使用了 dfs bfs 都不行,看答案,才发现需要使用动态规划。。。 太坑了
package main import "fmt" func main() { x, y := 0, 0 n := 0 fmt.Scan(&x, &y, &n) var dp [31][31]int // dp[i][j] 表示从0,0到i,j的所有路径数 bossLocation := [][2]int{} // 该数组用于bfs和dfs算法,dp算法可忽略 bx := 0 by := 0 for i := 0; i < n; i++ { fmt.Scan(&bx, &by) bossLocation = append(bossLocation, [2]int{bx, by}) dp[bx][by] = -1 } for i := 0; i <= x; i++ { dp[i][0] = 1 // 很明显,从0,0 到 x,0的路径数均为1 } for i := 0; i <= y; i++ { dp[0][i] = 1 // 很明显,从0,0 到 0,y的路径数均为1 } for i := 1; i <= x; i++ { for j := 1; j <= y; j++ { if dp[i][j] == -1 { continue } if dp[i][j-1] != -1 { dp[i][j] += dp[i][j-1] } if dp[i-1][j] != -1 { dp[i][j] += dp[i-1][j] } } } fmt.Println(dp[x][y]) //count := 0 //dfs(0, 0, &count, bossLocation, x, y) //fmt.Println(count) //ret := bfs(bossLocation, x, y) //fmt.Println(ret) } func dfs(x, y int, count *int, bossLocation [][2]int, targetX, targetY int) { if x > targetX || y > targetY { return } if x == targetX && y == targetY { *count++ return } if !isBossLocation(x+1, y, bossLocation) { dfs(x+1, y, count, bossLocation, targetX, targetY) } if !isBossLocation(x, y+1, bossLocation) { dfs(x, y+1, count, bossLocation, targetX, targetY) } } func isBossLocation(x, y int, bossLocation [][2]int) bool { for i := 0; i < len(bossLocation); i++ { if x == bossLocation[i][0] && y == bossLocation[i][1] { return true } } return false } func bfs(bossLocation [][2]int, x, y int) int { queue := [][2]int{} queue = append(queue, [2]int{0, 0}) // 初始坐标入库 count := 0 for len(queue) > 0 { front := queue[0] queue = queue[1:] if front[0] == x && front[1] == y { count++ } if !isBossLocation(front[0]+1, front[1], bossLocation) && front[0]+1 <= x { queue = append(queue, [2]int{front[0] + 1, front[1]}) } if !isBossLocation(front[0], front[1]+1, bossLocation) && front[1]+1 <= y { queue = append(queue, [2]int{front[0], front[1] + 1}) } } return count }