prufer序列

用途: 将带标号的树用唯一的整数序列表示出来，证明凯莱公式。

构造方法：每次选择一个编号最小的叶结点并删掉它，然后在序列中记录下它连接到的那个结点。重复\(n-2\)次后就只剩下两个结点，算法结束。

举个栗子(本图来自baoziwu2,侵删)

显然可以有一个用堆做的方法,时间复杂度\(O(n\log n)\)

\(code:\)

点此查看代码

#include<bits/stdc++.h>
#include<bits/extc++.h>
// using namespace __gnu_pbds;
// using namespace __gnu_cxx;
using namespace std;
#define infile(x) freopen(x,"r",stdin)
#define outfile(x) freopen(x,"w",stdout)
#define errfile(x) freopen(x,"w",stderr)
using ll=long long;using ull=unsigned long long;
char *p1,*p2,buf[1<<20];
#define gc() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<20,stdin),p1==p2)?EOF:*p1++)
#ifdef linux
#define pc putchar_unlocked
#else
#define pc putchar
#endif
namespace IO{
    template<typename T>inline bool read(T &x){x=0;char s=gc();bool f=true;for(;(s<'0'||'9'<s);s=gc()) {if(s=='-') f=false;if(s==EOF)return false;}for(;'0'<=s&&s<='9';s=gc()) x=(x<<1)+(x<<3)+(s^48);if(!f) x=~x+1;return true;}
    inline bool read(double &x){x=0.0;char s=gc();bool f=true;for(;(s<'0'||'9'<s);s=gc()) {if(s=='-') f=false;if(s==EOF)return false;}for(;'0'<=s&&s<='9';s=gc()) x=(x*10)+(s^48);if(s!='.'){return true;}double res=0.1;s=gc();for(;'0'<=s&&s<='9';res/=10,s=gc()) x+=(s^48)*res;x=f?x:-x;return true;}
    inline bool read(long double &x){x=0.0;char s=gc();bool f=true;for(;(s<'0'||'9'<s);s=gc()) {if(s=='-') f=false;if(s==EOF)return false;}for(;'0'<=s&&s<='9';s=gc()) x=(x*10)+(s^48);if(s!='.'){return true;}double res=0.1;s=gc();for(;'0'<=s&&s<='9';res/=10,s=gc()) x+=(s^48)*res;x=f?x:-x;return true;}
    inline bool read(float &x){x=0.0;char s=gc();bool f=true;for(;(s<'0'||'9'<s);s=gc()) {if(s=='-') f=false;if(s==EOF)return false;}for(;'0'<=s&&s<='9';s=gc()) x=(x*10)+(s^48);if(s!='.'){return true;}double res=0.1;s=gc();for(;'0'<=s&&s<='9';res/=10,s=gc()) x+=(s^48)*res;x=f?x:-x;return true;}
    inline bool read(string &str){string ().swap(str);char s=gc();for(;s==' '||s=='\n';s=gc());if(s==EOF) return false; for(;s!=' '&&s!='\n'&&s!=EOF;s=gc())str.push_back(s);return true;}
    inline bool read_line(string &str){string ().swap(str);char s=gc();for(;s==' '||s=='\n';s=gc());if(s==EOF) return false;for(;s!='\n'&&s!=EOF;s=gc()){str.push_back(s);}return true;}
    inline bool read_line(char *str){int len=0;char s=gc();for(;s==' '||s=='\n';s=gc());if(s==EOF) return false;for(;s!='\n'&&s!=EOF;s=gc()){str[len]=s;len++;}str[len]='\0';return true;}
    inline bool read(char &s){char x=gc();for(;x==' '||x=='\n';x=gc());if(x==EOF||x==' '||x=='\n')return false;s=x;return true;}
    inline bool read(char *s){int len=0;char x=gc();for(;x==' '||x=='\n';x=gc());if(x==EOF)return false;for(;x!=' '&&x!='\n'&&x!=EOF;x=gc())s[len++]=x;s[len]='\0';return true;}
    template<class T,class... Args> inline bool read(T &x,Args&... args){return (read(x)&&read(args...));}
    template<class T>inline void write(T x){static T st[45];int top=0;if(x<0)x=~x+1,pc('-');do{st[top++]=x%10;}while(x/=10);while(top)pc(st[--top]^48);}
    inline void write(char x){pc(x);}
    inline void write(string s){for(int i=0;s[i];++i) pc(s[i]);}
    inline void write(char *s){int len=strlen(s);for(int i=0;i<len;++i) pc(s[i]);}
    inline void write(const char *s){int len=strlen(s);for(int i=0;i<len;++i) pc(s[i]);}
    template<class T,class... Args> inline void write(T x,Args... args){write(x);write(args...);}
}using namespace IO;
const int N = 5e6 + 10;
int fa[N],d[N],pru[N];
priority_queue<int,vector<int>,greater<int> > q;
signed main(){
    #ifndef ONLINE_JUDGE
        infile("in.in");outfile("out.out");
    #else
    #endif
    int n,m;read(n,m);
    ll ans = 0;
    if(m == 1){
        for(int i = 1,x;i < n; ++i) read(x),++d[x],fa[i] = x;
        for(int i = 1;i <= n; ++i)
            if(!d[i]) q.push(i);
        int tot = 0;
        for(int i = 1;i <= n - 2; ++i){
            pru[i] = fa[q.top()];q.pop();
            if(!--d[pru[i]]) q.push(pru[i]);
        }
        for(int i = 1;i <= n - 2; ++i) ans ^= 1ll * i * pru[i];
    }
    else{
        for(int i = 1,x;i <= n - 2; ++i) read(x),pru[i] = x,d[x]++;
        for(int i = 1;i < n; ++i) fa[i] = n;
        for(int i = 1;i <= n; ++i) if(!d[i]) q.push(i);
        for(int i = 1;i <= n - 2; ++i){
            fa[q.top()] = pru[i],q.pop();
            if(!--d[pru[i]]) q.push(pru[i]);
        }
        for(int i = 1;i <= n - 1; ++i) ans ^= 1ll * i * fa[i];
    }
    write(ans);
}

