Jumping Frog

题目连接: Jumping Frog

大致题意:

一只青蛙从原点出发, 路途中有些点不可达, 每次最远跳跃d, 询问最少跳多少次可以到达终点

解题思路:

暴力dfs深搜

AC代码:

#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
int c, d; char s[1000];
int res = 0x7fffffff;
bool dfs(int index = 1, int num = 0) {if (index + d >= c) { res = num + 1; return 1; } bool flag = 0; //最先找到的一定是最优的, 因为我们先递归的是较远的路程for (int i = d; i >= 1; --i) {if (s[index + i] != 'X') flag = dfs(index + i, num + 1);if (flag) return 1;}return 0;
}
int main(void)
{int t; cin >> t;for (int T = 1; T <= t; ++T) {scanf("%d %d", &c, &d); scanf("%s", s + 1); d++;res = 0x7fffffff; dfs();printf("Day #%d\n", T);printf("%d %d\n", c, d - 1);printf("%s\n", s + 1);res == 0x7fffffff ? printf("%d\n", 0) : printf("%d\n", res);if (T != t) printf("\n");}return 0;
}

END

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Jumping Frog

题目连接: Jumping Frog

大致题意:

解题思路:

AC代码:

END

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