简单水池&&迷宫问题

#include <iostream>
#include <stdio.h>
#include <cstring>
using namespace std;
int M[101][101],flag[101][101];
int n,m;
int cnt;
void pool(int x,int y)
{
    flag[x][y]=1;
    M[x][y]=cnt;
    if(x-1>=1&&M[x-1][y]!=0&&flag[x-1][y]==0)  pool(x-1,y);
    if(x+1<=n&&M[x+1][y]!=0&&flag[x+1][y]==0)  pool(x+1,y);
    if(y-1>=1&&M[x][y-1]!=0&&flag[x][y-1]==0)  pool(x,y-1);
    if(y+1<=m&&M[x][y+1]!=0&&flag[x][y+1]==0)  pool(x,y+1);
}
int main()
{
    int i,j,t;
    cin>>t;
    while(cin>>n>>m&&t--)
    {
        cnt=0;
    for(i=1;i<=n;i++)
        for(j=1;j<=m;j++)
            cin>>M[i][j];
            //cout<<endl;

           // memset(m,0,sizeof(m));
            memset(flag,0,sizeof(flag));
    for(i=1;i<=n;i++)
        for(j=1;j<=m;j++)
        if(M[i][j]==1&&flag[i][j]==0)
        {
            cnt++;pool(i,j);
        }
        printf("%d\n",cnt);
         // for(i=1;i<=n;cout<<endl,i++)
         //       for(j=1;j<=m;j++)
         //       cout<<M[i][j]<<" ";
    }
    return 0;
    }
//这一道题是有关搜索的，看看就行

#include <iostream>
#include <cstring>
using namespace std;
int a,b,c,d,sum,cnt;
int M[9][9]={
 {1,1,1,1,1,1,1,1,1},
 {1,0,0,1,0,0,1,0,1},
 {1,0,0,1,1,0,0,0,1},
 {1,0,1,0,1,1,0,1,1},
 {1,0,0,0,0,1,0,0,1},
 {1,1,0,1,0,1,0,0,1},
 {1,1,0,1,0,1,0,0,1},
 {1,1,0,1,0,0,0,0,1},
 {1,1,1,1,1,1,1,1,1}};
void dfs(int x,int y,int cnt)
{
    if(x==c&&y==d)  {if(cnt<sum) sum=cnt;}
    else{
    if(!M[x-1][y])  {M[x-1][y]=1; dfs(x-1,y,cnt+1); M[x-1][y]=0;}
    if(!M[x+1][y])  {M[x+1][y]=1; dfs(x+1,y,cnt+1); M[x+1][y]=0;}
    if(!M[x][y-1])  {M[x][y-1]=1; dfs(x,y-1,cnt+1); M[x][y-1]=0;}
    if(!M[x][y+1])  {M[x][y+1]=1; dfs(x,y+1,cnt+1); M[x][y+1]=0;}
    }
}
int main()
{
    int t;
    cin>>t;
    while(t--&&cin>>a>>b>>c>>d)
    {
        cnt=0,sum=81;
        dfs(a,b,cnt);
        cout<<sum<<endl;
    }
    return 0;
}

这道题的代码，一开始不是这样写的，我另外定义一个flag，把每次访问的都标记，不过后面觉得这个是多此一举，可以在原来的数组上面，每次访问过后，都标记为墙1，这样就ok了，还有一个就是求最小的值，这个在dfs函数里面开始就给一个判断，其实大概的思路就是先找到一条路，记下步数，然后每次回溯，还原被标记的点，这样继续求出其他路径的值。。。。

#include <iostream>
#include <cstring>
using namespace std;
int M[9][9]={
 {1,1,1,1,1,1,1,1,1},
 {1,0,0,1,0,0,1,0,1},
 {1,0,0,1,1,0,0,0,1},
 {1,0,1,0,1,1,0,1,1},
 {1,0,0,0,0,1,0,0,1},
 {1,1,0,1,0,1,0,0,1},
 {1,1,0,1,0,1,0,0,1},
 {1,1,0,1,0,0,0,0,1},
 {1,1,1,1,1,1,1,1,1}},flag[100][100];
int n,m,cnt=0;
int a,b,c,d;
void maze(int x,int y)
{
    if(x==c&&y==d)  return;

    flag[x][y]=1;
    if(x-1>=0&&M[x-1][y]==0&&flag[x-1][y]==0)  {cnt++;  maze(x-1,y);  cnt--;flag[x-1][y]=0;}
    if(x+1<=8&&M[x+1][y]==0&&flag[x+1][y]==0)  {cnt++;  maze(x+1,y);  cnt--;flag[x+1][y]=0;}
    if(y-1>=0&&M[x][y-1]==0&&flag[x][y-1]==0)  {cnt++;  maze(x,y-1);  cnt--;flag[x][y-1]=0;}
    if(y+1<=8&&M[x][y+1]==0&&flag[x][y+1]==0)  {cnt++;  maze(x,y+1);  cnt--;flag[x][y+1]=0;}
}
int main()
{
    //int i,j;
    cin>>a>>b>>c>>d;
    memset(flag,0,sizeof(flag));
    //for(i=a;i<=8;i++)
     //   for(j=b;j<=8;j++)
            maze(a,b);
            cout<<cnt<<endl;
    return 0;
}