还有一个\(O(n)\)的构造方法

考虑删完一个点后，下一次改删哪个点，有两种情况，一是本来度数就为1的点，二是刚刚删完的点的父节点。

维护一个指针，初始为1，每遇到一个度数为1的节点\(y\)，就将其父节点\(x\)记录进\(prufer\)序列中，\(x\)的度数减一，若\(x\)的度数为1，则分类讨论。

若\(x>y\)，继续自增。

若\(x<y\)，另外处理\(x\)，删除它，记录\(x\)的父节点

重复以上操作。

\(code:\)

点此查看代码

#include<bits/stdc++.h>
#include<bits/extc++.h>
// using namespace __gnu_pbds;
// using namespace __gnu_cxx;
using namespace std;
#define infile(x) freopen(x,"r",stdin)
#define outfile(x) freopen(x,"w",stdout)
#define errfile(x) freopen(x,"w",stderr)
using ll=long long;using ull=unsigned long long;
char *p1,*p2,buf[1<<20];
#define gc() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<20,stdin),p1==p2)?EOF:*p1++)
#ifdef linux
#define pc putchar_unlocked
#else
#define pc putchar
#endif
namespace IO{
    template<typename T>inline bool read(T &x){x=0;char s=gc();bool f=true;for(;(s<'0'||'9'<s);s=gc()) {if(s=='-') f=false;if(s==EOF)return false;}for(;'0'<=s&&s<='9';s=gc()) x=(x<<1)+(x<<3)+(s^48);if(!f) x=~x+1;return true;}
    inline bool read(double &x){x=0.0;char s=gc();bool f=true;for(;(s<'0'||'9'<s);s=gc()) {if(s=='-') f=false;if(s==EOF)return false;}for(;'0'<=s&&s<='9';s=gc()) x=(x*10)+(s^48);if(s!='.'){return true;}double res=0.1;s=gc();for(;'0'<=s&&s<='9';res/=10,s=gc()) x+=(s^48)*res;x=f?x:-x;return true;}
    inline bool read(long double &x){x=0.0;char s=gc();bool f=true;for(;(s<'0'||'9'<s);s=gc()) {if(s=='-') f=false;if(s==EOF)return false;}for(;'0'<=s&&s<='9';s=gc()) x=(x*10)+(s^48);if(s!='.'){return true;}double res=0.1;s=gc();for(;'0'<=s&&s<='9';res/=10,s=gc()) x+=(s^48)*res;x=f?x:-x;return true;}
    inline bool read(float &x){x=0.0;char s=gc();bool f=true;for(;(s<'0'||'9'<s);s=gc()) {if(s=='-') f=false;if(s==EOF)return false;}for(;'0'<=s&&s<='9';s=gc()) x=(x*10)+(s^48);if(s!='.'){return true;}double res=0.1;s=gc();for(;'0'<=s&&s<='9';res/=10,s=gc()) x+=(s^48)*res;x=f?x:-x;return true;}
    inline bool read(string &str){string ().swap(str);char s=gc();for(;s==' '||s=='\n';s=gc());if(s==EOF) return false; for(;s!=' '&&s!='\n'&&s!=EOF;s=gc())str.push_back(s);return true;}
    inline bool read_line(string &str){string ().swap(str);char s=gc();for(;s==' '||s=='\n';s=gc());if(s==EOF) return false;for(;s!='\n'&&s!=EOF;s=gc()){str.push_back(s);}return true;}
    inline bool read_line(char *str){int len=0;char s=gc();for(;s==' '||s=='\n';s=gc());if(s==EOF) return false;for(;s!='\n'&&s!=EOF;s=gc()){str[len]=s;len++;}str[len]='\0';return true;}
    inline bool read(char &s){char x=gc();for(;x==' '||x=='\n';x=gc());if(x==EOF||x==' '||x=='\n')return false;s=x;return true;}
    inline bool read(char *s){int len=0;char x=gc();for(;x==' '||x=='\n';x=gc());if(x==EOF)return false;for(;x!=' '&&x!='\n'&&x!=EOF;x=gc())s[len++]=x;s[len]='\0';return true;}
    template<class T,class... Args> inline bool read(T &x,Args&... args){return (read(x)&&read(args...));}
    template<class T>inline void write(T x){static T st[45];int top=0;if(x<0)x=~x+1,pc('-');do{st[top++]=x%10;}while(x/=10);while(top)pc(st[--top]^48);}
    inline void write(char x){pc(x);}
    inline void write(string s){for(int i=0;s[i];++i) pc(s[i]);}
    inline void write(char *s){int len=strlen(s);for(int i=0;i<len;++i) pc(s[i]);}
    inline void write(const char *s){int len=strlen(s);for(int i=0;i<len;++i) pc(s[i]);}
    template<class T,class... Args> inline void write(T x,Args... args){write(x);write(args...);}
}using namespace IO;
const int N = 5e6+10;
int fa[N],d[N],pru[N];
signed main(){
    #ifndef ONLINE_JUDGE
        infile("in.in");outfile("out.out");
    #else
    #endif
    int n,m;read(n,m);
    ll ans = 0;
    if(m == 1){
        for(int i = 1,x;i < n; ++i) read(x),fa[i] = x,++d[x];
        for(int i = 1,p = 1,x;i <= n - 2; ++i,++p){
            while(d[p]) ++p;
            pru[i] = fa[x = p];
            while(i <= n - 2 && !--d[pru[i]] && pru[i] < p) pru[++i] = fa[x = fa[x]];
        }
        for(int i = 1;i <= n - 2; ++i) ans ^= 1ll * i * pru[i];
    }
    else{
        for(int i = 1,x;i <= n - 2; ++i) read(x),pru[i] = x,++d[x];
        pru[n - 1] = n;
        for(int i = 1,p = 1,x;i <= n - 1; ++i,++p){
            while(d[p]) ++p;
            fa[x = p] = pru[i];
            while(i <= n - 1 && !--d[pru[i]] && pru[i] < p) fa[x = fa[x]] = pru[++i];
        }
        for(int i = 1;i <= n - 1; ++i) ans ^= 1ll * i * fa[i];
    }
    write(ans);
}

