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| Time Limit: 4000MS | Memory Limit: 65536K | |
| Total Submissions: 3709 | Accepted: 1422 |
Description
Input
Output
Sample Input
3 4 2 1 4 2 1 3 2 2 4
Sample Output
4
Source
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<string>
using namespace std;
int n,m,mp[21][21];
int dp[(1<<20)+2];
int main()
{
scanf("%d%d",&n,&m);
for(int i=0;i<n;i++)
{
int x,y;
scanf("%d",&x);
while(x--)
{
scanf("%d",&y);
y--;
mp[i+1][y]=1;
}
}
if(n>m)
{
printf("0\n");
return 0;
}
dp[0]=1;
for(int i=0;i<n;i++)
{
for(int j=(1<<m)-1;j>=0;j--)
{
if(dp[j]==0)
continue;
for(int k=0;k<m;k++)
{
if((j&(1<<k))!=0)
continue;
if(mp[i+1][k]==0)
continue;
dp[j|(1<<k)]+=dp[j];
}
dp[j]=0;
}
}
int ans=0;
for(int i=0;i<(1<<m);i++)
ans+=dp[i];
printf("%d\n",ans);
return 0;
}
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原文地址:http://www.cnblogs.com/water-full/p/4488723.html