題目描述
如果A,B是C的父母親,則A,B是C的parent,C是A,B的child,如果A,B是C的(外)祖父,祖母,則A,B是C的grandparent,C是A,B的grandchild,如果A,B是C的(外)曾祖父,曾祖母,則A,B是C的great-grandparent,C是A,B的great-grandchild,之後再多一輩,則在關系上加一個great-。
輸入
輸入包含多組測試用例,每組用例首先包含2個整數n(0<=n<=26)和m(0
great-grandparent
AC代碼
#include<cstdio>
#include<iostream>
using namespace std;
const int INF=;
int map[][];
void Floyd() {
for(int k = ; k <= ; k ++)
for(int i = ; i <= ; i ++)
for(int j = ; j <= ; j ++)
if(map[i][j] > map[i][k] + map[k][j])
map[i][j] = map[i][k] + map[k][j];
}
int minn(int x, int y) {
return x < y ? x : y;
}
int main () {
int n,m,i,j,a,b,c;
char str[];
while(scanf("%d %d",&n,&m) && n + m) {
for(i = ; i <= ; i ++)
for(j = ; j <= ; j ++)
if(i == j)map[i][j] = ;
else map[i][j] = INF;
for(i = ; i <= n ; i ++) {
scanf("%s",str);
a = str[];
b = str[];
c = str[];
// cout<<a<<b<<c<<endl;
map[a][b] = map[a][c] = ;
}
Floyd();
for(i = ; i <= m ; i ++) {
scanf("%s",str);
a = str[];
b = str[];
c = minn(map[a][b],map[b][a]);
if(c == INF || !c) {
printf("-\n");
continue;
}
if(c == ) {
if(map[a][b] == )
printf("child\n");
else
printf("parent\n");
} else {
if(map[a][b] != INF) {
c -= ;
for(j = ; j <= c ; j ++)
printf("great-");
printf("grandchild\n");
} else {
c -= ;
for(j = ; j <= c ; j ++)
printf("great-");
printf("grandparent\n");
}
}
}
}
return ;
}