Idiomatic Phrases Game

Problem Description

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

5

5 12345978ABCD2341

5 23415608ACBD3412

7 34125678AEFD4123

15 23415673ACC34123

4 41235673FBCD2156

2

20 12345678ABCD

30 DCBF5432167D

0

Sample Output

17

-1

  1 #include<map>
  2 #include<queue>
  3 #include<cstdio>
  4 #include<string>
  5 #include<cstring>
  6 #include<algorithm>
  7 using namespace std;
  8
  9 struct edge
 10 {
 11     int from,to,time;
 12     edge(int u,int v,int w):from(u),to(v),time(w){};
 13 };
 14
 15 struct Node
 16 {
 17     int st;
 18     int ed;
 19     int val;
 20 }node[1005];
 21
 22 struct heapnode
 23 {
 24     int u,d;
 25     bool operator < (const heapnode &temp)const
 26     {
 27         return d>temp.d;
 28     }
 29 };
 30
 31 const int inf =0x3f3f3f3f;
 32 map<string,int>key;
 33 vector<edge>edges;
 34 vector<int>G[1005];
 35 int d[1005],vis[1005];
 36 int num,n;
 37
 38 void init()
 39 {
 40     for(int i=0;i<n;i++)
 41     G[i].clear();
 42     edges.clear();
 43     memset(vis,0,sizeof(vis));
 44     key.clear();
 45 }
 46
 47 void addedge(int u,int v,int w)
 48 {
 49     edges.push_back(edge(u,v,w));
 50     int m=edges.size();
 51     G[u].push_back(m-1);
 52 }
 53
 54 void dijkstra(int S)
 55 {
 56     priority_queue<heapnode>q;
 57     q.push((heapnode){S,0});
 58     for(int i=0;i<n;i++)
 59     d[i]=inf;
 60     d[S]=0;
 61     while(!q.empty())
 62     {
 63         int u=q.top().u;
 64         q.pop();
 65         if(vis[u])
 66         continue;
 67         vis[u]=1;
 68         for(int i=0;i<G[u].size();i++)
 69         {
 70             edge& e=edges[G[u][i]];
 71             if(d[e.to]>d[u]+e.time)
 72             {
 73                 d[e.to]=d[u]+e.time;
 74                 q.push((heapnode){e.to,d[e.to]});
 75             }
 76         }
 77     }
 78 }
 79
 80 int main()
 81 {
 82     //freopen("in.txt","r",stdin);
 83     char s[105],sub1[5],sub2[5];
 84     string temp1,temp2;
 85     int i,j,val,t,k;
 86     while(scanf("%d",&t),t)
 87     {
 88         j=num=0;
 89         n=t;
 90         init();
 91         while(t--)
 92         {
 93             scanf("%d %s",&val,s);
 94             int len=strlen(s);
 95             for(i=0;i<4;i++)
 96             sub1[i]=s[i];
 97             sub1[i]=‘\0‘;
 98             temp1=sub1;
 99             if(!key.count(temp1))
100             key[temp1]=num++;
101             for(k=0,i=len-4;i<=len;i++,k++)
102             sub2[k]=s[i];
103             temp2=sub2;
104             if(!key.count(temp2))
105             key[temp2]=num++;
106             node[j].st=key[temp1];
107             node[j].ed=key[temp2];
108             node[j].val=val;
109             for(int k=0;k<j;k++)
110             if(node[k].ed==node[j].st)
111             addedge(k,j,node[k].val);
112             else if(node[j].ed==node[k].st)
113             addedge(j,k,node[j].val);
114             else;
115             j++;
116         }
117         dijkstra(0);
118         if(d[n-1]==inf)
119         printf("-1\n");
120         else
121         printf("%d\n",d[n-1]);
122     }
123     return 0;
124 }