每日一题3.10.2

每日一题3.10.2

地下迷宫

解题思路： 递归+回溯。从起始位置开始，向上下左右四个方向分别前进一步，然后加上前进所要消耗的体力值，若某个位置为0，或下一个位置越界，或该位置已访问过，则跳出递归。

若到达出口位置，比较体力消耗值是否大于青蛙总共拥有的体力值，且同时小于目前最小的体力消耗值；若是，则更新最小体力消耗值和路径。
代码实现：

#include<iostream>
#include<vector>
using namespace std;
int n, m, p, step = 999999;
vector<int> res;
int a[10][10];
bool visit[10][10];

void escape(vector<int>&tmp, int sum, int i, int j) {
	if (i == 0 && j == m - 1) {
		if (sum <= p && sum <step){
			step = sum;
			res = tmp;
		}
		return;
	}

	if (i - 1 >= 0 && a[i - 1][j] == 1 && visit[i - 1][j] == 0) {
		sum += 3;
		tmp.push_back(i - 1);
		tmp.push_back(j);
		visit[i - 1][j] = 1;
		escape(tmp, sum, i - 1, j);
		sum -= 3;
		tmp.pop_back(); tmp.pop_back();
		visit[i - 1][j] = 0;
	}

	if (i + 1 <n && a[i + 1][j] == 1 && visit[i + 1][j] == 0) {
		tmp.push_back(i + 1);
		tmp.push_back(j);
		visit[i + 1][j] = 1;
		escape(tmp, sum, i + 1, j);
		tmp.pop_back(); tmp.pop_back();
		visit[i + 1][j] = 0;
	}

	if (j - 1 >= 0 && a[i][j - 1] == 1 && visit[i][j - 1] == 0) {
		sum += 1;
		tmp.push_back(i);
		tmp.push_back(j - 1);
		visit[i][j - 1] = 1;
		escape(tmp, sum, i, j - 1);
		sum -= 1;
		tmp.pop_back(); tmp.pop_back();
		visit[i][j - 1] = 0;
	}

	if (j + 1<m && a[i][j + 1] == 1 && visit[i][j + 1] == 0) {
		sum += 1;
		tmp.push_back(i);
		tmp.push_back(j + 1);
		visit[i][j + 1] = 1;
		escape(tmp, sum, i, j + 1);
		sum -= 1;
		tmp.pop_back(); tmp.pop_back();
		visit[i][j + 1] = 0;
	}

}

int main() {

	cin >> n >> m >> p;

	for (int i = 0; i<n; i++)
	for (int j = 0; j<m; j++)
		cin >> a[i][j];

	vector<int> tmp;
	int sum = 0;

	tmp.push_back(0); tmp.push_back(0);
	visit[0][0] = 1;

	escape(tmp, sum, 0, 0);

	if (step == 999999)
		cout << "Can not escape!";
	else {
		int size = res.size();

		for (int i = 0; i<size - 2; i += 2) {
			cout << "[" << res[i] << "," << res[i + 1] << "]" << ",";
		}
		cout << "[" << res[size - 2] << "," << res[size - 1] << "]";
	}
	system("pause");
	return 0;


}

参考答案：