后面想把路径打印出来，可是出不来，代码如下

#include <iostream>
#include <cstring>
using namespace std;
int a,b,c,d,sum,cnt,LJ1[1000],LJ2[1000],h,k;
int M[9][9]={
 {1,1,1,1,1,1,1,1,1},
 {1,0,0,1,0,0,1,0,1},
 {1,0,0,1,1,0,0,0,1},
 {1,0,1,0,1,1,0,1,1},
 {1,0,0,0,0,1,0,0,1},
 {1,1,0,1,0,1,0,0,1},
 {1,1,0,1,0,1,0,0,1},
 {1,1,0,1,0,0,0,0,1},
 {1,1,1,1,1,1,1,1,1}};
void dfs(int x,int y,int cnt)
{
    LJ1[k++]=x;LJ2[h++]=y;
    if(x==c&&y==d)  {if(cnt<sum) sum=cnt;}
    else{
    if(!M[x-1][y])  {M[x-1][y]=1; dfs(x-1,y,cnt+1); M[x-1][y]=0;LJ1[k-1]=0;LJ2[h-1]=0;}
    if(!M[x+1][y])  {M[x+1][y]=1; dfs(x+1,y,cnt+1); M[x+1][y]=0;LJ1[k-1]=0;LJ2[h-1]=0;}
    if(!M[x][y-1])  {M[x][y-1]=1; dfs(x,y-1,cnt+1); M[x][y-1]=0;LJ1[k-1]=0;LJ2[h-1]=0;}
    if(!M[x][y+1])  {M[x][y+1]=1; dfs(x,y+1,cnt+1); M[x][y+1]=0;LJ1[k-1]=0;LJ2[h-1]=0;}
    }
}
int main()
{
    int t,i;
    cin>>t;
    while(t--&&cin>>a>>b>>c>>d)
    {
        k=0;h=0;
        memset(LJ1,0,sizeof(LJ1));
        memset(LJ2,0,sizeof(LJ2));
        cnt=0,sum=81;
        dfs(a,b,cnt);
        cout<<sum<<endl;
        for(i=0;i<=100;i++)
            cout<<LJ1[i]<<","<<LJ2[i]<<" → ";
    }
    return 0;
}

//怎么把路径打印出来，感觉如果走一条路不通，就标记为0，可是又有回溯，就感觉全乱了，如果在每次dfs后，干脆数组全为0，k，h也从0开始应该可以。。。

#include <iostream>
#include <mem.h>
using namespace std;
char M[30][30];
bool Find;
int flag[30][30];
int n,m;
void ss(int x,int y)
{
    if(Find==true)  return ;

    cout<<x<<" "<<y<<" →";

    flag[x][y]=1;
    int a,b;
    int i;
    //上走
    a=x-1;
    b=y;
    if(a>=1&&M[a][b]!='w'&&flag[a][b]==0)
    {
        if(M[a][b]=='T')
        {
            Find=true;
            return ;
        }
        ss(a,b);
    }
    //下走
     a=x+1;
     b=y;
//cout<<a<<n<<endl;
    if(a<=n&&M[a][b]!='w'&&flag[a][b]==0)
    {
        if(M[a][b]=='T')
        {
            Find=true;
            return ;
        }
        ss(a,b);
    }
    //左走
     a=x;
    b=y-1;
    if(b>0&&M[a][b]!='w'&&flag[a][b]==0)
    {
        if(M[a][b]=='T')
        {
            Find=true;
            return ;
        }
        ss(a,b);
    }
    //右走
     a=x;
    b=y+1;
    if(b<=m&&M[a][b]!='w'&&flag[a][b]==0)
    {
        if(M[a][b]=='T')
        {
            Find=true;
            return ;
        }
        ss(a,b);
    }
}
int main()
{
    int j,i,x,y;
    cin>>n>>m;
    for(i=1;i<=n;i++)
        for(j=1;j<=m;j++)
    {
        cin>>M[i][j];
        if(M[i][j]=='s')  {x=i;y=j;}
    }
    memset(flag,0,sizeof(flag));
    ss(x,y);
    return 0;

}
//一个迷宫的问题，和水池问题是一个题型，不多讲了。。。。