98. 所有可达路径
题目链接: 98. 所有可达路径
思路
- 先创建邻接矩阵,再深搜
- 写代码是需要注意的是acm格式,输入的格式要转化为int,输出要转化为str,用map()实现。
dfs
def dfs(grid,node,n,path,res):if node == n:res.append(path[:])returnfor j in range(len(grid[0])):if grid[node-1][j] == 1:path.append(j+1)dfs(grid,j+1,n,path,res)path.pop()def main():# 构造邻接矩阵n,m = map(int,input().split())grid = [[0]*n for _ in range(n)]for _ in range(m):node1,node2 = map(int,input().split())grid[node1-1][node2-1] = 1res = []dfs(grid,1,n,[1],res)if not res:print(-1)else:for path in res:print(' '.join(map(str,path)))if __name__ == "__main__":main()