cf#33-B - String Problem-反向建图+最短路

http://codeforces.com/problemset/problem/33/B

给出字符串s,t；n个字母替换关系  x,y，w

题意：求使得给出两个字符串完全一致的代价，【可根据一个有向图替换字符，并消耗相应代价】

M - String Problem

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status

Description

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 A_i in one of the strings into arbitrary character B_i, but he has to pay for every move a particular sum of money, equal to W_i. 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 A_i and B_i (lower case Latin letters) and integer W_i (0 ≤ W_i ≤ 100), saying that it's allowed to change characterA_i into character B_i in any of the strings and spend sum of money W_i.

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.

Sample Input

Input

uayduxxd3a x 8x y 13d c 3

Output

21uxyd

Input

ab3a b 2a b 3b a 5

Output

2b

Input

abcab6a b 4a b 7b a 8c b 11c a 3a c 0

Output

-1

字母表比较小，预处理所有方案，（i-j替换成一致的）最优方案是反向建图的最短路,n^2 * n^2 (可做到n^3,n=26就没必要了)  ,然后O(n)遍历即可

反向建图的意思：  题目要求x,y的最短一致代价，即找出一条x联通y的最短路，但是由于给的边是有向的，可能最后的结果会是  x->k,y->k 即都替换成k,但是x,y并不直接联通，这样不方便用最短路，那么直接把边的方向反过来，枚举所有点作为起点，  求min(dis[x]+dis[y]) 得到的便是实际中最短的一条路了 

代码中直接n*n*(n^2)了，实际直接n*n*n也可。。。

 

思路很快好了，傻逼错误犯了不少。。最短路的 break写成return，外循环变量i与内部重复了，更新路径判断条件写错了。

注意: 图的权重存在为0的，应该初始化为-1而不是0！

#include <cstdio>#include <cmath>#include <cstring>#include <string>#include <algorithm>#include <iostream>#include <queue>#include <map>#include <set>#include <vector>#include<stack>using namespace std;  const double pi=acos(-1.0);const double eps=1e-6; const int inf=2147483647;char s1[100005];char tmp[100005];char s2[100005];int tm[30][30];int show[28];int check_dis[28][28];char check_ret[28][28];int get_dis(int st,int a,int b){int n=26;int i;int x=st; int dis[30];int vis[30];memset(dis,0,sizeof(dis));memset(vis,0,sizeof(vis));for (i=1;i<=n;i++){if (tm[x][i]!=-1)dis[i]=tm[x][i];elsedis[i]=inf;}vis[x]=1;dis[x]=0;for (int k=1;k<n;k++) //外循环变量i，与内循环变量重复，导致出现问题{int min_edge=inf;int minp=-1;for (i=1;i<=n;i++){if (!vis[i]&&dis[i]<min_edge){min_edge=dis[i];minp=i;}}if (minp==-1) break;//最短路写傻逼了。直接return -1;vis[minp]=1;for (i=1;i<=n;i++){if (!vis[i]&&tm[minp][i]!=-1&&dis[i]>min_edge+tm[minp][i])//*手误写错了dis[i]=min_edge+tm[minp][i];}}if (dis[a]==inf||dis[b]==inf) return -1;elsereturn dis[a]+dis[b];}int get_min_dis(char a,char b,char &tt){int x=a-'a'+1;int y=b-'a'+1;int i;int mini=-1;int ans=inf;for (i=1;i<=26;i++){if (!show[i]) continue;int ret=get_dis(i,x,y);if (ret==-1) continue;else{if (ret<ans) {mini=i;ans=ret;}}}if (ans==inf)return -1;tt=mini+'a'-1;return ans;}int main(){int i,j;int n;scanf("%s%s",s1,s2);cin>>n;getchar();char xx,yy;int x,y,w; memset(tm,-1,sizeof(tm));//权重存在为0,所以需要初始化为-1！！for (i=1;i<=n;i++){scanf("%c %c %d",&xx,&yy,&w);getchar();x=xx-'a'+1;y=yy-'a'+1;show[x]=1;show[y]=1;if (-1==tm[y][x])tm[y][x]=w;elseif (w<tm[y][x]) tm[y][x]=w;}int len1=strlen(s1);int len2=strlen(s2);if (len1!=len2){printf("-1\n"); return 0;}for (i=1;i<=26;i++){for (j=i+1;j<=26;j++){char tt;int ret=get_min_dis(i+'a'-1,j+'a'-1,tt);check_dis[i][j]=check_dis[j][i]=ret;check_ret[i][j]=check_ret[j][i]=tt;} }strcpy(tmp,s1);int ans=0;for (i=0;i<len1;i++){if (s1[i]==s2[i]) continue;char tt=check_ret[s1[i]-'a'+1][s2[i]-'a'+1];int ret=check_dis[s1[i]-'a'+1][s2[i]-'a'+1];if (ret==-1){printf("-1\n"); return 0;}tmp[i]=tt; ans+=ret;}printf("%d\n%s\n",ans,tmp);return 0;}