HDU 1546 idiomatic phrases game

Link:

Http://acm.hdu.edu.cn/showproblem.php? PID = 1, 1546

Question:

Idiomatic phrases game

Time Limit: 2000/1000 MS (Java/others) memory limit: 65536/32768 K (Java/Others)
Total submission (s): 969 accepted submission (s): 300


Problem descriptiontom 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 shoshould 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 shocould 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
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

 

Question:

N "idioms" are given, which are composed of at least three "Chinese characters". The so-called "Chinese characters" refer to four consecutive hexadecimal numbers (1 ~ 9, ~ F ).

Take the first idiom as the starting point, and the last one as the ending point. You need to find a sequence, the last "Chinese character" of the first idiom in this sequence is the same as that of the first "Chinese character" of the last idiom. It takes at least time.

Analysis and Summary:

After creating a picture, it is the most basic and most short-circuit problem.

During input, you only need to save the first and last Chinese characters for each idiom.

Code:

#include<iostream>#include<cstdio>#include<cstring>#include<queue>#include<utility>using namespace std;typedef pair<int,int>pii;priority_queue<pii,vector<pii>,greater<pii> >q;const int INF = 0x7fffffff;const int VN  = 1005;struct Edge{int v,next,w;}E[VN*VN];struct Idiom{    char beg[5], end[5];}arr[VN];int n,size;int head[VN];int d[VN];int cost[VN];void init(){    size=0;    memset(head, -1, sizeof(head));    while(!q.empty())q.pop();}void addEdge(int u,int v,int w){    E[size].v=v, E[size].w=w;    E[size].next = head[u];    head[u] = size++;}void Dijkstra(int src){    for(int i=1; i<=n; ++i) d[i] = INF;    d[src] = 0;    q.push(make_pair(d[src],src));    while(!q.empty()){        pii x = q.top(); q.pop();        int u = x.second;        if(d[u] != x.first) continue;        for(int e=head[u]; e!=-1; e=E[e].next){            int tmp = d[u] + E[e].w;            if(d[E[e].v] > tmp){                d[E[e].v] = tmp;                q.push(make_pair(tmp, E[e].v));            }        }    }}int main(){    int w;    char str[100];    while(scanf("%d",&n)&&n){        init();        for(int i=1; i<=n; ++i){            scanf("%d %s",&cost[i],str);            for(int j=0; j<5; ++j)                 arr[i].beg[j]=str[j];            int len = strlen(str);            for(int j=len-4,k=0; j<len; ++j,++k)                 arr[i].end[k]=str[j];            arr[i].end[4] = arr[i].beg[4] = 0;        }        for(int i=1; i<n; ++i){            for(int j=1; j<=n; ++j){                if(strcmp(arr[i].end, arr[j].beg)==0)                    addEdge(i,j,cost[i]);             }        }        Dijkstra(1);        if(d[n]!=INF) printf("%d\n", d[n]);        else  puts("-1");    }    return 0;}

 
  
-- The meaning of life is to give it meaning.
  

   Original
   Http://blog.csdn.net/shuangde800
   , By d_double (reprinted, please mark)