uoj#352. 新年的五维几何(概率期望+爆搜)

传送门

我还以为这是个五维半平面交呢……结果没看数据范围……

题解

//minamoto
#include<bits/stdc++.h>
#define R register
#define fp(i,a,b) for(R int i=a,I=b+1;i<I;++i)
#define fd(i,a,b) for(R int i=a,I=b-1;i>I;--i)
#define go(u) for(int i=head[u],v=e[i].v;i;i=e[i].nx,v=e[i].v)
using namespace std;
const int N=7;
int l[N],r[N],a[N][N],vis[N],x[N],to[N],f[(1<<5)+5];
int n,tot,ans,lim,sum,res;
void solve(){
	fp(i,1,n)fp(j,1,n){
		if(x[i]-x[j]-a[i][j]<0)return;
		if(x[i]-x[j]-a[i][j]>0)continue;
		if(l[i]==r[i]&&l[j]!=r[j])return;
	}
	memset(f,0,sizeof(f));
	memset(to,0,sizeof(to));
	fp(i,1,n)if(!vis[i]){
		fp(j,1,n)if(!vis[j]&&i!=j){
			if(x[i]-x[j]-a[i][j]==0)to[j-1]|=(1<<(i-1));
		}
	}
	f[0]=1;
	fp(i,1,(1<<n)-1)fp(j,0,n-1)if((i>>j&1)&&(i&to[j])==to[j])f[i]+=f[i^(1<<j)];
	ans+=f[lim];
}
void dfs(int pos){
	if(pos==n+1)return solve();
	if(vis[pos])dfs(pos+1);
	else{
		fp(i,l[pos],r[pos]-1)x[pos]=i,dfs(pos+1);
	}
}
int main(){
//	freopen("testdata.in","r",stdin);
	scanf("%d",&n);
	fp(i,1,n){
		scanf("%d%d",&l[i],&r[i]);
		if(l[i]==r[i])vis[i]=1,x[i]=l[i];
	}
	fp(i,1,n)fp(j,1,n)scanf("%d",&a[i][j]);
	fp(i,1,n)if(a[i][i]>0)return puts("0"),0;
	fp(i,1,n)if(!vis[i])lim|=(1<<(i-1));
	dfs(1);
	sum=1;
	fp(i,1,n)if(l[i]!=r[i])sum*=r[i]-l[i],++res,sum*=res;
	printf("%.10lf
",1.0*ans/sum);
	return 0;
}
原文地址:https://www.cnblogs.com/bztMinamoto/p/10248740.html