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<n<=6000) indicating the number of people in MagicStar. The following n lines each containing an ID

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

41 kingkong2 dinfkong3 kinfkong4 kinfkonq221 23 411 2

样例输出

70

#include<stdio.h>#include<string.h>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;}