描述
给一个20×20的迷宫、起点坐标和终点坐标,问从起点是否能到达终点。
输入
多个测例。输入的第一行是一个整数n,表示测例的个数。接下来是n个测例,每个测例占21行,第一行四个整数x1,y1,x2,y2是起止点的位置(坐标从零开始),(x1,y1)是起点,(x2,y2)是终点。下面20行每行20个字符,’.’表示空格;’X’表示墙。
输出
每个测例的输出占一行,输出Yes或No。
输入样例
2
0 0 19 19
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XXXXXXXXXXXXXXXXXXXX
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0 0 19 19
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XXXXXXXXXXXXXXXXXXX.
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.XXXXXXXXXXXXXXXXXXX
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XXXXXXXXXXXXXXXXXXX.
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.XXXXXXXXXXXXXXXXXXX
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XXXXXXXXXXXXXXXXXXX.
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.XXXXXXXXXXXXXXXXXXX
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XXXXXXXXXXXXXXXXXXX.
XXXXXXXXXXXXXXXXXXX.
XXXXXXXXXXXXXXXXXXX.
XXXXXXXXXXXXXXXXXXX.
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.XXXXXXXXXXXXXXXXXXX
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0 0 19 19
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XXXXXXXXXXXXXXXXXXXX
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0 0 19 19
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XXXXXXXXXXXXXXXXXXX.
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.XXXXXXXXXXXXXXXXXXX
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XXXXXXXXXXXXXXXXXXX.
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.XXXXXXXXXXXXXXXXXXX
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XXXXXXXXXXXXXXXXXXX.
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.XXXXXXXXXXXXXXXXXXX
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XXXXXXXXXXXXXXXXXXX.
XXXXXXXXXXXXXXXXXXX.
XXXXXXXXXXXXXXXXXXX.
XXXXXXXXXXXXXXXXXXX.
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.XXXXXXXXXXXXXXXXXXX
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输出样例
No
Yes
基本思路:没什么好说的了= =。直接贴代码吧。。。
Yes
基本思路:没什么好说的了= =。直接贴代码吧。。。
#include <iostream>
using namespace std;
int n, x1, x2, y1, y2;
int key ;
int a[21][21];
void search(int row, int col);
int main()
{
char x;
cin >> n;
for(int i = 0; i < n ; i++)
{
cin >> x1 >> y1 >> x2 >> y2;
for(int j = 0; j < 20; j++)
{
for(int k = 0; k < 20; k++)
{
cin >> x;
if(x == '.') a[j][k] = 0;
else a[j][k] = 1;
}
}
/*for(int j = 0; j < 20; j++)
{
for(int k = 0; k < 20; k++)
{
cout << a[j][k];
}
cout << endl;
}
cout << endl;*/
search(x1, y1);
if(1 == key)
{
cout << "Yes" << endl;
}
else
{
cout << "No" << endl;
}
key = 0;
//if(key == 1)cout << "Yes" << endl;
//else cout << "No" << endl;
}
}
void search(int row, int col)
{
if(row == x2 && col == y2)
{
key = 1;
}
else
{
a[row][col] = 2;
if((row - 1) >= 0 && a[row - 1][col] != 1 && a[row - 1][col] != 2)
search(row - 1, col);
if((row + 1) <= 19 && a[row + 1][col] != 1 && a[row+1][col] != 2)
search(row + 1, col);
if((col - 1) >= 0 && a[row][col - 1] != 1 && a[row][col - 1] != 2)
search(row, col - 1);
if((col + 1) <= 19 && a[row][col + 1] != 1 && a[row][col + 1] != 2)
search(row, col + 1);
}
}