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：\(f_{i,j}=f_{k,j-1}+a_i\bigotimes a_k\)
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\(n\cdot v\cdot K\)的做法。
期望得分 60.
其实可以乱搞 比如取 前255小的\(a_i\bigotimes a_k\)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\(n\cdot V\cdot K\)
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\n",x)
#define putl(x) printf("%lld\n",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;
}
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\n",x)
#define putl(x) printf("%lld\n",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;
}
