Description
详见 啊哈算法 解救小哈章节
标准的dfs求最短路径,dfs其实就是暴力,这道题对我来说有1个难点。第一个是return的用法,return的时候,返回到调用return的那个函数上,然后就会执行book[i]=0,然后就可以在这个点进行其他位置的搜索。思路其实很简单,也是典型的DFS例题
1 #include<iostream> 2 using namespace std; 3 int n, m, mins = 99999; 4 int w[20][20], book[20][20]; 5 int x1, y1, x2, y2; 6 int x, y; 7 int nextX[4] = {1, 0, -1}; 8 int nextY[4] = {1, 0, -1}; 9 void dfs(int a, int b, int step) { 10 int tx, ty; 11 if(a == x2 && b == y2) { 12 //cout << "等于啦"; 13 if(step <= mins) { 14 mins = step; 15 //cout << "step" << mins; 16 } 17 return; 18 } else { 19 20 for(int i = 0; i < 3; i++) { 21 for(int j = 0; j < 3; j++) { 22 tx = a+nextX[i]; 23 ty = b+nextY[j]; 24 if(tx<1 || tx>m || ty<1 || ty>n) { 25 } else { 26 if(w[tx][ty]==0 && book[tx][ty]==0) 27 { 28 book[tx][ty] = 1; 29 dfs(tx, ty, ++step); 30 book[tx][ty] = 0; 31 } 32 } 33 } 34 } 35 } 36 } 37 int main() 38 { 39 cin >> n >> m; 40 for(int i = 1; i <= n; i++) { 41 for(int j = 1; j <= m; j++) { 42 int x; 43 cin >> x; 44 w[i][j] = x; 45 } 46 } 47 cin >> x1 >> y1 >> x2 >> y2; 48 book[x1][y1] = 1; 49 dfs(x1, y1, 0); 50 cout << mins; 51 } 52 /* 53 5 4 54 0 0 1 0 55 0 0 0 0 56 0 0 1 0 57 0 1 0 0 58 0 0 0 1 59 1 1 4 3 60 */
原文:https://www.cnblogs.com/wzy-blogs/p/9158750.html