基础实验6-2.3-拯救007-编程题

解题代码

#include<stdio.h>
#include<math.h>
#define MAXN 100
#define DIA 15
#define SIDE 100
typedef enum { false, true } bool;
bool CanJumpToSide(int index);
bool FirstJump(int index);
bool CanJump(int index1, int index2);
bool DFS(int index);
struct Data {
	int x, y;
}A[MAXN];
int B[MAXN], N, D;
int main()
{
	int i, flag = 0;
	scanf("%d %d", &N, &D);
	for (i = 0; i < N; i++) scanf("%d %d", &A[i].x, &A[i].y);
	if (CanJumpToSide(-1)) flag = 1;
	else {
		for (i = 0; i < N; i++) {
			if (!B[i] && FirstJump(i)) {
				B[i] = 1;
				if (DFS(i)) flag = 1;
			}
		}
	}
	if (flag) printf("Yes");
	else printf("No");
	return 0;
}
bool CanJumpToSide(int index) {
	if (index == -1) {
		if ((double)DIA / 2 + D >=(double) SIDE / 2) return true;
		else return false;
	}
	else {
		int x = A[index].x;
		int y = A[index].y;
		if ((x >= SIDE / 2 - D )||( x <= D - SIDE / 2 )||( y >= SIDE / 2 - D )||( y <= D - SIDE / 2)) return true;
		else return false;
	}
}
bool FirstJump(int index) {
	double distance = sqrt(pow(A[index].x, 2) + pow(A[index].y, 2));
	if (distance <= (double)DIA / 2 + D) return true;
	else return false;
}
bool CanJump(int index1, int index2) {
	int x1 = A[index1].x, y1 = A[index1].y;
	int x2 = A[index2].x, y2 = A[index2].y;
	double distance = sqrt(pow(x1 - x2, 2) + pow(y1 - y2, 2));
	if (distance <= D) return true;
	else return false;
}
bool DFS(int index) {
	if (CanJumpToSide(index)) return true;
	for (int i = 0; i < N; i++) {
		if (!B[i] && CanJump(i, index)) {
			B[i] = 1;
			if (DFS(i)) return true;
		}
	}
	return false;
}

测试结果

问题整理

1.还是DFS的return的问题阻碍了输出，在递归函数中最重要的便是return的定义和循环终止条件。