迷宫问题
Time Limit: 1000MS | | Memory Limit: 65536K |
Total Submissions: 10783 | | Accepted: 6435 |
Description
定义一个二维数组:
int maze[5][5] = {
0, 1, 0, 0, 0,
0, 1, 0, 1, 0,
0, 0, 0, 0, 0,
0, 1, 1, 1, 0,
0, 0, 0, 1, 0,
};
它表示一个迷宫,其中的1表示墙壁,0表示可以走的路,只能横着走或竖着走,不能斜着走,要求编程序找出从左上角到右下角的最短路线。
Input
一个5 × 5的二维数组,表示一个迷宫。数据保证有唯一解。
Output
左上角到右下角的最短路径,格式如样例所示。
Sample Input
0 1 0 0 0
0 1 0 1 0
0 0 0 0 0
0 1 1 1 0
0 0 0 1 0
Sample Output
(0, 0)
(1, 0)
(2, 0)
(2, 1)
(2, 2)
(2, 3)
(2, 4)
(3, 4)
(4, 4)
#include <iostream>
#include <cstdio>
#include <queue>
#include <vector>
#include <algorithm>
using namespace std;
typedef pair<int, int> P;
const int INF = 10000000;
int s[5][5], d[5][5];
int dx[4] = {-1, 1, 0, 0}, dy[4] = {0, 0, -1, 1};
struct node
{
int x, y;
};
node e, pre[5][5];
void bfs()
{
queue <P> que;
fill(d[0], d[0] + 25, INF);
que.push(P(0, 0));
d[0][0] = 0;
while(! que.empty())
{
P p = que.front(); que.pop();
if(p.first == 4 && p.second == 4) break;
for(int i = 0; i < 4; i++)
{
int x = p.first + dx[i], y = p.second + dy[i];
if(x >= 0 && x < 5 && y >= 0 && y < 5 && s[x][y] == 0 && d[x][y] == INF)
{
que.push(P(x, y));
d[x][y] = d[p.first][p.second] + 1;
e.x = p.first, e.y = p.second;
pre[x][y] = e;
}
}
}
vector <node> path;
e.x = 4, e.y = 4;
path.push_back(e);
for(int i = 4, j = 4; i || j; i = e.x, j = e.y)
{
e = pre[i][j];
path.push_back(e);
}
reverse(path.begin(), path.end());
for(vector <node> :: iterator pp = path.begin(); pp != path.end(); pp++)
printf("(%d, %d)\n", pp -> x, pp -> y);
}
int main()
{
for(int i = 0; i < 5; i++)
for(int j = 0; j < 5; j++)
scanf("%d", &s[i][j]);
bfs();
return 0;
}