HDU1044 Collect More Jewels(BFS+DFS+地图压缩)

                                                   Collect More Jewels

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3548    Accepted Submission(s): 704

Problem Description

It is written in the Book of The Lady: After the Creation, the cruel god Moloch rebelled against the authority of Marduk the Creator.Moloch stole from Marduk the most powerful of all the artifacts of the gods, the Amulet of Yendor, and he hid it in the dark cavities of Gehennom, the Under World, where he now lurks, and bides his time.

Your goddess The Lady seeks to possess the Amulet, and with it to gain deserved ascendance over the other gods.

You, a newly trained Rambler, have been heralded from birth as the instrument of The Lady. You are destined to recover the Amulet for your deity, or die in the attempt. Your hour of destiny has come. For the sake of us all: Go bravely with The Lady!

If you have ever played the computer game NETHACK, you must be familiar with the quotes above. If you have never heard of it, do not worry. You will learn it (and love it) soon.

In this problem, you, the adventurer, are in a dangerous dungeon. You are informed that the dungeon is going to collapse. You must find the exit stairs within given time. However, you do not want to leave the dungeon empty handed. There are lots of rare jewels in the dungeon. Try collecting some of them before you leave. Some of the jewels are cheaper and some are more expensive. So you will try your best to maximize your collection, more importantly, leave the dungeon in time.

 

Input

Standard input will contain multiple test cases. The first line of the input is a single integer T (1 <= T <= 10) which is the number of test cases. T test cases follow, each preceded by a single blank line.

The first line of each test case contains four integers W (1 <= W <= 50), H (1 <= H <= 50), L (1 <= L <= 1,000,000) and M (1 <= M <= 10). The dungeon is a rectangle area W block wide and H block high. L is the time limit, by which you need to reach the exit. You can move to one of the adjacent blocks up, down, left and right in each time unit, as long as the target block is inside the dungeon and is not a wall. Time starts at 1 when the game begins. M is the number of jewels in the dungeon. Jewels will be collected once the adventurer is in that block. This does not cost extra time.

The next line contains M integers，which are the values of the jewels.

The next H lines will contain W characters each. They represent the dungeon map in the following notation:
> [*] marks a wall, into which you can not move;
> [.] marks an empty space, into which you can move;
> [@] marks the initial position of the adventurer;
> [<] marks the exit stairs;
> [A] - [J] marks the jewels.

 

Output

Results should be directed to standard output. Start each case with "Case #:" on a single line, where # is the case number starting from 1. Two consecutive cases should be separated by a single blank line. No blank line should be produced after the last test case.

If the adventurer can make it to the exit stairs in the time limit, print the sentence "The best score is S.", where S is the maximum value of the jewels he can collect along the way; otherwise print the word "Impossible" on a single line.

 

Sample Input

   
   
   
   

    
    
    
    3  4 4 2 2 100 200 **** *@A* *B<* ****  4 4 1 2 100 200 **** *@A* *B<* ****  12 5 13 2 100 200 ************ *B.........* *.********.* *@...A....<* ************
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    Case 1: The best score is 200.  Case 2: Impossible  Case 3: The best score is 300.
   
   
   
   

 

Source

Asia 2005, Hangzhou (Mainland China), Preliminary

 

Recommend

JGShining



解题思路：本题需要做地图压缩，要是直接（BFS+DFS）搜索的话，思路会很混乱，还会超时。
          咋地一看题目，第一感觉就是，BFS+DFS，也就是说BFS求出所有可以走得地方到终点的距离，然后再DFS全图，果断超时了。后面想到地图压缩，直接BFS全图，求出各宝石、起点、终点两两之间的距离，然后用数组存储形成新的地图。然后再用DFS搜索新的地图得出答案（相关基础题目（链接）：HDU2614 Beat）。
        主要问题有二：
        （一）宝石的定位与定价问题，该问题无法解决将导致BFS无法进行，无法形成新地图。直接用结构体数组存储（x,y,val）即可，用的时候，直接记录下标即可找到宝石的地图位置和价值（需要把起点和终点处理下，分别放到数组的第一个和最后一个下标的存储内容中（相对于W））。
         （二）DFS搜索序问题，该问题无法解决将导致答案错误。若再该题代码中无法理解，请到基础题链接，这里不分析。
        剪枝：只要剪掉无法到达终点和已经遍历过的的枝节即可（BFS中的inmap(next.x,next.y)&&!flag[next.x][next.y]&&map[next.x][next.y]!='*'；DFS中的：宝石没捡过（不会重捡），时间还可以走到终点（不会超时），宝石捡完标记以及main函数中的，若起点就走不到终点，直接输出Impossible）

