1050: Find the immediate family

1050: Find out immediate family time limit: 1 Sec memory limit: MB
Submitted by: 321 Resolution: 264
Submitted State [Discussion Version] Title Description If a, B is the parent of C, then a A, a, B is the parent,c of C is a A, a A, if A is C (outside) grandfather, grandmother, then a B is the grandparent,c of C is a A, B is the grandchild, if A is C (outside) great grandfather, Once grandmother, then A, B is the great-grandparent,c of C is a, b great-grandchild, then more than one generation, then add a great-on the relationship. The input input contains multiple sets of test cases, each of which consists of 2 integers n (0<=n<=26) and M (0&LT;M&LT;50), each representing n kinship and M questions, followed by a string of n rows, such as ABC, which indicates that A's parents are B and C, respectively. If A's parental information is incomplete, then use-instead, for example A-c, then the M-line form of the FA string, which indicates the relationship between F and a.
When N and m are 0 o'clock end input. Output if the 2 persons questioned are immediate family members, please describe the relationship of output 2 according to the topic, if there is no direct relationship, please output-.
Refer to the example for the specific meaning and output format. Sample input

3 2ABCCDEEFGFABE0 0

Sample output

great-grandparent-

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
using namespace Std;
Char cmp (char A,char b) {
Return a<b?0:1;
}
int main () {
int m,n;
while (cin>>m>>n&& (m!=0&&n!=0)) {
String Str[m];
for (int i=0;i<m;i++) {
cin>>str[i];
}
Char wt[n][2];
for (int i=0;i<n;i++) {
for (int j=0;j<2;j++) {
cin>>wt[i][j];
}
}
for (int i=0;i<n;i++) {
int temp=0,index=0;
int j=m-1;
int count=0;
if (Wt[i][1]<wt[i][0]) {index=1;};
Sort (wt[i],wt[i]+2,cmp);
while (j>=0) {
if (Str[j].find (wt[i][0])!=string::npos) {
count++;
WT[I][0]=STR[J][0];
}
if (Wt[i][1]==str[j][0]) {temp=j;break;}
j--;
}
if (wt[i][1]!=str[temp][0]) count=0;
if (index==1) {
while (COUNT&GT;2) {
cout<< "great-";
count--;
}
if (count==2)
cout<< "Grandparent" <<endl;
if (count==1)
cout<< "Parent" <<endl;
if (count==0)
cout<< "-" <<endl;
}
else{
while (COUNT&GT;2) {
cout<< "great-";
count--;
}
if (count==2)
cout<< "grandchild" <<endl;
if (count==1)
cout<< "Child" <<endl;
if (count==0)
cout<< "-" <<endl;
}
}
}
return 0;
}

1050: Find the immediate family