4.2.7 Waiting ten thousand years for Love

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

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.

#include <cstdio>
#include <queue>
#include <cstring>
using namespace std;

struct qq
{
int x,y,step,p;
bool friend operator < (const qq &a,const qq &b)
{
if (a.step==b.step)
return a.p<b.p;
return a.step>b.step;
}
} s,ya;

priority_queue<qq> q;

const int dx[]={1,-1,0,0};
const int dy[]={0,0,1,-1};
bool f[82][82][82];
char map[82][82];
int power,t,n,m,ax,ay,i,cnt=0,tot=0;

bool pd()
{
if (map[ya.x][ya.y]=='#') return true; //也就是不能到达，直接退出
if (ya.x>=0 && ya.x<n && ya.y>=0 && ya.y<m && not f[ya.x][ya.y][ya.p])
return false;
return true;
}

void judge()
{
if (s.p>0)  //也就是说这一步是可以用能量的
for (i=0;i<4;i++)
{
ya.x=s.x+dx[i];
ya.y=s.y+dy[i];
ya.p=s.p-1; //能量减少1
ya.step=s.step;
if (not pd())
{
f[ya.x][ya.y][ya.p]=true;
ya.step++; //步数增加1
q.push(ya);
}
}
if (map[s.x][s.y]=='@') return; //在@的时候，如果不能移动，则挂了
for (i=0;i<4;i++)
{
ya.x=s.x+dx[i];
ya.y=s.y+dy[i];
ya.p=s.p;
ya.step=s.step; //这一句一定要加上，否则就未清零啊！
if (not pd() && ( map[ya.x][ya.y]=='.' || map[ya.x][ya.y]=='L' ))
{
f[ya.x][ya.y][ya.p]=true; //这句话把我整惨了！
ya.step+=2;
q.push(ya);
}
}
}

void bfs()
{
memset(f,false,sizeof(f));
while (!q.empty())
q.pop();
s.x=ax; s.y=ay; s.p=power; s.step=0;
f[s.x][s.y][s.step]=true;
q.push(s);
while (!q.empty())
{
s=q.top();
q.pop();
if (map[s.x][s.y]=='#') break;
if (map[s.x][s.y]=='L')
{
if (t>=s.step)
printf("Yes, Yifenfei will kill Lemon at %d sec.\n",s.step);
else printf("Poor Yifenfei, he has to wait another ten thousand years.\n");
return;
}
judge();
}
printf("Poor Yifenfei, he has to wait another ten thousand years.\n");
}

void init()
{
while (scanf("%d%d%d%d",&n,&m,&t,&power)!=EOF)
{
cnt++;
printf("Case %d:\n",cnt);
for(int i=0;i<n;i++)
{
scanf("%s",map[i]);
for(int j=0;j<m;j++)
if(map[i][j]=='Y')
{
ax=i;
ay=j;
}
}
//        printf("ax:%d ay:%d\n",ax,ay);
bfs();
}
}

int main ()
{
init();
return 0;
}