Idiomatic Phrases Game(最短路+注意坑点)-白红宇

Idiomatic Phrases Game(最短路+注意坑点)

阅读量：5009 次

发布时间：2019-06-12

本文共 3293 字，大约阅读时间需要 10 分钟。

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.

InputThe 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.

OutputOne 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 这个题的坑点在字符串的长度上 代码： vector版

#include
     
      #include
      
       #include
       
        #include
        
         #include
         
          #include
          
           #include
           
            #include
            #include
             
              #include
              
               #define Inf 0x3f3f3f3fconst int maxn=1e5+5;typedef long long ll;using namespace std;char str[1005][105];int w[1005];struct node{ int pos; int w; node (int x,int y) { pos=x; w=y; } bool friend operator < (node x,node y) { return x.w>y.w; }};vector
               
                vec[1005];int dis[1005];int vis[1005];int n,m;void init(){ for(int t=1;t<=n;t++) { dis[t]=Inf; } memset(vis,0,sizeof(vis));}void Dijkstra(int s){ priority_queue
                
                 q; q.push(node(s,0)); dis[s]=0; while(!q.empty()) { node now=q.top(); q.pop(); if(vis[now.pos])continue; vis[now.pos]=1; for(int t=0;t

邻接表版

#include
     
      #include
      
       #include
       
        #include
        
         #include
         
          #include
          
           #include
           
            #include
            
             #include
             #include
              
               #define Inf 0x3f3f3f3fconst int maxn=1e5+5;typedef long long ll;using namespace std;struct edge{ int u,v,w; int next;}Edge[10*maxn];int w[1005];char str[1005][105];struct node{ int pos,w; node(int x,int y) { pos=x; w=y; } bool friend operator < (node x,node y) { return x.w>y.w; }};int head[1005],dis[1005],vis[1005],cnt;void add(int u,int v,int w){ Edge[cnt].u=u; Edge[cnt].v=v; Edge[cnt].w=w; Edge[cnt].next=head[u]; head[u]=cnt++;}void Dijkstra(int s){ dis[s]=0; priority_queue
               
                q; q.push(node(s,0)); while(!q.empty()) { node now=q.top(); q.pop(); if(vis[now.pos])continue; vis[now.pos]=1; for(int i=head[now.pos];i!=-1;i=Edge[i].next) { if(dis[now.pos]+Edge[i].w

转载于:https://www.cnblogs.com/Staceyacm/p/11274685.html

你可能感兴趣的文章