codeforces 199D Jumping on Walls
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有左右两堵墙,高度都是n,每米为一个单位,每个单位是障碍后者是空地,你从左面的第一米开始爬,每秒可以上爬一米,或下爬一米,或者跳到对面墙k米之上,如果爬到n米或者更高就算是登顶,问登顶的最短时间。还有一个限制就是,洪水每秒一米往上,不能到洪水之下
最好的办法就是bfs,拓展出每秒所能达到的所有位置,因为bfs是按照时间顺序的,所以之前记录下达到每个格子的最短时间,一定是最优解,这题我用dfs也水过去了。。
#include <iostream>#include <cmath>#include <map>#include <cstring>#include <string>#include <cstdio>#include <algorithm>#define rep(i, j, k) for(ll i = j; i <= k; i++)#define ll long long#define maxn 500009#define inf 1e8using namespace std;int n, k, a[3][maxn], vis[3][maxn];void dfs (int x, int y){//printf ("dfs %d %d === %d\n", x, y, vis[x][y]);if (vis[x][y] == 0)return;if (y + k <= n && a[3 - x][y + k]){if (vis[3 - x][y + k] == 0 && vis[x][y] + 1 <= y + k)vis[3 - x][y + k] = vis[x][y] + 1, dfs (3 - x, y + k);elseif (vis[3 - x][y + k] && vis[x][y] + 1 <= y + k)vis[3 - x][y + k] = min (vis[3 - x][y + k], vis[x][y] + 1);}if (y < n && a[x][y + 1]){if (vis[x][y + 1] == 0 &&vis[x][y] + 1 <= y + 1)vis[x][y + 1] = vis[x][y] + 1, dfs (x, y + 1);elseif (vis[x][y + 1] && vis[x][y] + 1 <= y + 1)vis[x][y + 1] = min (vis[x][y + 1], vis[x][y] + 1);}if (y > 1 && a[x][y - 1]){//printf (" down %d %d == %d\n", x, y - 1, vis[x][y - 1]);if (vis[x][y - 1] == 0 && vis[x][y] + 1 <= y - 1)vis[x][y - 1] = vis[x][y] + 1, dfs (x, y - 1);elseif (vis[x][y - 1] && vis[x][y] + 1 <= y - 1)vis[x][y - 1] = min (vis[x][y - 1], vis[x][y] + 1);}}int main (){cin >> n >> k;rep (i, 1, 2){char s[maxn];scanf ("%s", s);rep (j, 1, n)a[i][j] = (s[j - 1] == '-');}memset (vis, 0, sizeof (-1));vis[1][1] = 1;if (a[1][1])dfs (1, 1);//rep (i, 1, 2) rep (j, 1, n) printf ("%d %d ==== %d\n", i, j, vis[i][j]);rep (i, 1, 2)rep (j, 1, n)if (vis[i][j] > 0 && j + k > n){printf ("YES\n");return 0;}cout << "NO" << endl;return 0;}
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