1671: [Usaco2005 Dec]Knights of Ni Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 175  Solved: 107[Submit][Status][Discuss] Description Bessie is in Camelot and has encountered a sticky situation: she needs to pass through the forest that is guarded b…

1671: [Usaco2005 Dec]Knights of Ni 骑士 Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 281  Solved: 180[Submit][Status][Discuss] Description Bessie is in Camelot and has encountered a sticky situation: she needs to pass through the forest that is guarde…

Description Bessie is in Camelot and has encountered a sticky situation: she needs to pass through the forest that is guarded by the Knights of Ni. In order to pass through safely, the Knights have demanded that she bring them a single shrubbery. Tim…

[Usaco2005 Dec]Knights of Ni 骑士 Description  贝茜遇到了一件很麻烦的事:她无意中闯入了森林里的一座城堡,如果她想回家,就必须穿过这片由骑士们守护着的森林．为了能安全地离开,贝茜不得不按照骑士们的要求,在森林寻找一种特殊的灌木并带一棵给他们．当然,贝茜想早点离开这可怕的森林,于是她必须尽快完成骑士们给的任务,贝茜随身带着这片森林的地图,地图上的森林被放入了直角坐标系,并按x,y轴上的单位长度划分成了W×H(1≤W,H≤1000)块,贝茜在地图上查出了她自…

1671: [Usaco2005 Dec]Knights of Ni 骑士 Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 254  Solved: 163[Submit][Status][Discuss] Description Bessie is in Camelot and has encountered a sticky situation: she needs to pass through the forest that is guarde…

http://www.lydsy.com/JudgeOnline/problem.php?id=1671 从骑士bfs一次,然后从人bfs一次即可. #include <cstdio> #include <cstring> #include <cmath> #include <string> #include <iostream> #include <algorithm> #include <queue> using na…

Description Bessie is in Camelot and has encountered a sticky situation: she needs to pass through the forest that is guarded by the Knights of Ni. In order to pass through safely, the Knights have demanded that she bring them a single shrubbery. Tim…

Description Bessie is in Camelot and has encountered a sticky situation: she needs to pass through the forest that is guarded by the Knights of Ni. In order to pass through safely, the Knights have demanded that she bring them a single shrubbery. Tim…

题目: https://www.lydsy.com/JudgeOnline/problem.php?id=1671 题解: 按题意分别从贝茜和骑士bfs然后meet_in_middle.. 把一个逗号打成分号调了20分钟... 代码: #include<bits/stdc++.h> using namespace std; ]={,-,,,},my[]={,,,-,}; ; int n,m,g[maxn][maxn],vis[maxn][maxn],f[maxn][maxn],sum,cnt,…

bfs预处理出每个点s和t的距离d1和d2(无法到达标为inf),然后在若干灌木丛格子(x,y)里取min(d1[x][y]+d2[x][y]) /* 0:贝茜可以通过的空地 1:由于各种原因而不可通行的区域 2:贝茜现在所在的位置 3:骑士们的位置 4:长着贝茜需要的灌木的土地 */ #include<iostream> #include<cstdio> #include<queue> #include<cstring> using namespace s…