#include<cstdio> #include<cstring> #include<queue> #include<algorithm> using namespace std; int n,m,l,w; char map[50][50]; int flag[50][50]; int time1[14][14]; int flag_jews[14]; int dir[4][2]={1,0,0,1,-1,0,0,-1};   //四个可以走的方向，用于bfs int f,max_valu; struct node    //用于bfs {     int x;     int y;     int t; }; struct node2  //存储宝石情况，用于dfs {     int x;     int y;     int val; }jews[12]; int inmap(int x,int y)//判断点(x,y)是否在地图内，用于bfs {     if(x>=0&&x<n&&y>=0&&y<m)         return 1;     return 0; } int bfs(int x,int y,int x1,int y1) //求点（x,y）到点(x1,y1)之间的距离 {     int i;     node first,next;     queue<node> q;     if(x==x1&&y==y1)         return 0;     memset(flag,0,sizeof(flag));     first.x=x;     first.y=y;     first.t=0;     flag[x][y]=1;     q.push(first);     while(!q.empty())     {         first=q.front();         q.pop();         for(i=0;i<4;i++)         {             next.x=first.x+dir[i][0];             next.y=first.y+dir[i][1];             next.t=first.t+1;             if(inmap(next.x,next.y)&&!flag[next.x][next.y]&&map[next.x][next.y]!='*')             {                 if(next.x==x1&&next.y==y1)                     return next.t;                 flag[next.x][next.y]=1;                 q.push(next);             }         }     }     return 10000000; } void dfs(int r,int ld,int w1,int val)//第r个宝石，剩下ld时间，还有W个宝石没有捡，当前已有宝石的价值 {     if(w==0)  //宝石捡完了         return ;     for(int i=1;i<w+1;i++)     {         if(!flag_jews[i]&&time1[i][w+1]<=ld-time1[r][i]) //宝石没捡过（不会重捡），时间还可以走到终点（不会超时）         {             val+=jews[i].val;             flag_jews[i]=1;             max_valu=val>max_valu?val:max_valu;             dfs(i,ld-time1[r][i],w1-1,val);             val-=jews[i].val;             flag_jews[i]=0;         }     } } int main() {     int t;     int i,j,k;     scanf("%d",&t);     for(k=1;k<=t;k++)     {         f=0;         max_valu=0;         scanf("%d%d%d%d",&m,&n,&l,&w);         jews[0].val=0;         jews[w+1].val=0;         for(i=1;i<=w;i++)             scanf("%d",&jews[i].val);         for(i=0;i<n;i++)  //地图输入处理         {             scanf("%s",map[i]);             for(j=0;j<m;j++)             {                 if(map[i][j]>='A'&&map[i][j]<='J')  //此处有宝石                 {                     jews[map[i][j]-'A'+1].x=i;                     jews[map[i][j]-'A'+1].y=j;                 }                 else if(map[i][j]=='@')    //起点                 {                     jews[0].x=i;                     jews[0].y=j;                     map[i][j]='.';                 }                 else if(map[i][j]=='<')    //终点                 {                     jews[w+1].x=i;                     jews[w+1].y=j;                     map[i][j]='.';                 }             }         }         for(i=0;i<w+2;i++)   //各宝石堆，起点，终点的距离             for(j=0;j<w+2;j++)                 time1[i][j]=time1[j][i]=bfs(jews[i].x,jews[i].y,jews[j].x,jews[j].y);         if(k!=1)             printf("\n");         printf("Case %d:\n",k);         if(time1[0][w+1]>l)          printf("Impossible\n");  //起点直接走到终点，超时         else         {             memset(flag_jews,0,sizeof(flag_jews));             flag_jews[0]=1;             dfs(0,l,w,0);             printf("The best score is %d.\n",max_valu);         }     }     return 0; }