1005: [HNOI2008]明明的烦恼
Time Limit: 1 Sec Memory Limit: 162 MB
Submit: 2248 Solved: 898
[Submit][Status]
Description
自从明明学了树的结构,就对奇怪的树产生了兴趣...... 给出标号为1到N的点,以及某些点最终的度数,允许在任意两点间连线,可产生多少棵度数满足要求的树?
Input
第一行为N(0 < N < = 1000),接下来N行,第i+1行给出第i个节点的度数Di,如果对度数不要求,则输入-1
Output
一个整数,表示不同的满足要求的树的个数,无解输出0
Sample Input
3
1
-1
-1
1
-1
-1
Sample Output
2
HINT
两棵树分别为1-2-3;1-3-2
Source
生成树prufer标号
首先选这棵树叶子中编号最小的点，将这个点删除，并且把它的邻接点加入一个数组中，例如第一个删除的节点为1，并且把5加入数组中。删除节点后形成一棵新的树，再在新树中删除最小的节点，并且把邻接点加入数组中，，这样重复以上步骤，知道树中最后剩余两个点的时候终止操作。这时候数组中的便是prufer编码。
生成树prufer标号的性质（觉得相当神奇啊）：
1.prufer序列和树一一对应。
2.序列的长度=树的节点个数-2。
3.树中该节点的度数=该节点编号在序列中出现的次数+1,叶子节点不会出现在序列中。
每一个标号有且仅有一个对应的生成树。
有些时候当计数问题不含加减，但含除法时，可通过分解质因数简化。
而且光靠想出来的公式很不靠谱的。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define MAXN 1010
#define MAXL 1000
#define VAL1 10000
class number//四位
{
public:
number()
{
clear();
}
bool is_odd()
{
return numb[]%==;
}
bool is_even()
{
return numb[]%==;
}
void lsh_bin()
{
int i;
for (i=topn;i>;i--)
{
if (numb[i]%==)
{
numb[i-]+=VAL1;
}
numb[i]/=;
}
numb[]/=;
while (topn&&!numb[topn])topn--;
}
bool equal_to(int x)
{
if (topn==)
{
return x==numb[];
}
if (topn==)
{
return x==numb[]+numb[]*VAL1;
}
return false;
}
int size()
{
return topn;
}
int length()
{
int x=numb[topn];
int ret=;
while (x)
{
ret++;
x/=;
}
int y=;
x=VAL1;
while (x)
{
y++;
x/=;
}
y--;
ret+=topn*y;
return ret;
}
void operator =(int x)//{{{
{
int now=;
clear();
numb[now]=x;
while (numb[now]>=VAL1)
{
numb[now+]+=numb[now]/VAL1;
numb[now]%=VAL1;
now++;
if (now>topn)topn=now;
}
}//}}}
void operator =(number num)//{{{
{
topn=num.topn;
memcpy((this->numb),num.numb,sizeof(num.numb[])*(topn+));
}//}}}
void operator +=(number &num)//{{{
{
int i;
topn=max(topn,num.topn);
for (i=;i<=topn;i++)
{
numb[i]+=num.numb[i];;
if (numb[i]>=VAL1)
{
numb[i+]+=numb[i]/VAL1;
numb[i]%=VAL1;
}
}
while (numb[topn+])
{
topn++;
numb[topn+]+=numb[topn]/VAL1;
numb[topn]%=VAL1;
}
}//}}}
void operator +=(int x)//{{{
{
int now=;
if (topn==-)topn=;
numb[now]+=x;
while (numb[now]>=VAL1)
{
numb[now+]+=numb[now]/VAL1;
numb[now]%=VAL1;
now++;
if (now>topn)topn=now;
}
}//}}}
void operator *=(int x)//{{{
{
int i;
for (i=;i<=topn;i++)
{
numb[i]*=x;
}
for (i=;i<=topn;i++)
{
if (numb[i]>=VAL1)
{
numb[i+]+=numb[i]/VAL1;
numb[i]%=VAL1;
}
}
while (numb[topn+])
{
topn++;
numb[topn+]+=numb[topn]/VAL1;
numb[topn]%=VAL1;
}
}//}}}
void operator *=(number &num)
{
number ret;
ret=(*this)*num;
(*this)=ret;
}
void operator -=(number &num)//{{{
{
if (*this<num)throw "Error!\n->void operator -=(number &num)\n";
int i;
for (i=;i<=topn;i++)
{
numb[i]-=num.numb[i];
}
for (i=;i<=topn;i++)
{
while (numb[i]<)
{
numb[i]+=VAL1;
numb[i+]--;
}
}
while (topn&&!numb[topn])topn--;
}//}}}
void operator /=(int x)
{
int i;
int tot=;
for (i=topn;i>=;i--)
{
tot=tot*VAL1+numb[i];
numb[i]=tot/x;
tot%=x;
}
while (topn&&!numb[topn])topn--;
}
void operator --(int)//{{{
{
if (topn==&&numb[]==)throw "Error!\n->void operator --(int)\n";
int now=;
numb[now]--;
while (numb[now]<)
{
numb[now+]--;
numb[now]+=VAL1;
}
while (topn&&!numb[topn])topn--;
}//}}}
private:
int numb[MAXL];
int topn;
void clear()
{
topn=;
memset(numb,,sizeof(numb)); }
friend bool operator <(number num1,number num2);
friend bool operator <=(number num1,number num2);
friend bool operator ==(number num1,number num2);
friend ostream& operator <<(ostream &out,number &num);
