HDU2653 BFS+优先队列

Waiting ten thousand years for Love

Time Limit: 10000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1057    Accepted Submission(s): 335

Problem Description

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.

Hint

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.

 

Author

lemon

 

Source

决战龙虎门

 

题意：

n行m列的地图，开始位置Y，结束位置L，#代表墙不能走，@代表陷阱必须飞过去连续飞两下才能通过一个陷阱，.代表路可以走过去也可以飞过去，位在时间t内能否到达并求最短时间，飞过去用1秒，走过去用2秒，飞一次消耗1点魔法值，共有P点魔法值。

代码：

//卡了很长时间，后来发现别人的题解vis是三维。
//显然要用到优先队列，本题的vis不是只访问一次，可以访问P次，因为以不同的能量值经过同一点的方案肯定是不同的。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
int n,m,t,p;
char map[82][82];
bool vis[82][82][82];
int dir[4][2]={1,0,-1,0,0,1,0,-1};
struct Lu
{
    int x,y,cnt,mag;
    friend bool operator<(Lu a,Lu b)
    {
        return a.cnt>b.cnt;
    }
};
int bfs(int sx,int sy)
{
    priority_queue<Lu>q;
    Lu L1,L2;
    L1.x=sx;L1.y=sy;L1.cnt=0;L1.mag=p;
    vis[sx][sy][p]=1;
    q.push(L1);
    while(!q.empty())
    {
        L2=q.top();
        q.pop();
        if(L2.cnt>t)
        return 10000000;
        if(map[L2.x][L2.y]=='L')
        return L2.cnt;
        for(int i=0;i<4;i++)
        {
            L1.x=L2.x+dir[i][0];
            L1.y=L2.y+dir[i][1];
            if(L1.x<0||L1.x>=n||L1.y<0||L1.y>=m) continue;
            if(map[L1.x][L1.y]=='#') continue;
            if(L2.mag>0&&!vis[L1.x][L1.y][L2.mag-1])
            {
                L1.cnt=L2.cnt+1;
                L1.mag=L2.mag-1;
                vis[L1.x][L1.y][L1.mag]=1;
                q.push(L1);
            }
            if(map[L1.x][L1.y]!='@'&&map[L2.x][L2.y]!='@'&&!vis[L1.x][L1.y][L2.mag])
            {
                L1.cnt=L2.cnt+2;
                L1.mag=L2.mag;
                vis[L1.x][L1.y][L2.mag]=1;
                q.push(L1);
            }
        }
    }
    return 10000000;
}
int main()
{
    int sx,sy,k=0;
    while(scanf("%d%d%d%d",&n,&m,&t,&p)!=EOF)
    {
        k++;
        for(int i=0;i<n;i++)
        {
            scanf("%s",map[i]);
            for(int j=0;j<m;j++)
            if(map[i][j]=='Y')
            {
                sx=i;sy=j;
            }
        }
        memset(vis,0,sizeof(vis));
        int ans=bfs(sx,sy);
        printf("Case %d:
",k);
        if(ans>t) printf("Poor Yifenfei, he has to wait another ten thousand years.
");
        else printf("Yes, Yifenfei will kill Lemon at %d sec.
",ans);
    }
    return 0;
}