【ZOJ2750】Idiomatic Phrases Game（最短路）

题目链接

Idiomatic Phrases Game

Time Limit: 2 Seconds Memory Limit: 65536 KB

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

【题意】

成语接龙，每个成语由4个16进制数组成，看是否能从第一个成语接到最后一个成语，接每个成语需要翻字典有耗时，如果能接到最后一个成语，则输出最少耗时，反之输出-1。

【解题思路】

将首尾能对应成语建边，权值为时间，然后用dijkstra跑一遍最短路。如果dis[n-1]=INF，说明没有能接到最后一个成语，所以输出-1，否则输出最短路。

【代码】

#include<bits/stdc++.h>
using namespace std;
const int INF=999999999;
const int maxn=1005;
int edge[maxn][maxn],vis[maxn],dis[maxn];
char s[maxn];
int n;
struct Node
{char a[5],b[5];int time;
}node[maxn];
void dijkstra()
{for(int i=0;i<n;i++)dis[i]=INF;dis[0]=0;while(1){int v=-1;for(int i=0;i<n;i++){if(!vis[i] && (v==-1 |dis[i]<dis[v]))v=i;}if(v==-1)break;vis[v]=1;for(int i=0;i<n;i++)dis[i]=min(dis[i],dis[v]+edge[v][i]);}
}
int main()
{while(~scanf("%d",&n) && n){for(int i=0;i<n;i++){scanf("%d %s",&node[i].time,s);int l=strlen(s);for(int k=0,j=l-1;k<4;j--,k++){node[i].a[k]=s[k];node[i].b[3-k]=s[j];}node[i].a[4]=node[i].b[4]='\0';}for(int i=0;i<n;i++){for(int j=0;j<n;j++){edge[i][j]=INF;if(i==j)continue;if(strcmp(node[i].b,node[j].a)==0)edge[i][j]=node[i].time;}}memset(vis,0,sizeof(vis));dijkstra();if(dis[n-1]==INF)printf("-1\n");else printf("%d\n",dis[n-1]);}return 0;
}

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