凯莱公式

内容:

完全图\(k_n\)有\(n^{n-2}\)棵生成树
证明：

任意一个长度为\(n-2\)的值域为\([1,n]\)的整数序列都可以通过\(Prufer\)序列双射对应一个生成树，于是方案数就是\(n^{n-2}\) 。

图连通方案数

问题：

一个\(n\)个点\(m\)条边的带标号无向图有\(k\)个连通块。添加\(k-1\)条边使得整个图连通。求方案数。
证明：

设\(s_i\)为第\(i\)个连通块的数量,\(d_i\)为第\(i\)个连通块的度数，由于构造\(prufer\)序列的方案数为

\[{k-2}\choose {d_1-1,d_2-1,...,d_k-1} \]
对于第\(i\)个连通块，它的连接方式有\(s_i^{d_i}\)种，因此方案数为

\[{{k-2}\choose {d_1-1,d_2-1,...,d_k-1}}\times\prod_{i=1}^ks_i^{d_i} \]
枚举\(d\)序列，方案数有

\[\sum_{d_i\ge 1,(\sum_{i=1}^kd_i)=2k-2}{{k-2}\choose {d_1-1,d_2-1,...,d_k-1}}\times\prod_{i=1}^ks_i^{d_i} \]
换元,令\(a_i=d_i-1\),原式为

\[\sum_{a_i\ge 0,(\sum_{i=1}^ka_i)=k-2}{{k-2}\choose {a_1,a_2,...,a_k}}\times\prod_{i=1}^ks_i^{a_i+1} \]
利用多元二项式定理

\[(x_1+x_2+\dots+x_n)^m=\sum\limits_{\sum_{i=1}^n d_i=m}\binom{m}{d_1,\dots,d_n}\prod_{i=1}^nx_i^{d_i} \]
化简得

\[(s_1+s_2+...+s_k)^{k-2}\times \prod_{i=1}^{k}s_i \]
所以为

\[n^{k-2}\times\prod_{i=1}^ks_i \]
~~学这玩意就是为了这个式子~~

posted @ 2024-07-04 18:36 CuFeO4 阅读(16) 评论(0) 编辑收藏举报

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