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?
标签：des style blog http color os
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?
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
一个整数,表示不同的满足要求的树的个数,无解输出0
两棵树分别为1-2-3;1-3-2
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,布布扣,bubuko.com
bzoj 1005: [HNOI2008]明明的烦恼 prufer编号&&生成树计数
标签：des style blog http color os
原文地址：http://www.cnblogs.com/mhy12345/p/3861663.html
