ACM:POJ-3009 Curling2.0

讲真的这个题目我现在觉得特别恶心,不是说难,弄了两种解决算法用JAVA写都没A,但是神奇的是用了c++写就可以直接A了,我把代码贴出来。

哦对,其中一个JAVA代码确实有问题,并且不知道BUG在哪里,是我的问题;

算法不再写了,贴完代码直接刷题去了。仍需努力啊!再哭一会。。

代码如下:

JAVA第一个代码,有BUG;

import java.util.Scanner;

public class Curling_2 {
    static Scanner input = new Scanner(System.in);

    static int[][] Room = new int[20][20];
    static int dir[][] = {{1,0},{-1,0},{0,1},{0,-1}}; //四个搜索方向
    static int move=0;
    static int x,y;

    public static void main(String[] args){

        int cou=1;

        while (true){
            //初始化矩形空间的 宽 高,以及函数引用的存放起点座标的变量
            x=input.nextInt();
            y=input.nextInt();

            if (x==0&&y==0)
                break;

            int m=0,n=0;

            for (int i=0;i<x;i++){
                for (int j=0;j<y;j++){

                    //初始化
                    Room[i][j]=input.nextInt();

                    //寻找起点
                    if (Room[i][j]==2){
                        m=i;
                        n=j;
                    }
                }
            }
            search(m,n,cou);
            System.out.println(move);
        }
    }

    static void search(int a,int b,int mo){
        if (mo>10){
            move = -1;
            return;
        }

        else {
            for (int i=0;i<4;i++){
                int h=a+dir[i][0];
                int l=b+dir[i][1];

                //如果选择方向以后第一次移动数不大于10并且是目标座标,返回移动数。
                if (Room[h][l]==3)
                    if (mo<=10){
                        move=mo;
                        break;
                    }

                //选择方向以后接着移动
                while (h>=0&&h<x&& l>=0&&l<y &&Room[h][l]==0){

                    h+=dir[i][0];
                    l+=dir[i][1];

                    if (Room[h][l]==1){
                        search(h-dir[i][0],l-dir[i][1],mo+1);
                        Room[h][l]=0;
                        return;

                    }

                    if (Room[h][l]==3)
                        if (mo<=10){
                            move=mo;
                            return;
                        }
                }
            }
            return;
        }
    }
}

JAVA第二个代码无BUG:

import java.util.Scanner;
import java.math.*;
import java.lang.*;

public class Curling2third {
   static int[][] s=new int[25][25];
   static int result;
   static int sx, sy;
   static int n, m;

   static void dfs(int x, int y, int step, int type) {
        if (step > 10) return;    //超过10步失败
        //到达目的地
        if (s[x][y] == 3) {
            if (result == -1) result = step;
            else result = Math.min(result, step);
            return;
        }
        //石头把墙撞碎
        switch (type) {
            case 1:
                s[x - 1][y] = 0;
                break; //north
            case 2:
                s[x + 1][y] = 0;
                break; //south
            case 3:
                s[x][y - 1] = 0;
                break;  //west
            case 4:
                s[x][y + 1] = 0;
                break;  //east
        }
        //四处搜索寻找终点
        for (int i = 1; i <= x; i++) {
            if (s[x - i][y] == 1) break;
            if (s[x - i][y] == 3) {
                dfs(x - i, y, step + 1, 0);
                break;
            }
        }
        for (int i = 1; i < m - x; i++) {
            if (s[x + i][y] == 1) break;
            if (s[x + i][y] == 3) {
                dfs(x + i, y, step + 1, 0);
                break;
            }
        }
        for (int i = 1; i <= y; i++) {
            if (s[x][y - i] == 1) break;
            if (s[x][y - i] == 3) {
                dfs(x, y - i, step + 1, 0);
                break;
            }
        }
        for (int i = 1; i < n - y; i++) {
            if (s[x][y + i] == 1) break;
            if (s[x][y + i] == 3) {
                dfs(x, y + i, step + 1, 0);
                break;
            }
        }
        //四处探索
        if (x >= 1 && s[x - 1][y] != 1)
            for (int i = 2; i <= x; i++)
                if (s[x - i][y] == 1) {
                    dfs(x - i + 1, y, step + 1, 1);
                    break;
                }
        if (x + 1 < m && s[x + 1][y] != 1)
            for (int i = 2; i < m - x; i++)
                if (s[x + i][y] == 1) {
                    dfs(x + i - 1, y, step + 1, 2);
                    break;
                }
        if (y >= 1 && s[x][y - 1] != 1)
            for (int i = 2; i <= y; i++)
                if (s[x][y - i] == 1) {
                    dfs(x, y - i + 1, step + 1, 3);
                    break;
                }
        if (y + 1 < n && s[x][y + 1] != 1)
            for (int i = 2; i < n - y; i++)
                if (s[x][y + i] == 1) {
                    dfs(x, y + i - 1, step + 1, 4);
                    break;
                }
        //墙壁复原
        switch (type) {
            case 1:
                s[x - 1][y] = 1;
                break;
            case 2:
                s[x + 1][y] = 1;
                break;
            case 3:
                s[x][y - 1] = 1;
                break;
            case 4:
                s[x][y + 1] = 1;
                break;
        }
    }

