Codeforces 33B. String Problem（Floyd最短路）

题目链接：http://codeforces.com/contest/33/problem/B

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

Copy

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

output

Copy

21
uxyd

input

Copy

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

output

Copy

2
b

input

Copy

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

output

Copy

-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。

代码：

#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<queue>
#include<stack>
#include<map>

using namespace std;

#define FOU(i,x,y) for(int i=x;i<=y;i++)
#define FOD(i,x,y) for(int i=x;i>=y;i--)
#define MEM(a,val) memset(a,val,sizeof(a))
#define PI acos(-1.0)

const double EXP = 1e-9;
typedef long long ll;
typedef unsigned long long ull;
const int INF = 0x3f3f3f3f;
const ll MINF = 0x3f3f3f3f3f3f3f3f;
const double DINF = 0xffffffffffff;
const int mod = 1e9+7;
const int N = 1e5+5;

char a[N],b[N],c[N];
int change[30][30];

int mp[30][30];
int d[30][30];

void Floyd()
{
    int n=25;
    for(int i=0;i<=n;i++)
        for(int j=0;j<=n;j++)
            d[i][j]=mp[i][j];

    for(int k=0;k<=n;k++)  //枚举以k为中间点的所有点的最短路
        for(int i=0;i<=n;i++)
            for(int j=0;j<=n;j++)
                d[i][j]=min(d[i][j],d[i][k]+d[k][j]);
//    for(int i=0;i<26;i++)
//    {
//        for(int j=0;j<26;j++)
//        {
//            printf("%c->%c=%d\n",'a'+i,'a'+j,d[i][j]);
//        }
//    }
}

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    std::ios::sync_with_stdio(false);
    int n,val;
    char x,y;
    scanf(" %s %s",a,b);
    scanf("%d",&n);
    for(int i=0;i<30;i++)
        for(int j=0;j<30;j++)
        {
            if(i==j)
                mp[i][j]=0;
            else
                mp[i][j]=INF;
        }
    for(int i=1;i<=n;i++)
    {
        scanf(" %c %c %d",&x,&y,&val);
        int t1=x-'a';
        int t2=y-'a';
        mp[t1][t2]=min(mp[t1][t2],val);
    }
    Floyd();
    int len1=strlen(a);
    int len2=strlen(b);
    if(len1!=len2)
    {
        printf("-1\n");
        return 0;
    }
    bool flag = false;
    int ans=0;
    for(int i=0;i<len1;i++)
    {
        int minn=INF;
        int t1=a[i]-'a';
        int t2=b[i]-'a';
        for(int j=0;j<26;j++)
        {
            int x1 = d[t1][j];
            int x2 = d[t2][j];
            //printf("%c\n",'a'+j);
            //printf("x1=%d\n",x1);
            //printf("x2=%d\n",x2);
            if(x1!=INF&&x2!=INF&&minn>x1+x2)
            {
                minn=x1+x2;
                c[i]='a'+j;
            }
        }
        //printf("%d=%d\n",i,minn);
        if(minn==INF)
        {
            flag=true;
            break;
        }
        ans+=minn;
    }
    if(flag)
        printf("-1\n");
    else
    {
        c[len1]='\0';
        printf("%d\n",ans);
        printf("%s\n",c);
    }
    return 0;
}