HDU 4771 (DFS+BFS)

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.

Input

　　There are several test cases. 　　In each test cases: 　　The first line are two integers N and M, meaning that the bank is a N × M grid(0<N,M <= 100). 　　Then a N×M matrix follows. Each element is a letter standing for a room. ‘#‘ means a indestructible room, ‘.‘ means a vulnerable room, and the only ‘@‘ means the vulnerable room from which Dudely starts to move. 　　The next line is an integer K ( 0 < K <= 4), indicating there are K Harry Potter‘s precious in the bank. 　　In next K lines, each line describes the position of a Harry Potter‘s precious by two integers X and Y, meaning that there is a precious in room (X,Y). 　　The input ends with N = 0 and M = 0

Output

　　For each test case, print the minimum number of steps Dudely must take. If Dudely can‘t get all Harry‘s things, print -1.

Sample Input

2 3

##@

#.#

1

2 2

4 4

#@##

....

####

....

2

2 1

2 4

0 0

Sample Output

-1 5

Source

2013 Asia Hangzhou Regional Contest

最近一直再做搜索的题目  姿势不知道有没有涨了许多  可以可以

这题 还是坑了很久的 刚开始是想不断bfs 找到最近的一个点 记录步数 然后更新起点 继续bfs 但是 有bug 有反例 gg

先找到k处宝藏与出发点之间的最短路 bfs处理  然后dfs 找到最短连接路

这个地方刚开始还想用并查集 但是题目的要求的联通是首尾相接的 gg

dfs+bfs

#include<bits/stdc++.h>
using namespace std;
char a[105][105];
int mp[105][105];
map<int,int>flag;
int mpp[105][105];
int shorpath[6][2];
int dis[4][2]={{1,0},{-1,0},{0,1},{0,-1}};
int n,m;
int k;
int s_x,s_y;
int parent[30];
int jishu=0;
int A[6][6];
int re=100000;
int sum;
struct node
{
    int x;
    int y;
    int step;
};
int Find(int n)
{
    if(n!=parent[n])
        n=Find(parent[n]);
    return n;
}
void unio( int ss,int bb)
{
    ss=Find(ss);
    bb=Find(bb);
    if(ss!=bb)
    parent[ss]=bb;
}
queue<node>q;
node N,now;
void init_()
{
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
           mpp[i][j]=mp[i][j];
}
int bfs(int a,int b,int c,int d)
{
    init_();
    while(!q.empty())
    {
        q.pop();
    }
    N.x=a;
    N.y=b;
    N.step=0;
    q.push(N);
    mpp[a][b]=0;
    while(!q.empty())
    {
        now=q.front();
        q.pop();
        if(now.x==c&&now.y==d)
            return now.step;
        for(int i=0;i<4;i++)
        {
            int aa=now.x+dis[i][0];
            int bb=now.y+dis[i][1];
            if(aa>0&&aa<=n&&bb>0&&bb<=m&&mpp[aa][bb])
            {
                mpp[aa][bb]=0;
                N.x=aa;
                N.y=bb;
                N.step=now.step+1;
                q.push(N);
            }
        }
    }
    return -1;
}
void init()
{
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
           mp[i][j]=0;
    for(int i=0;i<=30;i++)
         parent[i]=i;
    re=100000;
}
/*void kruskal()
{
    int re=0;
   for(int i=0;i<jishu;i++)
 {
     int qq=A[i].s;
     int ww=A[i].e;
     //cout<<re<<endl;
     if(Find(qq)!=Find(ww))
     {
         unio(qq,ww);
         re+=A[i].x;
     }
}
  printf("%d\n",re);
}*/
void dfs(int n,int ce)
{
    if(ce==k)
    {
        if(sum<re)
        {
            re=sum;
        }
        //printf("%d\n",re);
        return ;
    }
    for(int i=0;i<=k;i++)
    {
        if(i!=n&&flag[i]==0)
        {
            sum+=A[n][i];
            flag[i]=1;
            dfs(i,ce+1);
            sum-=A[n][i];
            flag[i]=0;
        }
    }
}
int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        if(n==0&&m==0)
            break;
        init();
        getchar();
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=m;j++)
            {
                scanf("%c",&a[i][j]);
                if(a[i][j]==‘@‘)
                {
                    s_x=i;
                    s_y=j;
                }
                if(a[i][j]==‘.‘||a[i][j]==‘@‘)
                    mp[i][j]=1;
            }
            getchar();
        }
            scanf("%d",&k);
            shorpath[0][0]=s_x;
            shorpath[0][1]=s_y;
            for(int i=1;i<=k;i++)
               scanf("%d%d",&shorpath[i][0],&shorpath[i][1]);
            jishu=0;
            for(int i=0;i<=k;i++)
                for(int j=0;j<=k;j++)
            {
                A[i][j]=bfs(shorpath[i][0],shorpath[i][1],shorpath[j][0],shorpath[j][1]);
            }
            sum=0;
            flag[0]=1;
            dfs(0,0);
            printf("%d\n",re);

     }

return 0;
}