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: