Waiting ten thousand years for Love

最新推荐文章于 2020-08-20 19:13:48 发布

原创最新推荐文章于 2020-08-20 19:13:48 发布 · 761 阅读

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Yifenfei为了复仇并拯救爱人，要在限定时间内穿越迷宫般的城堡找到恶魔Lemon。利用广度优先搜索结合优先队列算法，考虑走动与飞行的不同成本及魔法值限制，帮助Yifenfei制定最优路径。

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最后一道搜索题，关键是划分状态。分析这道题：

如果以（x,y）为状态，无法唯一确定状态。因为（‘.’ Indicate the road that Yifenfei can walk on it, or fly over it.）

所以必须要（x，y，p）为状态区分。

本题采用bfs+优先队列

It was ten thousand years, after Demon Lemon caught Yifenfei’s love. In order to revenge and save his love, Yifenfei have been practising sword all day long and his Kongfu skills becomes so powerful that he can kill Demon Lemon immediately. Recently, Yifenfei have found Lemon’s castle, and now he is going to kill Lemon. At the same time, hearing about the terrible news, Demon Lemon is now preparing for escaping...

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.

Input

Lots of test cases, please process to end of file. In each test case, firstly will have four integers N, M, T, P(1 <= N, M, P <= 80, 1 <= T <= 100000), indicates the size of map N * M, the Lemon’s leaving time(after T seconds, Lemon will disappeared) and Yifenfei’s magic power. Then an N * M two-dimensional array follows indicates the map.

Output

For each test case, first Print a line “Case C:”, C indicates the case number. Then, if Yifenfei can kill Demon Lemon successfully, Print “Yes, Yifenfei will kill Lemon at T sec.”, T indicates the minimum seconds he must cost. Otherwise, Print ”Poor Yifenfei, he has to wait another ten thousand years.”

Sample Input

2 3 2 2
Y@L
###
2 3 4 1
Y@L
###
2 3 4 0
Y.L
###
2 3 3 0
Y.L
###

Sample Output

Case 1:
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.



HintHint
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<queue>
using namespace std;
typedef struct node{
int i;
int j;
int t;
int p;


}Node;
struct qcmp
{
bool operator() (const Node a,const Node b)
{
return a.t>b.t;
}
};
char field[81][81];

int N,M,T,P;
bool isvisit[81][81][81];
bool cross(int a,int b)
{
if(a>=0&&a<N&&b>=0&&b<M&&field[a][b]!='#')
{
return true;
}
return false;

}
int dir[][2]={{0,1},{1,0},{-1,0},{0,-1}};

void bfs(int a,int b)
{
bool isfind=false;
int count=0;
Node start={a,b,0,P};
isvisit[a][b][P]=true;
priority_queue<Node,vector<Node>,qcmp> myqueue;
myqueue.push(start);
while(!myqueue.empty())
{
Node pre=myqueue.top();
myqueue.pop();
if(pre.t<=T&&field[pre.i][pre.j]=='L')
{
isfind=true;
count=pre.t;

break;
}
if(pre.t>T)
break;

for(int i=0;i<4;i++)
{
int curi=pre.i+dir[i][0];
int curj=pre.j+dir[i][1];
int curp=pre.p;
int curt=pre.t;
if(cross(curi,curj)&&pre.t<T)
{
if(field[pre.i][pre.j]!='@'&&field[curi][curj]!='@'&&!isvisit[curi][curj][curp])
{

isvisit[curi][curj][curp]=true;
int tempt=curt;
tempt+=2;
Node cur1={curi,curj,tempt,curp};
myqueue.push(cur1);

}
if(!isvisit[curi][curj][curp-1]&&curp>0)
{
isvisit[curi][curj][curp-1]=true;
curt+=1;
curp-=1;
Node cur2={curi,curj,curt,curp};
myqueue.push(cur2);

}

}



}


}



if(isfind)
{
printf("Yes, Yifenfei will kill Lemon at %d sec.\n",count);
}
else
{
printf("Poor Yifenfei, he has to wait another ten thousand years.\n");
}

}
int main()
{
//freopen("in.txt","r",stdin);
int icase=0;
int i,j;
while(scanf("%d%d%d%d",&N,&M,&T,&P)!=EOF)
{
int s1,s2;
for(i=0;i<N;i++)
{
for(j=0;j<M;j++)
{
cin>>field[i][j];
if(field[i][j]=='Y')
{
s1=i;
s2=j;

}

}
}
memset(isvisit,false,sizeof(isvisit));

cout<<"Case "<<++icase<<":"<<endl;
bfs(s1,s2);

}
return 0;
}