POJ3009-同样的算法为什么C++代码可以Accept但是JAVA返回错误？

[古德猫宁丶丶]

题目见POJ3009
很奇怪，这是用第二种思路写的结果了，但是C++可以通过，JAVA不可以。瞪了两个小时了，大家帮忙看看吧，代码如下：

C++代码：
#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;
}

JAVA代码：
import java.util.Scanner;
import java.math.*;

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);
}
}
}

PS：另外google的时候发现很早就有人提出相同的代码两种语言会有不同的运行结果，所以这里想好好讨论一下，最后总结总结看。