首先要知道一种prufer数列的东西...一个prufer数列和一颗树对应..然后树上一个点的度数-1是这个点在prufer数列中出现次数..这样就转成一个排列组合的问题了。算个可重集的排列数和组合数就行了...要写高精..
---------------------------------------------------------------------
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = 1009;
int deg[maxn], N, m, c;
int p[maxn], pn;
int cnt[maxn];
bool F[maxn];
void Init() {
m = 0;
scanf("%d", &N);
for(int i = 0; i < N; i++) {
scanf("%d", deg + i);
if(~deg[i])
m++, c += deg[i] - 1;
}
pn = 0;
memset(F, 0, sizeof F);
for(int i = 2; i <= N; i++) {
if(!F[i])
p[pn++] = i;
for(int j = 0; j < pn && i * p[j] <= N; j++) {
F[i * p[j]] = true;
if(i % p[j] == 0) break;
}
}
}
void Work(int V, bool t) {
for(int x = 2; x <= V; x++ )
for(int i = 0, v = x; i < pn && v != 1; i++)
for(; v % p[i] == 0; v /= p[i], t ? cnt[i]++ : cnt[i]--);
}
struct BigInt {
static const int MAXN = 10000;
static const int BASE = 10000;
static const int WID = 4;
int s[MAXN], n;
BigInt(int _n = 0) : n(_n) {
memset(s, 0, sizeof s);
}
BigInt operator = (int num) {
for(; num; s[n++] = num % BASE, num /= BASE);
return *this;
}
BigInt operator *= (const int &x) {
for(int i = 0; i < n; i++) s[i] *= x;
for(int i = 0; i < n; i++) if(s[i] >= BASE) {
s[i + 1] += s[i] / BASE;
s[i] %= BASE;
}
if(s[n]) n++;
return *this;
}
void WRITE() {
int buf[8], t;
for(int i = n; i--; ) {
t = 0;
for(int v = s[i]; v; buf[t++] = v % 10, v /= 10);
if(i + 1 != n) {
for(int j = WID - t; j; j--)
putchar('0');
}
while(t--)
putchar(buf[t] + '0');
}
}
};
int main() {
Init();
if(c > N - 2) {
puts("0"); return 0;
}
memset(cnt, 0, sizeof cnt);
for(int i = 0, v = N - m; i < pn && v != 1; i++)
for(; v % p[i] == 0; v /= p[i], cnt[i]++);
for(int i = 0, t = N - 2 - c; i < pn; i++)
if(cnt[i] > 0) cnt[i] *= t;
Work(N - 2, 1);
Work(N - c - 2, 0);
for(int i = 0; i < N; i++)
if(~deg[i]) Work(deg[i] - 1, 0);
BigInt ans; ans = 1;
for(int i = 0; i < pn; i++)
for(; cnt[i]--; ans *= p[i]);
ans.WRITE();
return 0;
}
---------------------------------------------------------------------
1005: [HNOI2008]明明的烦恼
Time Limit: 1 Sec Memory Limit: 162 MB
Submit: 3412 Solved: 1358
[Submit][Status][Discuss]
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