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
迪杰斯特拉算法，加了一些离散化。
算法核心：
1.建起点到每条边的边权INF，与起点相连的为初始边权。
2.每次取最近的点，标记访问情况。
3.将该点相邻的点更新与起点的距离。
执行以上知道所有点都访问过
这样以来数组d中储存的就是起点到其他所有点的最短距离。
1 #include<cstring> 2 #include<cstdio> 3 #include<algorithm> 4 #include<cstdio> 5 #include<vector> 6 #include<iostream> 7 #include<map> 8 #include<string> 9 10 using namespace std; 11 12 #define INF 0x3f3f3f3f 13 14 map<string,int>mp; 15 vector<int>in[1005]; 16 int cost[1005][1005]; 17 int vis[1005]; 18 int cur,aim; 19 int n; 20 21 void init() 22 { 23 cur=0; 24 for(int i=0;i<=1000;i++) vis[i]=0,in[i].clear(); 25 for(int i=0;i<=1000;i++) 26 for(int j=0;j<=1000;j++) 27 cost[i][j]=INF; 28 mp.clear(); 29 } 30 31 int D(int s,int e) 32 { 33 int d[1005]; 34 for(int i=0;i<cur;i++) d[i]=INF; 35 36 int k,maxn=INF; 37 38 for(int i=0;i<in[s].size();i++) 39 d[in[s][i]]=cost[s][in[s][i]]; 40 41 d[s]=0; 42 43 vis[s]=1; 44 45 while(1) 46 { 47 maxn=INF; 48 k=-1; 49 for(int i=0;i<cur;i++) 50 { 51 if(d[i]<maxn&&!vis[i]) 52 maxn=d[i],k=i; 53 } 54 55 if(k==-1) break; 56 vis[k]=1; 57 58 for(int i=0;i<in[k].size();i++) 59 if(!vis[in[k][i]]&&cost[k][in[k][i]]+maxn<d[in[k][i]]) 60 d[in[k][i]]=cost[k][in[k][i]]+maxn; 61 } 62 63 return d[e]; 64 } 65 66 int main() 67 { 68 while(scanf("%d",&n)!=EOF&&n) 69 { 70 init(); 71 int fee; 72 for(int i=0;i<n;i++) 73 { 74 75 int co; 76 string temp; 77 cin>>co>>temp; 78 79 string tm; 80 string tn; 81 82 for(int i=0;i<4;i++) tm+=temp[i]; 83 for(int i=temp.size()-4;i<temp.size();i++) tn+=temp[i]; 84 85 int s,t; 86 if(mp[tm]&&mp[tn]) 87 { 88 s=mp[tm],t=mp[tn]; 89 in[s].push_back(t); 90 } 91 else if(mp[tm]&&!mp[tn]) 92 { 93 mp[tn]=cur++; 94 s=mp[tm],t=mp[tn]; 95 in[s].push_back(t); 96 } 97 else if(!mp[tm]&&mp[tn]) 98 { 99 mp[tm]=cur++; 100 s=mp[tm],t=mp[tn]; 101 in[s].push_back(t); 102 } 103 else if(!mp[tm]&&!mp[tn]) 104 { 105 mp[tm]=cur++;mp[tn]=cur++; 106 s=mp[tm],t=mp[tn]; 107 in[s].push_back(t); 108 } 109 110 cost[s][t]=min(co,cost[s][t]); //此处一定要注意更新最小边权 111 112 if(i==n-1) fee=co,aim=mp[tm]; 113 } 114 115 int flag=D(0,aim); 116 if(flag>=INF) cout<<"-1"<<endl; 117 else cout<<flag<<endl; 118 } 119 }