Waiting ten thousand years for Love
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
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.
#include
#include
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
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
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
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
for(j=0;j
cin>>field[i][j];
if(field[i][j]=='Y')
{
s1=i;
s2=j;
}
}
}
memset(isvisit,false,sizeof(isvisit));
cout<<"Case "<<++icase<<":"<
}
return 0;
}