K - 迷宫问题
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)
AC代码
#include <cstdio>
#include <map>
#include <vector>
#include <queue>
#include <cstring>
#include <iostream>
#define ll long long
using namespace std;
const int maxn = 10;
const int INF = 0x3f3f3f3f;
/*
fill(1) 1
drop(1) 2
pour12 3
pour21 4
fill2 5
drop2 6
*/
int a, b, c, d;
vector<int> ans;
vector<int> arr;
vector<int> brr;
char mp[maxn][maxn];
bool vis[maxn][maxn];
int flag = false;
int mv[4][2] = {1,0,0,1,-1,0,0,-1};
struct Node
{
int x, y;
int num;
Node(int _x, int _y, int _num) {x=_x;y=_y;num=_num;}
};
void bfs()
{
queue<Node> que;
que.push(Node(0,0,-1));
int cnt = 0;
while (que.size())
{
Node nn = que.front(); que.pop();
int i = nn.x, j = nn.y;
vis[i][j] =true;
//printf("%d %d %d %d\n", nn.num,nn.x,nn.y,nn.z);
if(i == 4 && j == 4)
{
for(int i = nn.num; i != -1; i = arr[i])
{
brr.push_back(ans[i]);
}
int x = 0, y = 0;
puts("(0, 0)");
for(int i = brr.size()-1; i >=0; --i)
{
printf("(%d, %d)\n", x+=mv[brr[i]][0], y+=mv[brr[i]][1]);
}
return;
}
//
for(int i = 0; i < 4; ++i)
{
int nx = mv[i][0] + nn.x, ny = mv[i][1] + nn.y;
if(0<=nx&&nx<5&&0<=ny&&ny<5&&!vis[nx][ny]&&mp[nx][ny]=='0')
{
ans.push_back(i);
arr.push_back(nn.num);
que.push(Node(nx,ny,cnt++));
}
}
}
}
int main() {
for(int i = 0; i < 5; ++i)
{
for(int j = 0; j < 5; ++j)
{
mp[i][j] = getchar();
getchar();
}
}
bfs();
return 0;
}