8.10 NOI 模拟赛 whftll 分块 最短路 异或问题 结论

LINK：whftll

感觉这题很容易推出来 就差了临门一脚.

失误的地方: 没有多画几种情况观察一下性质.

固定一些变量 多画几种情况 有的时候就可以得到问题的性质 从而解题.

35分做法：前20分

考虑整个序列的异或和 要么为0 要么为给定数字.

后者可以直接判断 而前者需要找到一个前缀异或和等于给定数字.

带修改 查前缀异或和某个数字是否出现过 只能上分块.

可以bitset 每次修改暴力重构某个块.

subtask2 15分

直接dp 复杂度(nQk)

score 35

//#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 100000000000ll
#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 1000000007ll
#define gf(x) scanf("%lf",&x)
#define pf(x) ((x)*(x))
#define ull unsigned long long
#define ui unsigned
#define EPS 1e-10
#define sq sqrt
#define S second
#define F first
#define id(i,j) ((i-1)*m+j)
#define zz p<<1
#define yy p<<1|1
#define mod 1000000007
#define min(x,y) ((x)>(y)?(y):(x))
#define max(x,y) ((x)>(y)?(x):(y))
using namespace std;
char *fs,*ft,buf[1<<15];
inline char gc()
{
	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=gc();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=gc();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=gc();}
	return x*f;
}
const int MAXN=100010,maxn=1100000,M=100003,N=352;
int n,Q,B,ww,sj;
int a[MAXN],pos[MAXN],b[MAXN],s[N],f[maxn],vis[maxn];
int L[N],R[N],w[N];bitset<maxn>g[N];
inline void modify(int x,int y)
{
	int p=pos[x];
	rep(x,R[p],i)g[p][a[i]]=0;
	ww^=b[x];b[x]=y;ww^=b[x];
	if(x!=L[p])a[x]=a[x-1]^y;else a[x]=y;
	rep(x+1,R[p],i)a[i]=a[i-1]^b[i];
	rep(L[p],R[p],j)g[p][a[j]]=1;
	w[p]=a[R[p]];
}
inline int ask(int x,int s)
{
	if(x>n)return n+1;
	int p=pos[x],cc=0;
	rep(1,p-1,i)cc=cc^w[i];
	rep(x,R[p],i)if((a[i]^cc)==s)return i;
	cc=cc^w[p];
	rep(p+1,B,i)
	{
		if(g[i][cc^s])
		{
			rep(L[i],R[i],j)if((a[j]^cc)==s)return j;
		}
		cc=cc^w[i];
	}
	return n+1;
}
priority_queue<pii> q;
inline bool solve(int m)
{
	while(q.size())q.pop();
	++sj;q.push(mk(0,0));f[0]=0;
	while(q.size())
	{
		pii x=q.top();q.pop();
		int pos=-x.F;
		if(pos!=f[x.S])continue;
		rep(1,m,i)
		{
			int kk=s[i]^x.S;
			int cc=ask(pos+1,kk);
			if(cc==n+1)continue;
			if(vis[kk]!=sj)
			{
				if(kk==ww)return 1;
				vis[kk]=sj;
				f[kk]=cc;
				q.push(mk(-f[kk],kk));
			}
			else if(f[kk]>cc)f[kk]=cc,q.push(mk(-f[kk],kk));
		}
		
	}
	return 0;
}
int main()
{
	freopen("whftll.in","r",stdin);
	freopen("whftll.out","w",stdout);
	get(n);get(Q);
	B=(int)sqrt(n*1.0);
	int S=n/B;
	rep(1,B,i)L[i]=S*(i-1)+1,R[i]=S*i;
	R[B]=n;
	rep(1,B,i)
	{
		rep(L[i],R[i],j)
		{
			b[j]=a[j]=read();
			if(j!=L[i])a[j]^=a[j-1];
			g[i][a[j]]=1;
			pos[j]=i;
		}
		w[i]=a[R[i]];
		ww=ww^w[i];
	}
	rep(1,Q,i)
	{
		int get(op);
		if(op==1)
		{
			int get(x),get(y);
			modify(x,y);
		}
		else
		{
			//rep(1,n,j)cout<<b[j]<<' ';
			//cout<<ww<<endl;
			int get(k);
			rep(1,k,j)get(s[j]);
			sort(s+1,s+1+k);
			int num=0;
			rep(1,k,j)if(j==1||s[j]!=s[j-1])s[++num]=s[j];
			if(solve(num))puts("yes");
			else puts("no");
		}
	}
	return 0;
}

正解：

多画几种情况 就可以发现 如果有解 那么一定有答案分的段数不超过(2^k)的解.

