Idiomatic Phrases Game

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

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

其实弄懂这道题后，就会发现。。水题一枚。

给出n个“成语”， 这写成语至少由3个“汉字”组成，所谓的“汉字”，是指4个连续的16进制数字（1～9， A～F）。

以第一个成语作为起点，最后一个作为终点， 需要找出一个序列，这个序列的前一个成语的最后一个“汉字”与后一个成语的第一个“汉字”是相同的，求最少花费时间。

代码：

#include <stdio.h>
#include <string.h>
#define INF 1000000000
struct Node{
	char s[5],e[5] ;
	int t ;
}list[1100];
int graph[1100][1100] , dis[1100];

void dijkstra(int n)
{
	bool visited[1100] ;
	for(int i = 0 ; i < n ; ++i)
	{
		dis[i] = graph[0][i] ;
		visited[i] = false ;
	}
	dis[0] = 0 ;
	visited[0] = true ;
	for(int i = 1 ; i < n ; ++i)
	{
		int min = INF , index = -1 ;
		for(int j = 0 ; j < n ; ++j)
		{
			if(!visited[j] && min>dis[j])
			{
				index = j ;
				min = dis[j] ;
			}
		}
		if(index == -1)
		{
			return ;
		}
		visited[index] = true ;
		for(int j = 0 ; j < n ; ++j)
		{
			if(!visited[j] && dis[j]>min+graph[index][j])
			{
				dis[j] = min+graph[index][j] ;
			}
		}
	}
}

int main()
{
	int n ;
	while(scanf("%d",&n) && n)
	{
		for(int i = 0 ; i < n ; ++i)
		{
			for(int j = 0 ; j < n ; ++j)
			{
				graph[i][j] = INF ;
			}
		}
		char str[110] ;
		for(int i = 0 ; i < n ; ++i)
		{
			scanf("%d%s",&list[i].t,str) ;
			for(int j = 0 ; j < 4 ; ++j)
			{
				list[i].s[j] = str[j] ;
			}
			list[i].s[4] = '\0' ;
			int k = 0 ;
			for(int j = strlen(str)-4 ; j < strlen(str) ; ++j)
			{
				list[i].e[k++] = str[j] ;
			}
			list[i].e[4] = '\0' ;
		}
		for(int i = 0 ; i < n-1 ; ++i)
		{
			for(int j = 0 ; j < n ; ++j)
			{
				if(strcmp(list[i].e,list[j].s)==0)
				{
					graph[i][j] = list[i].t ;
				}
			}
		}
		dijkstra(n) ;
		if(dis[n-1] == INF)
			puts("-1") ;
		else
			printf("%d\n",dis[n-1]) ;
	}
	return 0 ;
}

与君共勉