13杭州现场赛Stealing Harry Potter's Precious(bfs+dfs)

题意:一个简单图,给你一个起点,图中最多有4个宝藏,问你收集这四个宝藏最多要多少步.

解题思路:很简单的一个题,因为宝藏比较少,所以先对宝藏进行一个全排,然后再一步一步算步数.

解题代码:

// File Name: d.cpp
// Author: darkdream
// Created Time: 2014年05月06日 星期二 19时03分57秒

#include<vector>
#include<set>
#include<deque>
#include<stack>
#include<bitset>
#include<algorithm>
#include<functional>
#include<numeric>
#include<utility>
#include<sstream>
#include<iostream>
#include<iomanip>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<ctime>

using namespace std;
int n , m;
struct node{
  int x, y ,s;
}LIST[10005];
int visit[104][104];
int MAP[104][104];
struct node1{
  int x, y ; 
}a[10];
int xadd[] = {0,0,1,-1};
int yadd[] = {1,-1,0,0};
int ok = 1;
int bfs(int bx, int by ,int ex,int ey)
{
    //printf("%d %d %d %d
",bx,by,ex,ey);
    LIST[1].x = bx ; 
    LIST[1].y = by ; 
    LIST[1].s = 0 ; 
    int l = 1, r= 1; 
    int tempok = 0 ;
    memset(visit,0,sizeof(visit));
    visit[bx][by] = 1; 
    while(l <= r )
    {
      //  printf("%d %d
",LIST[l].x,LIST[l].y);
        if(LIST[l].x == ex && LIST[l].y == ey)
        {
          tempok = 1; 
          break;
        }
        for(int i = 0 ; i <= 3;i ++)
        {
          int tempx = LIST[l].x + xadd[i];
          int tempy = LIST[l].y + yadd[i];
          if(!visit[tempx][tempy] && MAP[tempx][tempy])
          {
             visit[tempx][tempy] = 1;
             r ++ ;
             LIST[r].x = tempx; 
             LIST[r].y = tempy; 
             LIST[r].s = LIST[l].s +1 ; 
          }
        }
       l ++;
    }
    if(tempok == 0)
        ok = 0;
    else return LIST[l].s;

}
int main(){


      while(scanf("%d %d",&n,&m) != EOF)
      {
        if(n == 0  && m == 0)
            break;
        memset(MAP,0,sizeof(MAP));
        memset(a,0,sizeof(a));
        for(int i = 1;i <= n;i ++)
        {
          char str[104];
          scanf("%s",&str[1]);
          for(int j = 1;j <= m;j ++)
          {
             if(str[j] == '#')
             {
             }
             else 
             {
                 MAP[i][j] = 1;
                 if(str[j] == '@')
                 {
                 a[0].x = i ; 
                 a[0].y = j; 
                 }
             }
          }
        }
        int  k ; 
        scanf("%d",&k);
        for(int i = 1;i <= k;i ++)
            scanf("%d %d",&a[i].x,&a[i].y);
        int temp[6];
        for(int i =0 ;i < k;i ++)
          temp[i] = i + 1; 
        ok = 1; 
        int min = 1000000;
        do{
           int last = 0;
           int sum = 0 ; 
           for(int i = 0 ;i < k ;i ++)
           {
             sum +=  bfs(a[last].x,a[last].y,a[temp[i]].x,a[temp[i]].y); 
             last = temp[i];
           }
           if (sum <= min)
               min = sum;
        
          //  printf("%d
",sum);
        }while(next_permutation(temp,temp+k));    
        if(ok == 0 )
            printf("-1
");
        else 
        printf("%d
",min);
      }
      
return 0 ;
}