HDU 2128Tempter of the Bone II

Tempter of the Bone II

Time Limit : 10000/5000ms (Java/Other)   Memory Limit : 98304/32768K (Java/Other)

Total Submission(s) : 3   Accepted Submission(s) : 1

Problem Description

The doggie found a bone in an ancient maze, which fascinated him a lot. However, when he picked it up, the maze was changed and the way he came in was lost.He realized that the bone was a trap, and he tried desperately to get out of this maze.

The maze was a rectangle with the sizes of N by M. The maze is made up of a door,many walls and many explosives. Doggie need to reach the door to escape from the tempter. In every second, he could move one block to one of the upper, lower, left or right neighboring
blocks. And if the destination is a wall, Doggie need another second explode and a explosive to explode it if he had some explosives. Once he entered a block with explosives,he can take away all of the explosives. Can the poor doggie survive? Please help him.

 

Input

The input consists of multiple test cases. The first line of each test case contains two integers N, M,(2 <= N, M <= 8). which denote the sizes of the maze.The next N lines give the maze layout, with each line containing M characters. A character is one of
the following: 'X': a block of wall; 'S': the start point of the doggie; 'D': the Door; '.': an empty block; '1'--'9':explosives in that block. Note,initially he had no explosives. The input is terminated with two 0's. This test case is not to be processed.

 

Output

For each test case, print the minimum time the doggie need to escape if the doggie can survive, or -1 otherwise.

 

Sample Input

4 4
SX..
XX..
....
1..D
4 4
S.X1
....
..XX
..XD
0 0

 

Sample Output

-1
9

 

Author

XHD

 

Source

HDU 2007-10 Programming Contest

地图上有炸弹，可以炸掉墙壁。问最少几步能达到终点

状态压缩  地图除了起点和终点，有62个点，可以用状态压缩保存拿去炸弹和炸掉墙的状态

用容器标记，四维  状态，坐标，炸弹数

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<vector>
#include<cmath>
#define LL unsigned long long
using namespace std;
int n,m;
char Map[10][10];
int Mp[10][10];
int dir[4][2]= {0,1,0,-1,1,0,-1,0};
int len;
struct node
{
int x,y;
int t;
int boom;
unsigned long long v;
friend bool operator < (node a,node b)
{
return a.t>b.t;
}
bool cheak()
{
if(x>=0&&x<n&&y>=0&&y<m)
return true ;
return false ;
}
} st,ed;
vector<unsigned long long>v[8][8][9*8*8+1];//标记 当前点拥有的炸弹数的状态 出现过则舍弃

bool cheak(node a)
{
for(int i=0; i<v[a.x][a.y][a.boom].size(); i++)
if(a.v==v[a.x][a.y][a.boom][i])
return false ;
return true ;
}

void bfs()
{
priority_queue<node>q;
q.push(st);
for(int i=0; i<8; i++)
for(int j=0; j<8; j++)
for(int l=0; l<8*8*9+1; l++)
v[i][j][l].clear();
v[st.x][st.y][st.boom].push_back(st.v);
while(q.size())
{
st=q.top();
q.pop();
if(Map[st.x][st.y]=='D')
{
printf("%d\n",st.t);
return ;
}
for(int i=0; i<4; i++)
{
ed=st;
ed.t++;
ed.x+=dir[i][0];
ed.y+=dir[i][1];
if(ed.cheak())
{
if(Mp[ed.x][ed.y]>=0&&(ed.v&(1LL<<Mp[ed.x][ed.y]))==0)
{
if(Map[ed.x][ed.y]=='X')
{
if(ed.boom>0)
{
ed.boom--;
ed.t++;
ed.v|=(1LL<<Mp[ed.x][ed.y]);
if(cheak(ed))
{
v[ed.x][ed.y][ed.boom].push_back(ed.v);
q.push(ed);
}
}
}
else
{
ed.boom+=Map[ed.x][ed.y]-'0';
ed.v|=(1LL<<Mp[ed.x][ed.y]);
if(cheak(ed))
{
v[ed.x][ed.y][ed.boom].push_back(ed.v);
q.push(ed);
}
}
}
else
{
if(cheak(ed))
{
v[ed.x][ed.y][ed.boom].push_back(ed.v);
q.push(ed);
}
}
}
}
}
printf("-1\n");
return ;
}

int main()
{
while(cin>>n>>m)
{
if(n==0&&m==0)
break;
memset(Mp,-1,sizeof(Mp));
len=0;
for(int i=0; i<n; i++)
{
for(int j=0; j<m; j++)
{
cin>>Map[i][j];
if(Map[i][j]=='X'||Map[i][j]>='0'&&Map[i][j]<='9') Mp[i][j]=len++;
else if(Map[i][j]=='S')
{
st.x=i;
st.y=j;
st.t=0;
st.boom=0;
st.v=0;
}
}
}
bfs();
}
return 0;
}