zoj 2750 Idiomatic Phrases Game

ZOJ Problem Set - 2750

Idiomatic Phrases Game

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Time Limit: 1 Second      Memory Limit: 32768 KB

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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

Author: ZHOU, Ran

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Source: Zhejiang Provincial Programming Contest 2006
Submit    Status

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//1858447 2009-05-07 15:18:33 Wrong Answer 2750 C++ 530 4124 Wpl //1858474 2009-05-07 15:42:48 Accepted 2750 C++ 570 4124 Wpl //简单的Dijkstra,注意取最短边长的时候考虑可能不通的情况. //简单的成语接龙,很有意思. #include <iostream> #include <string> #include <queue> #define MAX 1002 using namespace std; typedef struct node //定义优先队列结点 { int di; int heap; node(){} node(int h,int d) { heap=h; di=d; } friend bool operator<(node a,node b) //用小堆 { return a.di>b.di; } }Point; priority_queue<Point>Q; struct nodeData { int time; string s; string e; }data[MAX]; Point temp; int n,dis[MAX],edges[MAX][MAX]; bool mark[MAX]; void Init() { int i,j,len; string str; for(i=1;i<=n;i++) { scanf("%d",&data[i].time); cin>>str; len=str.length(); data[i].s=str.substr(0,4); data[i].e=str.substr(len-4,4); } for(i=1;i<=n;i++) for(j=1;j<=n;j++) edges[i][j]=-1; for(i=1;i<=n;i++) for(j=1;j<=n;j++) { if(data[i].e==data[j].s) //单向的 edges[i][j]=data[i].time; } } void Dijkstra(int s) //s是开始点 { int i,tp,td,k,j; while(!Q.empty()) Q.pop(); for(i=1;i<=n;i++) { dis[i]=edges[s][i]; if(dis[i]!=-1) { Q.push(node(i,dis[i])); //将有效的边入队列 } mark[i]=false; } mark[s]=true; //dis[1]=0; for(i=1;i<n;i++) { k=-1; //k最好有初始化值 while(!Q.empty()) { temp=Q.top(); Q.pop(); if(!mark[temp.heap]) { k=temp.heap; break; } } if(k==-1) { printf("-1/n"); return ; } mark[k]=true; if(k==n) //这里已经可以返回,n是终点 { printf("%d/n",dis[n]); return ; } for(j=1;j<=n;j++) { if(mark[j]||edges[k][j]==-1) continue; if(dis[k]+edges[k][j]<dis[j]||dis[j]==-1) { dis[j]=dis[k]+edges[k][j]; Q.push(node(j,dis[j])); // } } } } int main() { while(scanf("%d",&n)!=EOF&&n!=0) { Init(); /* if(n==1) { if(edges[1][1]!=-1) printf("%d/n",data[1].time); else printf("-1/n"); continue; }*/ Dijkstra(1); } return 0; }