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hdu 2112 HDU Today

 
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点击打开链接hdu 2112


思路:最短路
分析:只要把名字映射成整数,然后利用整数去求解即可。
注意事项:
1 题目中的起点和终点可能相同,这个时候输出0。
2 用map映射的时候用char类型,由于string是个类效率比较低。
3 处理成无向图


代码:

/*SPFA*/
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<string>
#include<cstring>
#include<queue>
#include<map>
using namespace std;
#define MAXN 20010
#define INF 0XFFFFFFF

int n , cnt;
int begin , goal;
int first[MAXN] , next[MAXN];
int star[MAXN] , end[MAXN] , value[MAXN];
int dis[MAXN];
queue<int>q;
map<string , int>m;

void init(){
   cnt = 0;
   m.clear();
   for(int i = 1 ; i <= 200 ; i++){
      first[i] = -1;
      next[i] = -1;
   }
}

void SPFA(int s){
   while(!q.empty())
      q.pop();
   int vis[MAXN];
   memset(vis , 0 , sizeof(vis));
   for(int i = 1 ; i <= cnt ; i++)
      dis[i] = INF;
   dis[s] = 0;
   q.push(s);
   vis[s] = 1;
   while(!q.empty()){
       int x = q.front();
       q.pop();
       vis[x] = 0;
       for(int i = first[x] ; i != -1 ; i = next[i]){
          if(dis[end[i]] > dis[x] + value[i]){
             dis[end[i]] = dis[x] + value[i];
             if(!vis[end[i]]){
                vis[end[i]] = 1;
                q.push(end[i]);
             }
          }
       }
   }
}

void input(){
   int a , b , v;
   char str1[200] , str2[200];
   init();
   scanf("%s%s" , str1 , str2);
   /*这里判断是不是相同点*/
   a = m[str1];
   if(!a)
      m[str1] = a = ++cnt;
   b = m[str2];
   if(!b)
      m[str2] = b = ++cnt;
   begin = a;
   goal = b;

   for(int i = 0 ; i < n ; i++){
        scanf("%s%s%d" , str1 , str2 , &v);
        a = m[str1];
        if(!a)
           m[str1] = a = ++cnt;
        b = m[str2];
        if(!b)
           m[str2] = b = ++cnt;
        /*处理成无向图*/
        star[i] = a;
        end[i] = b;
        value[i] = v;
    
        star[i+n] = b;
        end[i+n] = a;
        value[i+n] = v;
          
        next[i] = first[star[i]];
        first[star[i]] = i;
        next[i+n] = first[star[i+n]];
        first[star[i+n]] = i+n;
    }
}

int main(){
   while(scanf("%d" , &n) && n != -1){
         input();
         SPFA(begin);
         if(dis[goal] != INF)
            printf("%d\n" , dis[goal]);
        else
            printf("-1\n");
   } 
   return 0;
}

/*floyd*/
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<string>
#include<cstring>
#include<queue>
#include<map>
using namespace std;
#define MAXN 200
#define INF 0xFFFFFFF

int n , star , end , cnt;
long long  dis[MAXN][MAXN];
map<string , int>m;

long long  min(long long a , long long  b){
   return a < b ? a : b;
}

void init(){
   cnt = 0;
   m.clear();
   for(int i = 1 ; i < MAXN ; i++){
      for(int j = 1 ; j < MAXN ; j++){
          if(i == j)
            dis[i][j] = 0;
          else
            dis[i][j] = INF;
      }
   }
}

void floyd(){
   for(int k = 1 ; k <= cnt ; k++){
      for(int i = 1 ; i <= cnt ; i++){
         for(int j = 1 ; j <= cnt ; j++)
            dis[i][j] = min(dis[i][k]+dis[k][j] , dis[i][j]);
      }
   }
}

void input(){
   int a , b , v;
   char str1[MAXN] , str2[MAXN];
   init();
   scanf("%s%s" , str1 , str2);
       
   a = m[str1];
   if(!a)
      m[str1] = a = ++cnt;
   b = m[str2];
   if(!b)
      m[str2] = b = ++cnt;
   star = a;
   end = b;
   
   for(int i = 0 ; i < n ; i++){
        scanf("%s%s%d" , str1 , str2 , &v);
        a = m[str1];
        b = m[str2];
        if(!a)
           m[str1] = a = ++cnt;
        if(!b)
           m[str2] = b = ++cnt;
        if(dis[a][b] > v)
           dis[a][b] = dis[b][a] = v;
    }
}

int main(){
    while(scanf("%d" , &n) && n != -1){
          input();
          floyd();
          if(dis[star][end] != INF)
             printf("%d\n" , dis[star][end]);
          else
             printf("-1\n");
    }
    return 0;
}


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