hdu 2416 Treasure of the Chimp Island（bfs求最短路）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2416

Treasure of the Chimp Island

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 306    Accepted Submission(s): 148

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 (ZM ²). ZM ² 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. ZM ² 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 ZM ² 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 ZM ².

 

Input

The input consists of multiple test cases. Each test case contains the map of ZM ² 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
  

 

题意：输入一个字符矩阵；

   入口：'A'-'Z'，'#'，'#'代表不带炸弹进入，'A'代表带一个炸弹进入，以此类推

   终点：'$';

   数字代表进过该点有两种方式，a)炸掉，不付出代价通过；b)，花费该数字的代价通过该点

   '.'代表不花费任何代价通过，'*'则代表不能通过

思路：用一个三位数组记录当前点的状态，因为每个数字点可以有两种方法通过，第三维表示当前点的炸弹数量，此题输入也值得注意

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#define INF 9999999
using namespace std;
struct data
{
    int x,y;
    int bomb;
    int cost;
};
int row;
int col;
char mp[105][105];
int vis[105][105][26];
int result = INF;
int dir[4][2]= {0,1,1,0,-1,0,0,-1};

/*
bool operator<(data a,data b)
{
    return a.cost>b.cost;
}
*/
void bfs()
{
    memset(vis,-1,sizeof(vis));
    int i,j;
    data f;
    queue <data> qu;
    //入口
    for( i =0; i<row; i++)
    {
        for(j=0; j<col; j++)
        {

            if(mp[i][j]=='#' || (mp[i][j]>='A' && mp[i][j]<='Z'))
            {
                f.x=i;
                f.y=j;
                f.cost=0;
                if(mp[i][j]=='#') f.bomb=0;
                else f.bomb = mp[i][j]-'A'+1;
                qu.push(f);
                vis[i][j][f.bomb]=0;
                mp[i][j]='*';
            }
        }
    }
    //搜索
    while(!qu.empty())
    {
        data e;
        f = qu.front();
        qu.pop();
        for(i=0; i<4; i++)
        {
            e.x = f.x+dir[i][0];
            e.y = f.y+dir[i][1];

            if(mp[e.x][e.y]=='*' || e.x<0 || e.y<0 || e.x>=row || e.y>= col)
                continue;

            if(mp[e.x][e.y]=='.') //直接走过，时间不变，炸弹数不变
            {
                e.bomb=f.bomb;
                if(vis[e.x][e.y][e.bomb] == -1 || f.cost < vis[e.x][e.y][e.bomb])
                {
                    e.cost = f.cost;
                    vis[e.x][e.y][e.bomb] = f.cost;
                    qu.push(e);
                }
            }
            else if(mp[e.x][e.y]>='1' && mp[e.x][e.y]<='9')
            {
                //不炸,炸弹不变，时间加上mp[e.x][e.y]
                if(vis[e.x][e.y][f.bomb]==-1 || f.cost+mp[e.x][e.y]-'0' < vis[e.x][e.y][f.bomb])
                {
                    e.bomb = f.bomb;
                    e.cost = f.cost + mp[e.x][e.y]-'0';
                    vis[e.x][e.y][e.bomb]=e.cost;
                    qu.push(e);
                }
                //炸，炸弹数-1，时间不变；
                if(f.bomb>0)
                {
                    if(vis[e.x][e.y][f.bomb-1]==-1 || f.cost < vis[e.x][e.y][f.bomb-1])
                    {
                        e.bomb = f.bomb - 1;
                        e.cost = f.cost;
                        vis[e.x][e.y][e.bomb]=e.cost;
                        qu.push(e);
                    }
                }
            }
            /*else if(mp[e.x][e.y]=='$'&&f.cost<result)
                result = f.cost;*/
            else if(mp[e.x][e.y]=='$' && result > f.cost) //这个地方我找了一下午= =，找到$后，花费应该是前一步的，如果改成e.cost，则结果始终是0；
                result = f.cost;
        }
    }
}
int main()
{
    while(1)
    {
        row=0;
        while(gets(mp[row]))
        {
            if(mp[row][0]=='-')
                return 0;
            if(strlen(mp[row])<1)
                break;
            row++;
        }
        col=strlen(mp[0]);
        result=INF;
        bfs();
        if(result==INF) printf("IMPOSSIBLE\n");
        else printf("%d\n",result);
    }
    return 0;
}