| Blood groups |
| Time Limit: 2000ms, Special Time Limit:5000ms, Memory Limit:131072KB |
| Total submit users: 2, Accepted users: 2 |
| Problem 13813 : No special judgement |
| Problem description |
| There are four possible blood groups for humans: AB, A, B and O, meaning that the red blood cells have antigens of types, respectively, A and B, only A, only B, and no antigen at all. Our blood group is determined by two alleles in our DNA. Each allele is of type either A, B or O. The following table lists the possible allele combinations someone may have for each blood group:
|
| Input |
| The first line contains two integers N and Q, representing respectively the number of parents (and alleles, and antigen types) and the number of queries (1 ≤ N ≤ 100 and 1 ≤ Q ≤ 40). Each of the next N lines describes the blood group of a parent. After that, each of the next Q lines describes a blood group to test. Antigen types are identified with distinct integers from 1 to N, not letters. Each line describing a blood group contains an integer B indicating the number of antigen types in the blood group (0 ≤ B ≤ N), followed by B different integers C1,C2,...,CB representing the antigen types present in the blood group (1 ≤ Ci ≤ N for i = 1,2,...,B). |
| Output |
| For each of the Q queries, output a line with the uppercase letter "Y" if the corresponding blood group is possible in the offspring of the given parents; otherwise output the uppercase letter "N". Write the results in the same order that the queries appear in the input. |
| Sample Input |
Sample input 1 2 1 2 2 1 1 2 0 Sample input 2 3 4 1 1 2 2 3 0 1 3 3 2 1 3 2 1 2 2 3 2 Sample input 3 4 3 4 2 1 3 4 4 2 1 3 4 1 1 1 2 1 3 2 2 1 0 |
| Sample Output |
Sample output 1 N Sample output 2 Y N Y N Sample output 3 Y Y N |
题意：某星球n个父母亲可以携带有某些基因，有显性基因和隐性基因，可以遗传给子女。每个父母亲每个可以遗传给子女一个基因，并且也只能遗传一个基因给子女。题目给出n个父母的基因和q次询问，如果给出的子女的基因是合法的（可以由给出的父母遗传给子女）输出Y,否则输出N。
题解：没想到的匹配，二分图的最大匹配。给出了父母的基因，每个父母可以提供显性和隐性基因，并且基因的长度是n（每个父母当且仅当提供一个）。因此用到二分图的最大匹配，父母的基因构成一个点集，子女的构成一个点集。这样就会考虑子女的需要的基因，需要的基因与父母可以给的基因连边（注意是n个基因），建图。然后进行二分匹配（匈牙利算法）就可以搞定了，如果得到的最大匹配等于n，则是每个子女的基因都被匹配，则输出Y。
详见代码：
#include <stdio.h>
#include <algorithm>
#include <math.h>
#include <string.h>
#include <vector>
#include <queue>
#include <stack>
#include <iostream>
#define pi acos(-1.0)
#define INF 0x3f3f3f3f
using namespace std;
#define ll long long
const int maxn=110;
int n;
int g[maxn][maxn];
int mach[maxn],ch[maxn];
bool dfs(int u)
{
for(int i=0;i<=n;i++)
if(!ch[i]&&g[u][i])
{
ch[i]=1;
if(mach[i]==-1||dfs(mach[i]))
{
mach[i]=u;
return true;
}
}
return false;
}
int maxmach()
{
int ans=0;
memset(mach,-1,sizeof(mach));
for(int i=1;i<=n;i++)
{
memset(ch,0,sizeof(ch));
if(dfs(i))
ans++;
}
return ans;
}
int mmp[maxn][maxn];
int main()
{
//freopen("C:\\Users\\Administrator\\Desktop\\a.txt","r",stdin);
//ios::sync_with_stdio(false);
//freopen("C:\\Users\\Administrator\\Desktop\\b.txt","w",stdout);
int q;
while(scanf("%d%d",&n,&q)!=EOF)
{
memset(mmp,0,sizeof(mmp));
for(int i=1;i<=n;i++)
{
int t,t1;
scanf("%d",&t);
if(t<n) mmp[i][0]=1;
while(t--)
{
scanf("%d",&t1);
mmp[i][t1]=1;
}
}
// cout<<"mmp"<<endl;
// for(int i=0;i<=n;i++)
// {
// for(int j=0;j<=n;j++)
// printf("%d ",mmp[i][j]);
// cout<<endl;
// }
int Ans=0;
for(int j=0;j<q;j++)
{
memset(g,0,sizeof g);
int t,t1;
scanf("%d",&t);
if(t==0)
{
for(int i=1;i<=n;i++)
{
if(mmp[i][0])
{
int kk=1;
while(kk<=n)
{
g[i][kk]=1;
kk++;
}
}
}
}
for(int k=1;k<=t;k++)
{
scanf("%d",&t1);
for(int i=1;i<=n;i++)
{
if(mmp[i][t1])
{
g[i][k]=1;
for(int kk=t+1;kk<=n;kk++)
g[i][kk]=1;
}
if(t<n&&mmp[i][0])
{
for(int kk=t+1;kk<=n;kk++)
g[i][kk]=1;
}
}
}
// cout<<"j "<<j<<endl;
// for(int i=0;i<=n;i++)
// {
// for(int j=0;j<=n;j++)
// printf("%d ",g[i][j]);
// cout<<endl;
// }
// for(int i=0;i<=n;i++)
// printf("vis %d ",vis[i]);
//cout<<"t "<<t<<endl;
int Ans=maxmach();
//printf("%d\n",Ans);
if(Ans==n) printf("Y\n");
else printf("N\n");
}
}
return 0;
}