ZOJ Problem Set - 2750

Idiomatic Phrases Game

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Time Limit: 2 Seconds      Memory Limit: 65536 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

55 12345978ABCD23415 23415608ACBD34127 34125678AEFD412315 23415673ACC341234 41235673FBCD2156220 12345678ABCD30 DCBF5432167D0

Sample Output

17-1

这道题目就是比较裸的dijkstra了，首先输入字符串，获取前4个字符和后4个字符，然后建图。之后从第1个字符开始来一遍dijkstra，来获得1是否可达n。

代码：

#include<cstdio>#include<iostream>#include<cstring>#define Maxn 1010using namespace std;struct word{    char pre[5],suf[5];    int w;}p[Maxn];char s[Maxn];int adj[Maxn][Maxn],dist[Maxn],vis[Maxn];const int inf=0x3f3f3f3f;void dijkstra(int u,int n){    bool flag=true;    for(int i=0;i<n;i++)        dist[i]=adj[u][i],vis[i]=0;    vis[u]=1;    for(int i=1;i<n;i++){        int minn=inf,v=0;        for(int j=0;j<n;j++)            if(!vis[j]&&dist[j]<minn)                minn=dist[j],v=j;        vis[v]=1;        for(int j=0;j<n;j++)            if(!vis[j]&&dist[v]+adj[v][j]<dist[j])                dist[j]=dist[v]+adj[v][j];    }    printf("%d\n",dist[n-1]!=inf?dist[n-1]:-1);}int main(){    int n;    while(scanf("%d",&n),n){        for(int i=0;i<n;i++){            scanf("%d%s",&p[i].w,s);            int len=strlen(s);            for(int j=0;j<4;j++)                p[i].pre[j]=s[j],p[i].suf[j]=s[len-4+j];            p[i].pre[4]=p[i].suf[4]='\0';        }        memset(adj,0x3f,sizeof adj);        for(int i=0;i<n;i++) adj[i][i]=0;        for(int i=0;i<n;i++)            for(int j=0;j<n;j++){                if(i==j) continue;                if(!strcmp(p[i].suf,p[j].pre))                    adj[i][j]=p[i].w;            }        dijkstra(0,n);    }return 0;}