Idiomatic Phrases Game
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3029 Accepted Submission(s):
982
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
Author
ZHOU, Ran
Source
Recommend
最短路的变形，只需保存每个字符串的前四个后四个进行比较，再使用最短路构造即可。
题意：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。时间花费为前面成语的时间。
附上代码：
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #define M 1005 5 #define MAX 0x3f3f3f3f 6 using namespace std; 7 struct node 8 { 9 char f[5],e[5]; //只需记录每个字符串的前四位和后四位比较即可 10 int t; 11 } ss[M]; 12 int map[M][M],dis[M],vis[M]; 13 int main() 14 { 15 int n,m,i,j; 16 char s[105]; 17 while(~scanf("%d",&n)&&n) 18 { 19 for(i=0; i<n; i++) 20 { 21 scanf("%d%s",&ss[i].t,s); 22 int len=strlen(s); 23 for(j=0; j<4; j++) 24 { 25 ss[i].f[j]=s[j]; //保存前四位 26 ss[i].e[3-j]=s[--len]; //保存后四位 27 } 28 ss[i].f[4]=ss[i].e[4]=' '; //字符串以' '结尾，很重要！ 29 } 30 for(i=0; i<n; i++) 31 for(j=0; j<n; j++) 32 { 33 if(i==j) map[i][j]=0; 34 else map[i][j]=MAX; 35 if(strcmp(ss[i].e,ss[j].f)==0) //后四个和前四个一样，则可以传递 36 map[i][j]=ss[i].t; 37 } 38 memset(vis,0,sizeof(vis)); 39 memset(dis,0,sizeof(dis)); 40 for(i=0; i<n; i++) 41 dis[i]=map[0][i]; 42 vis[0]=1; 43 int min,t; 44 for(i=0; i<n; i++) 45 { 46 min=MAX; 47 for(j=0; j<n; j++) 48 { 49 if(!vis[j]&&dis[j]<min) 50 { 51 min=dis[j]; 52 t=j; 53 } 54 } 55 vis[t]=1; 56 for(j=0; j<n; j++) 57 if(!vis[j]&&map[t][j]<MAX) 58 if(dis[j]>dis[t]+map[t][j]) 59 dis[j]=dis[t]+map[t][j]; 60 } 61 if(dis[n-1]<MAX) 62 printf("%d ",dis[n-1]); 63 else 64 printf("-1 "); 65 } 66 return 0; 67 }