Treasure of the Chimp Island

Treasure of the Chimp Island

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 207 Accepted Submission(s): 84

Problem Description

Bob Bennett, the young adventurer, has found the map to the treasure of the Chimp Island, where the ghost zombie pirate LeChimp, the infamous evil pirate of the Caribbeans has hidden somewhere inside the Zimbu Memorial Monument (ZM²). ZM² is made up of a number of corridors forming a maze. To protect the treasure, LeChimp has placed a number of stone blocks inside the corridors to block the way to the treasure. The map shows the hardness of each stone block which determines how long it takes to destroy the block. ZM² has a number of gates on the boundary from which Bob can enter the corridors. Fortunately, there may be a pack of dynamites at some gates, so that if Bob enters from such a gate, he may take the pack with him. Each pack has a number of dynamites that can be used to destroy the stone blocks in a much shorter time. Once entered, Bob cannot exit ZM² and enter again, nor can he walk on the area of other gates (so, he cannot pick more than one pack of dynamites).

The hardness of the stone blocks is an integer between 1 and 9, showing the number of days required to destroy the block. We neglect the time required to travel inside the corridors. Using a dynamite, Bob can destroy a block almost immediately, so we can ignore the time required for it too. The problem is to find the minimum time at which Bob can reach the treasure. He may choose any gate he wants to enter ZM².

 

Input

The input consists of multiple test cases. Each test case contains the map of ZM² viewed from the above. The map is a rectangular matrix of characters. Bob can move in four directions up, down, left, and right, but cannot move diagonally. He cannot enter a location shown by asterisk characters (*), even using all his dynamites! The character ($) shows the location of the treasure. A digit character (between 1 and 9) shows a stone block of hardness equal to the value of the digit. A hash sign (#) which can appear only on the boundary of the map indicates a gate without a dynamite pack. An uppercase letter on the boundary shows a gate with a pack of dynamites. The letter A shows there is one dynamite in the pack, B shows there are two dynamite in the pack and so on. All other characters on the boundary of the map are asterisks. Corridors are indicated by dots (.). There is a blank line after each test case. The width and the height of the map are at least 3 and at most 100 characters. The last line of the input contains two dash characters (--).

 

Output

For each test case, write a single line containing a number showing the minimum number of days it takes Bob to reach the treasure, if possible. If the treasure is unreachable, write IMPOSSIBLE.

 

Sample Input

*****#*********
*.1....4..$...*
*..***..2.....*
*..2..*****..2*
*..3..******37A
*****9..56....*
*.....******..*
***CA**********

*****
*$3**
*.2**
***#*

--

 

Sample Output

1
IMPOSSIBLE

 

Source

2006 Asia Regional Tehran

 

Recommend

lcy

 

/*
地宫的四周不是*的位置是地宫的入口，有炸  药数，地图中有相应的数字炸  药，可以选择炸还是不炸
取得宝藏的最小时间

遇到石头将炸还是不炸的情况都加到队列中；
然后爆搜
*/
#include<bits/stdc++.h>
#define N 105
#define INF 0x3f3f3f3f
using namespace std;
struct node
{
    int x,y,v,step;
    node(){}
    node(int a,int b,int c,int d)
    {
        x=a;
        y=b;
        v=c;
        step=d;
    }
    bool operator <(const  node &b) const
    {
        return step>b.step;
    }
};
char mapn[N][N];
int vis[N][N][];
int n,m;
int cur=INF;
int dir[][]={{,},{-,},{,-},{,}};
bool ok(int x,int y)
{
    if(x>=&&x<n&&y>=&&y<m&&mapn[x][y]!='*')
        return true;
    return false;
}
void bfs(priority_queue<node>q)
{
    memset(vis,-,sizeof vis);
    //cout<<q.size()<<endl;
    int c=;
    while(!q.empty())
    {
        //cout<<q.size()<<endl;
        node fr=q.top();
        q.pop();
        vis[fr.x][fr.y][fr.v]=fr.step;
        for(int i=;i<;i++)
        {
            int fx=fr.x+dir[i][];
            int fy=fr.y+dir[i][];
            if(!ok(fx,fy)) continue;
            if(mapn[fx][fy]=='.')//是路
            {
                if(vis[fx][fy][fr.v]==-||vis[fx][fy][fr.v]>fr.step)
                {
                    q.push(node(fx,fy,fr.v,fr.step));
                    vis[fx][fy][fr.v]=fr.step;
                }
            }
            else if(mapn[fx][fy]>=''&&mapn[fx][fy]<='')//数字
            {
                if(fr.v>&&(vis[fx][fy][fr.v-]==-||vis[fx][fy][fr.v-]>fr.step))//炸
                {
                    q.push(node(fx,fy,fr.v-,fr.step));
                    vis[fx][fy][fr.v-]=fr.step;
                }
                if(vis[fx][fy][fr.v]==-||vis[fx][fy][fr.v]>(fr.step+mapn[fx][fy]-''))//不炸
                {
                    q.push(node(fx,fy,fr.v,(fr.step+mapn[fx][fy]-'')));
                    vis[fx][fy][fr.v]=fr.step+mapn[fx][fy]-'';
                }
            }
            else if(mapn[fx][fy]=='$'&&fr.step<cur)//找到宝藏
            {
                cur=fr.step;

            }
        }
    }
    //cout<<"cur="<<cur<<endl;
}
int main()
{
    //freopen("C:\\Users\\acer\\Desktop\\in.txt","r",stdin);
    n=;
    priority_queue<node>q;
    while(gets(mapn[n++]))
    {
        //cout<<mapn[n-1]<<endl;
        for(int i=;mapn[n-][i];i++)
        {
            if(mapn[n-][i]=='#')
            {
                q.push(node(n-,i,,));
            }
            else if(mapn[n-][i]>='A'&&mapn[n-][i]<='Z')
            {
                q.push(node(n-,i,mapn[n-][i]-'A'+,));
            }
        }
        if(mapn[n-][]=='-')
            break;
        if(strcmp(mapn[n-],"")==)
        {
            cur=INF;
            m=strlen(mapn[]);
            n--;
            //cout<<n<<" "<<m<<endl;
            bfs(q);
            //cout<<q.size()<<endl;
            //cout<<"cur="<<cur<<endl;
            if(cur==INF)
                puts("IMPOSSIBLE");
            else
                printf("%d\n",cur);
            n=;
            while(!q.empty()) q.pop();
        }
    }
    return ;
}