hdu2653之BFS
Waiting ten thousand years for Love
Time Limit: 10000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 615 Accepted Submission(s): 205
Now Yifenfei has got the map of the castle.
Here are all symbols of the map:
Only one ‘Y’ Indicates the start position of Yifenfei.
Only one ‘L’ Indicates the position of Demon Lemon.
‘.’ Indicate the road that Yifenfei can walk on it, or fly over it.
‘#’ Indicate the wall that Yifenfei can not walk or flay through it.
‘@’ Indicate the trap that Yifenfei can not walk on it, but can fly over it.
Yifenfei can walk or fly to one of up, down, left or right four directions each step, walk costs him 2 seconds per step, fly costs him 1 second per step and 1 magic power. His magic power will not increased, and if his magic power is zero, he can not fly any more.
Now Yifenfei asks you for helping him kill Demon Lemon smoothly. At the same time, Demon Lemon will Leave the castle Atfer T seconds. If Yifenfei can’t kill Demon Lemon this time, he have to wait another ten thousand years.
Y@L
###
2 3 4 1
Y@L
###
2 3 4 0
Y.L
###
2 3 3 0
Y.L
###
Yes, Yifenfei will kill Lemon at 2 sec.
Case 2:
Poor Yifenfei, he has to wait another ten thousand years.
Case 3:
Yes, Yifenfei will kill Lemon at 4 sec.
Case 4:
Poor Yifenfei, he has to wait another ten thousand years.
Hint
Case 1: Yifenfei cost 1 second and 1 magic-power fly to ‘@’,
but he can not step on it, he must cost another 1 second
and 1 magic-power fly to ‘L’ and kill Lemon immediately.
Case 2: When Yifenfei Fly to ‘@’, he has no power to fly,
and is killed by trap.
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<queue>
#include<algorithm>
#include<map>
#include<iomanip>
#define INF 99999999
using namespace std; const int MAX=80+10;
char Map[MAX][MAX];
int mark[MAX][MAX][MAX];//在位置i,j剩余魔力k是否标记
int n,m,t,p;
int dir[4][2]={0,1,0,-1,1,0,-1,0}; struct Node{
int x,y,time,m;
Node(){}
Node(int X,int Y,int Time,int M):x(X),y(Y),time(Time),m(M){}
bool operator<(const Node &a)const{
return time>a.time;
}
}start; priority_queue<Node>q;
int BFS(int &flag){
while(!q.empty())q.pop();
Node next,oq;
q.push(start);
mark[start.x][start.y][start.m]=flag;
while(!q.empty()){
oq=q.top();
q.pop();
if(Map[oq.x][oq.y] == 'L')return oq.time;
if(oq.time>t)return INF;
for(int i=0;i<4;++i){
next=Node(oq.x+dir[i][0],oq.y+dir[i][1],oq.time+1,oq.m);
if(next.x<0 || next.y<0 || next.x>=n || next.y>=m)continue;
if(Map[next.x][next.y] == '#')continue;
if(Map[next.x][next.y] != '@' && Map[oq.x][oq.y] != '@' && mark[next.x][next.y][next.m] != flag){
++next.time;
q.push(next);
mark[next.x][next.y][next.m]=flag;
}
if(next.m>0 && mark[next.x][next.y][next.m-1] != flag){
--next.m;
next.time=oq.time+1;
q.push(next);
mark[next.x][next.y][next.m]=flag;
}
}
}
return INF;
} int main(){
int num=0;
while(cin>>n>>m>>t>>p){
for(int i=0;i<n;++i)cin>>Map[i];
for(int i=0;i<n;++i){
for(int j=0;j<m;++j){
if(Map[i][j] == 'Y')start.x=i,start.y=j;
}
}
start.time=0,start.m=p;
int temp=BFS(++num);
cout<<"Case "<<num<<":\n";
if(temp>t)cout<<"Poor Yifenfei, he has to wait another ten thousand years."<<endl;
else cout<<"Yes, Yifenfei will kill Lemon at "<<temp<<" sec."<<endl;
}
return 0;
}
hdu2653之BFS的更多相关文章
- HDU2653 BFS+优先队列
Waiting ten thousand years for Love Time Limit: 10000/2000 MS (Java/Others) Memory Limit: 32768/3 ...
