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Idiomatic Phrases Game

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

5
5 12345978ABCD2341
5 23415608ACBD3412
7 34125678AEFD4123
15 23415673ACC34123
4 41235673FBCD2156
2
20 12345678ABCD
30 DCBF5432167D
0

Sample Output

17
-1


題意：每個成語由至少三個字組成，每個字為四個16進制的數字（0~F)組成。兩個成語首尾相接，消耗Tom對前一個成語的查詢時間，求最小查詢時間。
思路：最短路問題，註意對建圖的處理。map將成語的首尾字以數字點的形式存儲。使用map前必須初始化。

#include<stdio.h>
#include<string.h>
#include<string>
#include<map>
#include<deque>
#include<vector>
#define MAX 2005
#define INF 0x3f3f3f3f
using
 
 namespace std;

map<string,int> mp;

struct Node{
    int v,w;
}node;
vector<Node> edge[MAX];
int dis[MAX],b[MAX],w[MAX],bg[MAX],ed[MAX];
int n;

void spfa(int k)
{
    int i;
    deque<int> q;
    for(i=1;i<=n;i++){
        dis[i]=INF;
    }
    memset(b,0,sizeof(b));
    b[k]=1;
    dis[k]=0;
    q.push_back(k);
    while(q.size()){
        int u=q.front();
        for(i=0;i<edge[u].size();i++){
            int v=edge[u][i].v;
            int w=edge[u][i].w;
            if(dis[v]>dis[u]+w){
                dis[v]=dis[u]+w;
                if(b[v]==0){
                    b[v]=1;
                    if(dis[v]>dis[u]) q.push_back(v);
                    else q.push_front(v);
                }
            }
        }
        b[u]=0;
        q.pop_front();
    }
}
int main()
{
    int m,u,v,i,j;
    char s[200],s1[200],s2[200];
    while(scanf("%d",&n)&&n!=0){
        m=0;
        for(i=1;i<=2002;i++){
            edge[i].clear();
            mp.clear();    //！！
        }
        for(i=1;i<=n;i++){
            scanf("%d%s",&w[i],s);
            strncpy(s1,s,4);
            strncpy(s2,s+strlen(s)-4,4);
            s1[4]=‘\0‘;s2[4]=‘\0‘;
            if(!mp[s1]) mp[s1]=++m;
            if(!mp[s2]) mp[s2]=++m;
            u=mp[s1];v=mp[s2];
            bg[i]=u;ed[i]=v;
        }
        for(i=1;i<=n;i++){
            for(j=1;j<=n;j++){
                if(i!=j&&ed[i]==bg[j]){
                    node.v=j;
                    node.w=w[i];
                    edge[i].push_back(node);
                }
            }
        }
        spfa(1);
        if(n==1||dis[n]==INF) printf("-1\n");
        else printf("%d\n",dis[n]);
    }
    return 0;
}

Idiomatic Phrases Game 成語接龍SPFA+map