HDU

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1546点击打开链接

Idiomatic Phrases Game

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3852    Accepted Submission(s): 1310

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

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

 

Sample Output

17-1

 

Author

ZHOU, Ran

 

成语接龙 对于每个串只需要记录前四个和后四个

然后注意第一个和最后一个不能变 从第二个开始搜

#include<iostream>#include<algorithm>#include<string.h>#include<cstdio>#include<queue>#include<map>#include<math.h>#include<limits.h>#include<vector>#include <set>#include <string>using namespace std;int n,t;int tt;int cnt=1;int bbegin,eend;map<string ,int > mapp;struct xjy{    int num;    int dis;    bool operator < (const xjy &r) const    {        return dis>r.dis;    }};xjy dis[1111];int mmap[1111][1111];int book[1111];priority_queue<xjy>q;void dijkstra(int begin){    xjy mid;    dis[begin].dis=0;    dis[begin].num=begin;    mid=dis[begin];    q.push(mid);    while(!q.empty())    {        mid=q.top();        q.pop();        for(int i=1;i<=cnt-1;i++)        {            if(!book[i]&&dis[i].dis>mmap[mid.num][i]+dis[mid.num].dis&&mmap[mid.num][i]<INT_MAX)            {                book[mid.num]=1;                dis[i].dis=mmap[mid.num][i]+dis[mid.num].dis;                dis[i].num=i;                q.push(dis[i]);            }        }    }}int main(){        while(~scanf("%d",&n),n)    {        for(int i=0;i<=1100;i++)        {            for(int j=0;j<=1100;j++)                mmap[i][j]=INT_MAX;            dis[i].dis=INT_MAX;            book[i]=0;        }        mapp.clear();        cnt=1;        for(int i=0;i<n;i++)        {            scanf("%d",&t);            string s;            cin >> s;            string s1;            string s2;            s1+=s[0];s1+=s[1];s1+=s[2];s1+=s[3];            s2+=s[s.length()-4];s2+=s[s.length()-3];s2+=s[s.length()-2];s2+=s[s.length()-1];            if(!mapp[s1])                mapp[s1]=cnt++;            if(!mapp[s2])                mapp[s2]=cnt++;            if(mmap[mapp[s1]][mapp[s2]]>t)                mmap[mapp[s1]][mapp[s2]]=t;            if(i==0)            {                bbegin=mapp[s2];                tt=t;            }            if(i==n-1)                eend=mapp[s1];        }        dijkstra(bbegin);        if(dis[eend].dis==INT_MAX)            cout << "-1" << endl;        else            cout << dis[eend].dis+tt << endl;    }}