
	rep(2,K,j)
	{
		memset(g,0x3f,sizeof(g));
		rep(1,n,i)
		{
			int w1=a[i]>>8,w2=a[i]&maxx;
			f[i][j]=INF;
			rep(0,maxx,k)f[i][j]=min(f[i][j],g[k][w1]+(w2^k));
			rep(0,maxx,k)g[w2][k]=min(g[w2][k],f[i][j-1]+((k^w1)<<8));
		}
	}
	rep(K,n,i)printf("%d ",f[i][K]);
	return 0;
}