HDU1546 Idiomatic Phrases Game 最短路徑(Dijkstra演算法)

Problem DescriptionTom 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.

InputThe 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.

OutputOne 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 Input5 5 12345978ABCD2341 5 23415608ACBD3412 7 34125678AEFD4123 15 23415673ACC34123 4 41235673FBCD2156 2 20 12345678ABCD 30 DCBF5432167D 0

Sample Output17 -1參考代碼：

#include<iostream>using namespace std;#define INF 1000000000#define MAXN 1000struct idiom{char front[5],back[5];//記錄成語的頭尾兩個字int T;};idiom dic[MAXN];//字典int Edge[MAXN][MAXN];//鄰接矩陣int dist[MAXN];//可以說是記錄權值int S[MAXN];//判斷是否加入頂點集合中int N;int main(){int i,j,k;char s[100];int len;while(cin>>N){if(N==0)break;for(k=0;k<N;k++){cin>>dic[k].T>>s;len=strlen(s);for(i=0,j=len-1;i<4;i++,j--){dic[k].front[i]=s[i];dic[k].back[3-i]=s[j];}dic[k].front[4]=dic[k].back[4]='\0';}for(i=0;i<N;i++)//建圖{for(j=0;j<N;j++){Edge[i][j]=INF;if(i==j)continue;if(strcmp(dic[i].back,dic[j].front)==0)Edge[i][j]=dic[i].T;}}//Dijskal演算法的開始for(i=0;i<N;i++)//初始化{dist[i]=Edge[0][i];S[i]=0;}S[0]=1;dist[0]=0;//把頂點加入到頂點集合中for(i=0;i<N-1;i++)//確定N-1條路徑{int min=INF;//選擇當前集合T中具有最短路徑的頂點uint u=0;for(j=0;j<N;j++){if(!S[j]&&dist[j]<min){u=j;min=dist[j];}}S[u]=1;//將u加入到頂點集合中for(k=0;k<N;k++)//修改集合T中的dist數組元素值{if(!S[k]&&Edge[u][k]<INF&&(dist[u]+Edge[u][k])<dist[k]){dist[k]=dist[u]+Edge[u][k];}}}if(dist[N-1]==INF)cout<<-1<<endl;elsecout<<dist[N-1]<<endl;}return 0;}