friend istream& operator >>(istream &in,number &num);
friend number operator *(number &num1,number &num2);
friend number operator *(number num,int x);
friend number operator +(number num1,number num2);
friend number operator +(number num1,int x);
friend number operator -(number num1,number num2);
//a=a+b远没有a+=b快
};
bool operator <(number num1,number num2)//{{{
{
if (num1.topn!=num2.topn)
{
return num1.topn<num2.topn;
}
int i;
for (i=num1.topn;i>=;i--)
{
if (num1.numb[i]!=num2.numb[i])
{
return num1.numb[i]<num2.numb[i];
}
}
return false;
}//}}}
bool operator <=(number num1,number num2)//{{{
{
if (num1.topn!=num2.topn)
{
return num1.topn<num2.topn;
}
int i;
for (i=num1.topn;i>=;i--)
{
if (num1.numb[i]!=num2.numb[i])
{
return num1.numb[i]<num2.numb[i];
}
}
return true;
}//}}}
bool operator ==(number num1,number num2)//{{{
{
if (num1.topn!=num2.topn)return false;
for (int i=;i<=num1.topn;i++)
{
if (num1.numb[i]!=num2.numb[i])return false;
}
return true;
}//}}}
ostream& operator <<(ostream &out,number &num)//{{{
{
int i;
out<<num.numb[num.topn];
for (i=num.topn-;i>=;i--)
{
//压六位时
// if (num.numb[i]<100000)out<<"0";
// if (num.numb[i]<10000)out<<"0";
if (num.numb[i]<)out<<"";
if (num.numb[i]<)out<<"";
if (num.numb[i]<)out<<"";
out<<num.numb[i];
}
return out;
}//}}}
istream& operator >>(istream &in,number &num)//{{{
{
string str;
in>>str;
int i;
num.clear();
for (i=(int)str.length()-,num.topn=;i>=;i-=,num.topn++)
{
if (i-<str.length())
{
num.numb[num.topn]=(str[i]-'')+*(str[i-]-'')+*(str[i-]-'')+*(str[i-]-'');
}else
{
if (i-<str.length())num.numb[num.topn]+=*(str[i-]-'');
if (i-<str.length())num.numb[num.topn]+=*(str[i-]-'');
if (i <str.length())num.numb[num.topn]+=(str[i]-'');
}
}
num.topn--;
return in;
}//}}}
number operator *(number num,int x)//{{{
{
number ret;
ret=num;
ret*=x;
return ret;
}//}}}
number operator * (number &num1,number &num2)
{
int i,j;
number ret;
ret.topn=num1.topn+num2.topn;
for (i=;i<=num1.topn;i++)
{
for (j=;j<=num2.topn;j++)
{
ret.numb[i+j]+=num1.numb[i]*num2.numb[j];
if (ret.numb[i+j]>=VAL1)
{
ret.numb[i+j+]+=ret.numb[i+j]/VAL1;
ret.numb[i+j]%=VAL1;
}
}
}
for (i=;i<=ret.topn;i++)
{
if (ret.numb[i]>=VAL1)
{
ret.numb[i+]+=ret.numb[i]/VAL1;
ret.numb[i]%=VAL1;
}
}
while (ret.numb[ret.topn+])
{
ret.topn++;
ret.numb[ret.topn+]+=ret.numb[ret.topn]/VAL1;
ret.numb[ret.topn]%=VAL1;
}
return ret;
}
number operator +(number num1,number num2)//{{{
{
number ret;
ret=num1;
ret+=num2;
return ret;
}//}}}
number operator +(number num1,int x)//{{{
{
number ret;
ret=num1;
ret+=x;
return ret;
}//}}}
number operator -(number num1,number num2)//{{{
{
number ret;
ret=num1;
ret-=num2;
return ret;
}//}}}
number c (int x,int y)
{
number ret;
ret=;
int i;
for (i=y+;i<=x;i++)
ret*=i;
for (i=;i<=(x-y);i++)
ret/=i;
return ret;
}
int main()
{
freopen("input.txt","r",stdin);
int x;
int n;
scanf("%d",&n);
number ans;
number temp1,temp2,temp3;
// number temp;
// temp=c(1000,2);
// cout<<temp<<endl;
ans=;
int r=n-;
int i,j;
j=;
int totu=;
for (i=;i<n;i++)
{
scanf("%d",&x);
if (x==)
{
printf("0\n");
return ;
}
if (x!=-)
{
if (!r&&x-)
{
printf("0\n");
return ;
}
temp1=c(r,x-);
ans*=temp1;
r-=x-;
if (r<)
{
printf("0\n");
return ;
}
}else
{
totu++;
}
}
if (r&&totu<=)
{
printf("0\n");
return ;
}
if (r)
{
for (i=;i<=r;i++)
{
ans=ans*totu;
}
}
cout<<ans<<endl;
}