Floyd---String Problem

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本文介绍了一个字符串转换游戏的算法实现，玩家需将两个字符串通过一系列字符替换操作变为相同，目标是最小化操作成本。文章详细解释了如何使用Floyd算法预处理字符间转换的代价，以找到最低成本的解决方案。

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Boy Valera likes strings. And even more he likes them, when they are identical. That's why in his spare time Valera plays the following game. He takes any two strings, consisting of lower case Latin letters, and tries to make them identical. According to the game rules, with each move Valera can change one arbitrary character Ai in one of the strings into arbitrary character Bi, but he has to pay for every move a particular sum of money, equal to Wi. He is allowed to make as many moves as he needs. Since Valera is a very economical boy and never wastes his money, he asked you, an experienced programmer, to help him answer the question: what minimum amount of money should Valera have to get identical strings.

Input

The first input line contains two initial non-empty strings s and t, consisting of lower case Latin letters. The length of each string doesn't exceed 105. The following line contains integer n (0 ≤ n ≤ 500) — amount of possible changings. Then follow n lines, each containing characters Ai and Bi (lower case Latin letters) and integer Wi (0 ≤ Wi ≤ 100), saying that it's allowed to change character Ai into character Bi in any of the strings and spend sum of money Wi.

Output

If the answer exists, output the answer to the problem, and the resulting string. Otherwise output -1 in the only line. If the answer is not unique, output any.

Examples

Input

uayd
uxxd
3
a x 8
x y 13
d c 3

Output

21
uxyd

Input

a
b
3
a b 2
a b 3
b a 5

Output

2
b

Input

abc
ab
6
a b 4
a b 7
b a 8
c b 11
c a 3
a c 0

Output

-1

题目大意：

给你a，b两个字符串，n中变化，接下来n行，每行表示从字符1到字符2要花费代价val，问将a，b变成一样所需花费代价最小值，不能输出-1。

思路：将26个字符之间的变化代价用Floyd跑出来，求最小即可。

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
const int maxn=1e6;
char s1[maxn];
char s2[maxn];
int dis[30][30];
int main()
{

    int n;
    while(cin>>s1>>s2>>n)
    {
        int flag=1;
        int len1=strlen(s1);
        int len2=strlen(s2);
        if(len1!=len2)
        {
            flag=0;
        }
        for(int i=0; i<26; i++)
        {

            for(int j=0; j<26; j++)
            {
                dis[i][j]=1001;
            }
        }
        for(int i=0; i<n; i++)

        {

            char c1,c2;
            cin>>c1>>c2;
            int temp;
            cin>>temp;
            dis[c1-'a'][c2-'a']=min(temp, dis[c1-'a'][c2-'a']);
        }

        for(int i=0;i<26;i++)
            dis[i][i]=0;

        for(int k=0;k<26;k++)
            for(int i=0;i<26;i++)
                for(int j=0;j<26;j++)
                    if(dis[i][j]>dis[i][k]+dis[k][j])
                        dis[i][j]=dis[i][k]+dis[k][j];

        int sum=0;

        if(flag)
        {
            for(int i=0;i<len1;i++)
            {
                if(s1[i]!=s2[i])
                {
                    int id;
                    int temp=1001;

                    for(int j=0;j<26;j++)
                    {
                        if(dis[s1[i]-'a'][j]!=1001&&dis[s2[i]-'a'][j]!=1001&&temp>dis[s1[i]-'a'][j]+dis[s2[i]-'a'][j])
                        {
                            id=j;
                            temp=dis[s1[i]-'a'][j]+dis[s2[i]-'a'][j];
                        }

                    }
                    if(temp==1001)
                    {
                        flag=0;
                        break;
                    }
                    sum+=temp;
                    s1[i]=s2[i]=id+'a';
                }
            }
        }

    if(flag)
        cout<<sum<<endl<<s1<<endl;
    else
        cout<<-1<<endl;
    }
    return 0;
}