BZOJ 3139: [Hnoi2013]比赛

这题嘛，我们可以首先考虑枚举一共n*(n-1)/2次比赛的结果，然后判断一下就好了

其实有一点是显然的，分数序列的顺序不影响答案

所以我们用最小表示法来表示分数序列，然后记忆化搜索一下就好了。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<map>
#include<algorithm>
using namespace std;
typedef long long ll;
const int p=(1e9)+7;
map<ll,ll>mp;
int n;
struct state{
	int a[11];
	ll hash(){
		ll ans=0;
		for(int i=0;i<=n;i++)ans=ans*28+a[i];
		return ans;
	}
	void minexp(){
		sort(a+a[0],a+n+1);
	}
}st,ed;
ll dfs(int i,state now){
	if(now.a[0]==n)return mp[now.hash()]=-1;
	if(now.a[now.a[0]]>3*(n-i+1))return -1;
	if(i>n){
		now.a[0]++;
		now.minexp();
		if(mp[now.hash()])return mp[now.hash()];
		return dfs(now.a[0]+1,now);
	}
	ll ans=0,tmp;
	int idx=now.a[0];
	if(now.a[idx]>=3){
		now.a[idx]-=3;
		tmp=dfs(i+1,now);
		if(tmp!=-1)(ans+=tmp)%=p;
		now.a[idx]+=3;
	}
	if(now.a[idx]&&now.a[i]){
		now.a[idx]--;now.a[i]--;
		tmp=dfs(i+1,now);
		if(tmp!=-1)(ans+=tmp)%=p;
		now.a[idx]++;now.a[i]++;
	}
	if(now.a[i]>=3){
		now.a[i]-=3;
		tmp=dfs(i+1,now);
		if(tmp!=-1)(ans+=tmp)%=p;
		now.a[i]+=3;
	}
	ans=ans?ans:-1;
	if(i==idx+1)mp[now.hash()]=ans;
	return ans;
}
int main(){
	//freopen("a.in","r",stdin);
	//freopen("a.out","w",stdout);
	scanf("%d",&n);
	for(int i=1;i<=n;i++)scanf("%d",st.a+i);
	st.a[0]=1;
	st.minexp();
	ed.a[0]=n;mp[ed.hash()]=1;
	printf("%lld\n",dfs(2,st));
	return 0;
}