- 图的遍历(搜索)算法(深度优先算法DFS和广度优先算法BFS)
图的遍历的定义: 从图的某个顶点出发访问遍图中所有顶点,且每个顶点仅被访问一次.(连通图与非连通图) 深度优先遍历(DFS): 1.访问指定的起始顶点: 2.若当前访问的顶点的邻接顶点有未被访问的,则 ...
- 【BZOJ-1656】The Grove 树木 BFS + 射线法
1656: [Usaco2006 Jan] The Grove 树木 Time Limit: 5 Sec Memory Limit: 64 MBSubmit: 186 Solved: 118[Su ...
- POJ 3278 Catch That Cow(bfs)
传送门 Catch That Cow Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 80273 Accepted: 25 ...
- POJ 2251 Dungeon Master(3D迷宫 bfs)
传送门 Dungeon Master Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 28416 Accepted: 11 ...
- Sicily 1215: 脱离地牢(BFS)
这道题按照题意直接BFS即可,主要要注意题意中的相遇是指两种情况:一种是同时到达同一格子,另一种是在移动时相遇,如Paris在(1,2),而Helen在(1,2),若下一步Paris到达(1,1),而 ...
- Sicily 1048: Inverso(BFS)
题意是给出一个3*3的黑白网格,每点击其中一格就会使某些格子的颜色发生转变,求达到目标状态网格的操作.可用BFS搜索解答,用vector储存每次的操作 #include<bits/stdc++. ...
- Sicily 1444: Prime Path(BFS)
题意为给出两个四位素数A.B,每次只能对A的某一位数字进行修改,使它成为另一个四位的素数,问最少经过多少操作,能使A变到B.可以直接进行BFS搜索 #include<bits/stdc++.h& ...
- Sicily 1051: 魔板(BFS+排重)
相对1150题来说,这道题的N可能超过10,所以需要进行排重,即相同状态的魔板不要重复压倒队列里,这里我用map储存操作过的状态,也可以用康托编码来储存状态,这样时间缩短为0.03秒.关于康托展开可以 ...
随机推荐
- Julia - If 条件语句
Julia 中使用 if,elseif,else 进行条件判断 格式: if expression1 statement1 elseif expression2 statement2 else sta ...
- python装饰器注意事项
内容: 1.装饰器基本结构复习 2.装饰器注意事项 python装饰器详细内容:http://www.cnblogs.com/wyb666/p/8748102.html 1.装饰器基本结构复习 装饰器 ...
- 导出文件名带时间信息的dmp文件
exp system/orcl@orcl owner=aixm file=d:\aixm%date:~0,4%%date:~5,2%%date:~8,2%_%time:~0,2%%time:~3,2% ...
- svn关键词BASE, HEAD, COMMITTED, PREV的深入理解
svn关键词BASE, HEAD, COMMITTED, PREV可以很方便用于日常操作中,但是很多人对他们的工作原理和方式不是太了解. 在这里我将使用用例,诠释他们的作用和意图. 先给出svn手册中 ...
- DBCA Does Not Display ASM Disk Groups In 11.2
DBCA Does Not Display ASM Disk Groups In 11.2 https://oraclehowto.wordpress.com/2011/08/15/dbca-does ...
- git之GitHub Pages
git之GitHub Pages GitHub Pages是github的一项很实用的功能,它可以让我们将github里面的静态网站的代码在线上展示出来,可以用来做项目展示和个人博客的载体. 1.将 ...
- C#中有关数组和string引用类型或值类型的判断
直接来一段测试代码 class value_ref_type { public static void DEMO1() { ] { }; double[] location_new; string s ...
- leetcode704
public class Solution { public int Search(int[] nums, int target) { var len = nums.Length; ; ; if (t ...
- java byte to hex
String str; byte[] bs = null; bs =str.getBytes(); bs =str.getBytes("utf-8") java byte to ...
- delphi IOS 后台状态保存
FormSaveState procedure TFrm.FormSaveState(Sender: TObject);begin end; http://stackoverflow.com/ques ...