原因是至多只有(2^k)种前缀和 如果存在两个前缀和一样 那么显然可以合并起来.

而且考虑i处前缀和 如果和j处前缀和相同 那么如果(i<j) 那么j能转移的状态i也可以 所以j一定比i优.

所以决策点最多只有(2^k)个.

那么从0的位置开始每次暴力转移到下一个状态 且要保证某个状态的端点尽量靠前.

这样可以利用dij来做 状态数为(2^kcdot k)

然后 查以某个端点开始的前缀异或和 利用上述暴力的分块做法即可.

sol

//#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 100000000000ll
#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 1000000007ll
#define gf(x) scanf("%lf",&x)
#define pf(x) ((x)*(x))
#define ull unsigned long long
#define ui unsigned
#define EPS 1e-10
#define sq sqrt
#define S second
#define F first
#define id(i,j) ((i-1)*m+j)
#define zz p<<1
#define yy p<<1|1
#define mod 1000000007
#define min(x,y) ((x)>(y)?(y):(x))
#define max(x,y) ((x)>(y)?(x):(y))
using namespace std;
char *fs,*ft,buf[1<<15];
inline char gc()
{
	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=gc();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=gc();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=gc();}
	return x*f;
}
const int MAXN=100010,maxn=1100000,M=100003,N=352;
int n,Q,B,ww,sj;
int a[MAXN],pos[MAXN],b[MAXN],s[N],f[maxn],vis[maxn];
int L[N],R[N],w[N];bitset<maxn>g[N];
inline void modify(int x,int y)
{
	int p=pos[x];
	rep(x,R[p],i)g[p][a[i]]=0;
	ww^=b[x];b[x]=y;ww^=b[x];
	if(x!=L[p])a[x]=a[x-1]^y;else a[x]=y;
	rep(x+1,R[p],i)a[i]=a[i-1]^b[i];
	rep(L[p],R[p],j)g[p][a[j]]=1;
	w[p]=a[R[p]];
}
inline int ask(int x,int s)
{
	if(x>n)return n+1;
	int p=pos[x],cc=0;
	rep(1,p-1,i)cc=cc^w[i];
	rep(x,R[p],i)if((a[i]^cc)==s)return i;
	cc=cc^w[p];
	rep(p+1,B,i)
	{
		if(g[i][cc^s])
		{
			rep(L[i],R[i],j)if((a[j]^cc)==s)return j;
		}
		cc=cc^w[i];
	}
	return n+1;
}
priority_queue<pii> q;
inline bool solve(int m)
{
	while(q.size())q.pop();
	++sj;q.push(mk(0,0));f[0]=0;
	while(q.size())
	{
		pii x=q.top();q.pop();
		int pos=-x.F;
		if(pos!=f[x.S])continue;
		rep(1,m,i)
		{
			int kk=s[i]^x.S;
			int cc=ask(pos+1,kk);
			if(cc==n+1)continue;
			if(vis[kk]!=sj)
			{
				if(kk==ww)return 1;
				vis[kk]=sj;
				f[kk]=cc;
				q.push(mk(-f[kk],kk));
			}
			else if(f[kk]>cc)f[kk]=cc,q.push(mk(-f[kk],kk));
		}
		
	}
	return 0;
}
int main()
{
	freopen("whftll.in","r",stdin);
	freopen("whftll.out","w",stdout);
	get(n);get(Q);
	B=(int)sqrt(n*1.0);
	int S=n/B;
	rep(1,B,i)L[i]=S*(i-1)+1,R[i]=S*i;
	R[B]=n;
	rep(1,B,i)
	{
		rep(L[i],R[i],j)
		{
			b[j]=a[j]=read();
			if(j!=L[i])a[j]^=a[j-1];
			g[i][a[j]]=1;
			pos[j]=i;
		}
		w[i]=a[R[i]];
		ww=ww^w[i];
	}
	rep(1,Q,i)
	{
		int get(op);
		if(op==1)
		{
			int get(x),get(y);
			modify(x,y);
		}
		else
		{
			//rep(1,n,j)cout<<b[j]<<' ';
			//cout<<ww<<endl;
			int get(k);
			rep(1,k,j)get(s[j]);
			sort(s+1,s+1+k);
			int num=0;
			rep(1,k,j)if(j==1||s[j]!=s[j-1])s[++num]=s[j];
			if(solve(num))puts("yes");
			else puts("no");
		}
	}
	return 0;
}