HDU 2416 Treasure of the Chimp Island

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Treasure of the Chimp Island

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 231 Accepted Submission(s): 108

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

Source

2006 Asia Regional Tehran

Recommend

lcy

这道题还是很不错的练习搜索的题。

题意：给你一张地图，找出找到宝藏的最少时间。

刚开始的状态表示没有用三维，只是考虑到从不同的门BFS一次，然后比较得出

最小值。刚开始把路过的石头的状态用数组记录在结点里，直接导致TLE。

后来也看了些题解，搜索的时候把身上带的炸药个数考虑进去，对于每个点，将

炸与不炸分开讨论，最后比较得出最小值。

AC代码（内附注释）

#include<stdio.h>
#include<iostream>
#include<queue>
#include<string.h>
#include<algorithm>
#include<stdlib.h>
#include<cstring>
#include<string>
#define N 20000000
using namespace std;
int n,m,ss;//ss用于存最小时间
int a[105][105][30];//状态数组，第3维是带的炸药个数，前2个是坐标。
string map[105];
int dir[4][2]={{1,0},{0,1},{-1,0},{0,-1}};//用于变换坐标
struct node
{
	int x,y;
	int k,time;//k是携带的炸药包，time是到这里所花的时间
	//重载<运算符，用于优先队列的实现
	friend bool operator < (const node a,const node b)
	{
		return a.time>b.time;
	}
}w,v;
priority_queue<node> q;
bool check(int x,int y)//边境检查
{
	if(x>=0&&x<n&&y>=0&&y<m&&map[x][y]!='*')
		return 1;
	return 0;
}
void bfs()
{
	memset(a,-1,sizeof(a));
	for(int i=0;i<n;i++)//找出所有的门并加入队列
	{
		for(int j=0;j<m;j++)
		{
			if(map[i][j]=='#')
			{
				w.x=i;w.y=j;
				w.time=0;w.k=0;
				map[i][j]='*';
				q.push(w);
			}
			else if(map[i][j]>='A'&&map[i][j]<='Z')
			{
				w.x=i;w.y=j;
				w.time=0;
				w.k=map[i][j]-'A'+1;
				map[i][j]='*';
				q.push(w);
			}
		}
	}
	while(!q.empty())
	{
		v=q.top();
		q.pop();
		for(int i=0;i<4;i++)
		{
			w.x=v.x+dir[i][0];
			w.y=v.y+dir[i][1];
			if(!check(w.x,w.y))
				continue;
			if(map[w.x][w.y]=='.')//如果是路，直接走过去
			{
				w.k=v.k;
				//如果携带w.k个炸药在x,y点的状态没有检查过
				//或者以前的走法没有现在的走法优
				if(a[w.x][w.y][w.k]==-1||a[w.x][w.y][w.k]>v.time)
				{
					w.time=v.time;
					a[w.x][w.y][w.k]=w.time;
					q.push(w);
				}
			}
			else if(map[w.x][w.y]>='1'&&map[w.x][w.y]<='9')
			{
				//炸这个石头
				if(v.k>0&&(a[w.x][w.y][v.k-1]==-1||a[w.x][w.y][v.k-1]>v.time))
				{
					w.k=v.k-1;
					w.time=v.time;
					a[w.x][w.y][w.k]=w.time;
					q.push(w);
				}
				//不炸这个石头
				if(a[w.x][w.y][v.k]==-1||a[w.x][w.y][v.k]>v.time+map[w.x][w.y]-'0')
				{
					w.k=v.k;
					w.time=v.time+map[w.x][w.y]-'0';
					a[w.x][w.y][w.k]=w.time;
					q.push(w);
				}
			}
			else if(map[w.x][w.y]=='$'&&v.time<ss)
			{
				ss=v.time;
			}
		}
	}
}
int main()
{
	while(getline(cin,map[0])&&map[0][0]!='-')
	{
		int i=1;
		while(getline(cin,map[i])&&map[i].size()!=0)
			i++;
		n=i;
		m=map[0].size();
        ss=N;
		bfs();
		if(ss==N)
			printf("IMPOSSIBLE\n");
		else
			printf("%d\n",ss);
	}
	return 0;
}