Idiomatic Phrases Game

Tom is playing a game called Idiomatic Phrases Game. An idiom consists of several Chinese characters and has a certain meaning. This game will give Tom two idioms. He should build a list of idioms and the list starts and ends with the two given idioms. For every two adjacent idioms, the last Chinese character of the former idiom should be the same as the first character of the latter one. For each time, Tom has a dictionary that he must pick idioms from and each idiom in the dictionary has a value indicates how long Tom will take to find the next proper idiom in the final list. Now you are asked to write a program to compute the shortest time Tom will take by giving you the idiom dictionary.

Input

The input consists of several test cases. Each test case contains an idiom dictionary. The dictionary is started by an integer N (0 < N < 1000) in one line. The following is N lines. Each line contains an integer T (the time Tom will take to work out) and an idiom. One idiom consists of several Chinese characters (at least 3) and one Chinese character consists of four hex digit (i.e., 0 to 9 and A to F). Note that the first and last idioms in the dictionary are the source and target idioms in the game. The input ends up with a case that N = 0. Do not process this case.

Output

One line for each case. Output an integer indicating the shortest time Tome will take. If the list can not be built, please output -1.

Sample Input

55 12345978ABCD23415 23415608ACBD34127 34125678AEFD412315 23415673ACC341234 41235673FBCD2156220 12345678ABCD30 DCBF5432167D0

Sample Output

17-1

这个题难点还是在边的创建，给我们一堆点，问我们最短的路径，然后我们没有什么好办法，只能两个for循环，然后符合条件的创建边。

#include<stdio.h>#include<string.h>#define INF 0x3f3f3f3fusing namespace std;int n;struct//每个点的权值，和字符{    int w;    char str[100];}node[1000];int mp[1000][1000];int dis[1000];int vis[1000];void dijkstra(){    memset(vis,0,sizeof(vis));    memset(dis,INF,sizeof(dis));    dis[0]=0;    for(int i=0;i<=n;i++)    {        int temp=INF;int m;        for(int j=0;j<=n;j++)        {            if(!vis[j]&&dis[j]<temp)            {                temp=dis[j];                m=j;            }        }        vis[m]=1;        for(int j=0;j<=n;j++)        {            if(dis[j]>dis[m]+mp[m][j])                dis[j]=dis[m]+mp[m][j];        }    }}int main(){    while(~scanf("%d",&n)&&n)    {        memset(mp,INF,sizeof(mp));        for(int i=0;i<n;i++)        {            scanf("%d %s",&node[i].w,node[i].str);        }        for(int i=0;i<n;i++)        {            for(int j=0;j<n;j++)            {                if(i!=j)                {                    int len=strlen(node[i].str);                    if(node[i].str[len-4]==node[j].str[0]&&node[i].str[len-3]==node[j].str[1]&&node[i].str[len-2]==node[j].str[2]&&node[i].str[len-1]==node[j].str[3])                        mp[i][j]=node[i].w;//标定u,v,w                }            }        }        dijkstra();        if(dis[n-1]==INF)            printf("-1\n");        else            printf("%d\n",dis[n-1]);    }    return 0;}