Succession

Description

The king in Utopia has died without an heir. Now several nobles in the country claim
the throne. The country law states that if the ruler has no heir, the person who is most
related to the founder of the country should rule.
To determine who is most related we measure the amount of blood in the veins of a
claimant that comes from the founder. A person gets half the blood from the father and
the other half from the mother. A child to the founder would have 1/2 royal blood, that
child's child with another parent who is not of royal lineage would have 1/4 royal blood,
and so on. The person with most blood from the founder is the one most related.

Input

The rst line contains two integers,N(2<=N<=50) and M(2<=M<=50).
The second line contains the name of the founder of Utopia.
Then followsNlines describing a family relation. Each such line contains three names,
separated with a single space. The rst name is a child and the remaining two names are
the parents of the child.
Then followsMlines containing the names of those who claims the throne.
All names in the input will be between 1 and 10 characters long and only contain the
lowercase English letters 'a'-'z'. The founder will not appear among the claimants, nor
be described as a child to someone else.

Output

A single line containing the name of the claimant with most blood from the founder. The
input will be constructed so that the answer is unique.
The family relations may not be realistic when considering sex, age etc. However, every
child will have two unique parents and no one will be a descendent from themselves. No
one will be listed as a child twice.

Sample Input

9 2

edwardi
charlesi edwardi diana
philip charlesi mistress
wilhelm mary philip
matthew wilhelm helen
edwardii charlesi laura
alice laura charlesi
helen alice bernard
henrii edwardii roxane
charlesii elizabeth henrii
charlesii
matthew

4 5

andrew
betsy andrew flora
carol andrew betsy
dora andrew carol
elena andrew dora
carol
dora
elena
flora
gloria

Sample Output

matthew

elena

HINT

分析：纯模拟n遍即可，每次模拟，更新一次数据。

 

#include<iostream>
#include<stdio.h>
#include<cstring>
using namespace std;
struct in
{
    char a[11];
    char b[11];
}s[55];
struct name
{
    char na[11];
    double ro;
};
name num[55],temp[55];
double find(char *p,int n)
{
    //printf("-->%s  ",p);
    for(int i=0;i<=n;i++)
        if(strcmp(num[i].na,p)==0)
        {
            //printf("%s %lf\n",num[i].na,num[i].ro);
            return num[i].ro;
        }
    return 0;
}
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(int i=0;i<=n;i++)
            num[i].ro=0;
        scanf("%s",&num[0].na);
        num[0].ro=1<<30;//赋值为1后面小数会很小，超出范围
        for(int i=1;i<=n;i++)
            scanf("%s%s%s",&num[i].na,&s[i].a,&s[i].b);
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                //printf("s[j].a=%s s[j].b=%s\n",s[j].a,s[j].b);
                num[j].ro=(find(s[j].a,n)+find(s[j].b,n))/2;
                //printf("num[%d]=%.6lf\n",j,num[j].ro);
            }
        }
        double max=0;
        int ans;
        for(int i=0;i<m;i++)
        {
            scanf("%s",&temp[i].na);
            double flag=find(temp[i].na,n);
            //printf("flag=%lf\n",flag);
            if(max<flag)
            {
                max=flag;
                ans=i;
            }
        }
        printf("%s\n",temp[ans].na);
    }
    return 0;
}