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
Sample Output
2
HINT
两棵树分别为1-2-3;1-3-2
Source
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[0]%2==1;
}
bool is_even()
{
return numb[0]%2==0;
}
void lsh_bin()
{
int i;
for (i=topn;i>0;i--)
{
if (numb[i]%2==1)
{
numb[i-1]+=VAL1;
}
numb[i]/=2;
}
numb[0]/=2;
while (topn&&!numb[topn])topn--;
}
bool equal_to(int x)
{
if (topn==0)
{
return x==numb[0];
}
if (topn==1)
{
return x==numb[0]+numb[1]*VAL1;
}
return false;
}
int size()
{
return topn;
}
int length()
{
int x=numb[topn];
int ret=0;
while (x)
{
ret++;
x/=10;
}
int y=0;
x=VAL1;
while (x)
{
y++;
x/=10;
}
y--;
ret+=topn*y;
return ret;
}
void operator =(int x)//{{{
{
int now=0;
clear();
numb[now]=x;
while (numb[now]>=VAL1)
{
numb[now+1]+=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[0])*(topn+1));
}//}}}
void operator +=(number &num)//{{{
{
int i;
topn=max(topn,num.topn);
for (i=0;i<=topn;i++)
{
numb[i]+=num.numb[i];;
if (numb[i]>=VAL1)
{
numb[i+1]+=numb[i]/VAL1;
numb[i]%=VAL1;
}
}
while (numb[topn+1])
{
topn++;
numb[topn+1]+=numb[topn]/VAL1;
numb[topn]%=VAL1;
}
}//}}}
void operator +=(int x)//{{{
{
int now=0;
if (topn==-1)topn=0;
numb[now]+=x;
while (numb[now]>=VAL1)
{
numb[now+1]+=numb[now]/VAL1;
numb[now]%=VAL1;
now++;
if (now>topn)topn=now;
}
}//}}}
void operator *=(int x)//{{{
{
int i;
for (i=0;i<=topn;i++)
{
numb[i]*=x;
}
for (i=0;i<=topn;i++)
{
if (numb[i]>=VAL1)
{
numb[i+1]+=numb[i]/VAL1;
numb[i]%=VAL1;
}
}
while (numb[topn+1])
{
topn++;
numb[topn+1]+=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=0;i<=topn;i++)
{
numb[i]-=num.numb[i];
}
for (i=0;i<=topn;i++)
{
while (numb[i]<0)
{
numb[i]+=VAL1;
numb[i+1]--;
}
}
while (topn&&!numb[topn])topn--;
}//}}}
void operator /=(int x)
{
int i;
int tot=0;
for (i=topn;i>=0;i--)
{
tot=tot*VAL1+numb[i];
numb[i]=tot/x;
tot%=x;
}
while (topn&&!numb[topn])topn--;
}
void operator --(int)//{{{
{
if (topn==0&&numb[0]==0)throw "Error!\n->void operator --(int)\n";
int now=0;
numb[now]--;
while (numb[now]<0)
{
numb[now+1]--;
numb[now]+=VAL1;
}
while (topn&&!numb[topn])topn--;
}//}}}
private:
int numb[MAXL];
int topn;
void clear()
{
topn=0;
memset(numb,0,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>=0;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>=0;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=0;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-1;i>=0;i--)
{
//压六位时
// if (num.numb[i]<100000)out<<"0";
// if (num.numb[i]<10000)out<<"0";
if (num.numb[i]<1000)out<<"0";
if (num.numb[i]<100)out<<"0";
if (num.numb[i]<10)out<<"0";
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()-1,num.topn=0;i>=0;i-=4,num.topn++)
{
if (i-3<str.length())
{
num.numb[num.topn]=(str[i]-‘0‘)+10*(str[i-1]-‘0‘)+100*(str[i-2]-‘0‘)+1000*(str[i-3]-‘0‘);
}else
{
if (i-2<str.length())num.numb[num.topn]+=100*(str[i-2]-‘0‘);
if (i-1<str.length())num.numb[num.topn]+=10*(str[i-1]-‘0‘);
if (i <str.length())num.numb[num.topn]+=(str[i]-‘0‘);
}
}
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=0;i<=num1.topn;i++)
{
for (j=0;j<=num2.topn;j++)
{
ret.numb[i+j]+=num1.numb[i]*num2.numb[j];
if (ret.numb[i+j]>=VAL1)
{
ret.numb[i+j+1]+=ret.numb[i+j]/VAL1;
ret.numb[i+j]%=VAL1;
}
}
}
for (i=0;i<=ret.topn;i++)
{
if (ret.numb[i]>=VAL1)
{
ret.numb[i+1]+=ret.numb[i]/VAL1;
ret.numb[i]%=VAL1;
}
}
while (ret.numb[ret.topn+1])
{
ret.topn++;
ret.numb[ret.topn+1]+=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=1;
int i;
for (i=y+1;i<=x;i++)
ret*=i;
for (i=1;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=1;
int r=n-2;
int i,j;
j=1;
int totu=0;
for (i=0;i<n;i++)
{
scanf("%d",&x);
if (x==0)
{
printf("0\n");
return 0;
}
if (x!=-1)
{
if (!r&&x-1)
{
printf("0\n");
return 0;
}
temp1=c(r,x-1);
ans*=temp1;
r-=x-1;
if (r<0)
{
printf("0\n");
return 0;
}
}else
{
totu++;
}
}
if (r&&totu<=0)
{
printf("0\n");
return 0;
}
if (r)
{
for (i=1;i<=r;i++)
{
ans=ans*totu;
}
}
cout<<ans<<endl;
}
bzoj 1005: [HNOI2008]明明的烦恼 prufer编号&&生成树计数
时间： 2024-10-13 11:19:43