WOJ1043-Magic Star

There is a story that people who have the similar names will get along well with each other.So when finding a girl friend, you can

take this point into consideration ^_^.

In MagicStar , people have a way to count the comparability between two names : the comparability between "kinfkong" and "kingkong"

is 3 because they have the same prefix "kin", while it is 0 between the names "kinfkong" and "dingkong" which have not common prefix.

Every person has an ID and a name. Dr.Longge want to count the sum of the comparabilities of q pairs of persons in the same set.

Can you help him?

输入格式

In the first line there is n (1

and a name for a person. The length of the names is at least one but no more than 1000.IDs are in the range of 1 and n.
No two IDs are the same. Next line is a number m. Then m sets follow.Each set starts with a number q in a line, meaning there are q queries
in this set. Every query is made up of two IDs in a line. (m*q <= 10^6) .

输出格式

You task is to output m numbers, one in a line . The ith number is total comparabilities of the ith set.

样例输入

4
1 kingkong
2 dinfkong
3 kinfkong
4 kinfkonq
2
2
1 2
3 4
1
1 2

样例输出

7
0

#include
#include
char name[6001][1001];
char n1[1001],n2[1002];
int n,m,q;
int main(){
	int ans,i,j,k;
	scanf("%d",&n);
	while(n--){
		scanf("%d",&i);
		scanf("%s",&name[i]);
	}
	scanf("%d",&m);
	while(m--){
		scanf("%d",&q);
		ans=0;
		while(q--){		
			scanf("%d %d",&i,&j);
			for(k=0;name[i][k]!='\0'&&name[j][k]!='\0';k++){
				if(name[i][k]!=name[j][k])
				break;
			}
			ans+=k;
		}
		printf("%d\n",ans);
	}
	return 0;
}