hdu 4771 （2013acmicpc 亞洲區域賽杭州站 B）

http://acm.hdu.edu.cn/showproblem.php?pid=4771

Stealing Harry Potter's Precious

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 277    Accepted Submission(s): 141

Problem Description

　　Harry Potter has some precious. For example, his invisible robe, his wand and his owl. When Hogwarts school is in holiday, Harry Potter has to go back to uncle Vernon's home. But he can't bring his precious with him. As you know, uncle Vernon never allows such magic things in his house. So Harry has to deposit his precious in the Gringotts Wizarding Bank which is owned by some goblins. The bank can be considered as a N × M grid consisting of N × M rooms. Each room has a coordinate. The coordinates of the upper-left room is (1,1) , the down-right room is (N,M) and the room below the upper-left room is (2,1)..... A 3×4 bank grid is shown below:



　　Some rooms are indestructible and some rooms are vulnerable. Goblins always care more about their own safety than their customers' properties, so they live in the indestructible rooms and put customers' properties in vulnerable rooms. Harry Potter's precious are also put in some vulnerable rooms. Dudely wants to steal Harry's things this holiday. He gets the most advanced drilling machine from his father, uncle Vernon, and drills into the bank. But he can only pass though the vulnerable rooms. He can't access the indestructible rooms. He starts from a certain vulnerable room, and then moves in four directions: north, east, south and west. Dudely knows where Harry's precious are. He wants to collect all Harry's precious by as less steps as possible. Moving from one room to another adjacent room is called a 'step'. Dudely doesn't want to get out of the bank before he collects all Harry's things. Dudely is stupid.He pay you $1,000,000 to figure out at least how many steps he must take to get all Harry's precious.

題目大意：其實就是求從起點出發途徑所有寶藏所需的最小步數。

思路：首先將起點和各寶藏之間的最短路徑求出來，若有不可達的點直接輸出-1，求出來後，一開始理所當然的想到了狀壓，但是後來一看寶藏最多才4個，我們暴力一下途徑寶藏的順序就行了，一共就4！=24種。。。這樣簡單一些。取最小的步數即可。。。

代碼如下：

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <algorithm>
#include <queue>
#define inf 2100000000
using namespace std;
char map[110][110];
int dir[4][2]={1,0,0,1,-1,0,0,-1};
int po[5][2];
int mp[5][5];
int dist[110][110];
int n,m,k;
void getin()
{
    int i,j;
    for(i=1;i<=n;i++)
        {
            for(j=1;j<=m;j++)
            {
                if(map[i][j]=='@')
                {
                    po[0][0]=i;
                    po[0][1]=j;
                    map[i][j]='.';
                    return;
                }
            }
        }
}
bool check(int x,int y)
{
    if(x<1||y<1||x>n||y>m||map[x][y]=='#'||dist[x][y]!=inf)
    return false;
    return true;
}
void bfs(int now)
{
    int i,j;
    for(i=0;i<=n;i++)
    {
        for(j=0;j<=m;j++)
        dist[i][j]=inf;
    }
      queue<int> q;
      int x=po[now][0],y=po[now][1];
      dist[x][y]=0;
      q.push(x);
      q.push(y);
      while(!q.empty())
      {
          x=q.front();q.pop();
          y=q.front();q.pop();
          for(i=0;i<4;i++)
          {
              int xx=x+dir[i][0],yy=y+dir[i][1];
              if(check(xx,yy))
              {
                  dist[xx][yy]=dist[x][y]+1;
                  q.push(xx);
                  q.push(yy);
              }
          }
      }
      for(i=0;i<=k;i++)
      {
          mp[now][i]=dist[po[i][0]][po[i][1]];
      }
}
int vis[5];
int tmp,ans;
void dfs(int now,int num)
{
    if(num==k)
    {
        ans=min(ans,tmp);
        return;
    }
    for(int i=1;i<=k;i++)
    {
        if(!vis[i])
        {
            vis[i]=1;
            tmp+=mp[now][i];
            dfs(i,num+1);
            vis[i]=0;
            tmp-=mp[now][i];
        }
    }
}
int main()
{
   // freopen("dd.txt","r",stdin);
    while(scanf("%d%d",&n,&m)&&(n+m))
    {
        int i,j;
        for(i=1;i<=n;i++)
        scanf("%s",map[i]+1);
        scanf("%d",&k);
        memset(vis,0,sizeof(vis));
        for(i=1;i<=k;i++)
        scanf("%d%d",&po[i][0],&po[i][1]);
        getin();
        for(i=0;i<5;i++)
        for(j=0;j<5;j++)
        mp[i][j]=inf;
        for(i=0;i<=k;i++)
        {
            bfs(i);
        }
        int tru=1;
        for(i=0;i<=k;i++)
        {
            if(mp[0][i]==inf)
            {
                tru=0;
                break;
            }
        }
        ans=inf;
        if(!tru)
        printf("%d\n",-1);
        else
        {
            tmp=0;
            for(i=1;i<=k;i++)
            {
                tmp+=mp[0][i];
                vis[i]=1;
                dfs(i,1);
                vis[i]=0;
                tmp-=mp[0][i];
            }
            printf("%d\n",ans);
        }
    }
    return 0;
}