    public static void main(String[] args){
        Scanner input=new Scanner(System.in);
        //freopen("3009.in","r",stdin);
        while (true)  //m行n列
        {
            m=input.nextInt();
            n=input.nextInt();
            if (n == 0 && m == 0) break;
            result = -1;
            for (int i = 0; i < m; i++)
                for (int j = 0; j < n; j++) {
                    s[i][j]=input.nextInt();
                    if (s[i][j] == 2) {
                        sx = i;
                        sy = j;
                    }
                }
            dfs(sx, sy, 0, 0);
            System.out.println(result);
        }
        return;
    }
}

C++代码,直接A了:

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
int s[25][25];
int result;
int sx, sy;
int n, m;
void dfs(int x, int y, int step, int type)
{
  if(step > 10)   return;    //超过10步失败 
  //到达目的地    
  if(s[x][y] == 3)
  {
    if(result == -1)  result = step;
    else  result = min(result, step);
    return;
  }
  //石头把墙撞碎 
  switch(type)
  {
      case 1:  s[x-1][y] = 0;  break; //north
    case 2:  s[x+1][y] = 0;  break; //south
    case 3:  s[x][y-1] = 0;  break;  //west
    case 4:  s[x][y+1] = 0;  break;  //east
  }
  //四处搜索寻找终点        
  for(int i=1; i<=x; i++)
  {   
      if(s[x-i][y]==1)    break;
      if(s[x-i][y]==3)  
        {
            dfs(x-i, y, step+1, 0);
            break;
        }
  }
  for(int i=1; i<m-x; i++)
  {
    if(s[x+i][y]==1)    break;
    if(s[x+i][y]==3)  
    {
        dfs(x+i, y, step+1, 0);
        break;
    }
  }              
  for(int i=1; i<=y; i++)
  {
      if(s[x][y-i]==1)    break;
    if(s[x][y-i]==3)  
    {
        dfs(x, y-i, step+1, 0);
        break;
    }
  }
  for(int i=1; i<n-y; i++)
  {
      if(s[x][y+i]==1)    break;
    if(s[x][y+i]==3) 
    {
        dfs(x, y+i, step+1, 0);
        break;
    }
  } 
  //四处探索 
  if(x>=1 && s[x-1][y]!=1)
        for(int i=2; i<=x; i++)
            if(s[x-i][y]==1)  
            {
                dfs(x-i+1, y, step+1, 1);
                break;
            }
  if(x+1<m && s[x+1][y]!=1)
      for(int i=2; i<m-x; i++)
          if(s[x+i][y]==1)  
            {
                dfs(x+i-1, y, step+1, 2);
                break;
            }    
  if(y>=1 && s[x][y-1]!=1)
        for(int i=2; i<=y; i++)
        if(s[x][y-i]==1)  
            {
                dfs(x, y-i+1, step+1, 3);
                break;
            }
    if(y+1<n && s[x][y+1]!=1)
        for(int i=2; i<n-y; i++)
        if(s[x][y+i]==1)  
            {
                dfs(x, y+i-1, step+1, 4);    
                break;
            }                                
  //墙壁复原 
  switch(type)
  {    
        case 1:  s[x-1][y] = 1;  break;
    case 2:  s[x+1][y] = 1;  break;
    case 3:  s[x][y-1] = 1;  break;
    case 4:  s[x][y+1] = 1;  break;
  }
}
int main(void)
{
    //freopen("3009.in","r",stdin);
    while(scanf("%d%d",&n,&m))  //m行n列
  {
    if(n==0 && m==0)  return 0;
    result = -1;
    for(int i=0; i<m; i++)
      for(int j=0; j<n; j++)
      {
        scanf("%d",&s[i][j]);
        if(s[i][j]==2) {sx=i; sy=j;}
      }
    dfs(sx, sy, 0, 0);
    printf("%d\n",result);
  }
    return 0;
}

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