Tempter of the Bone II
Time Limit:5000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
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
题解
走迷宫问最短时间，不同的是这道题有炸弹这个因素，同样的BFS，不过节点要多加地图的保存和标志数组多加一个用来标志炸弹个数即可
代码
#include <cstdio>
#include <queue>
#include <cstring>
#include <iostream>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <string>
#include <set>
#include <ctime>
#include <cmath>
#include <cctype>
using namespace std;
#define MAX 100000
#define LL long long
int cas=1,T;
struct Node
{
int x,y;
int num; //炸弹个数
int step; //步数
char mapp[8][8];
bool operator<(const Node &p)const
{
return step>p.step;
}
}node;
int net[4][2] = {{1,0},{0,1},{-1,0},{0,-1}};
char map[8][8];
int vis[8][8][577];
int main()
{
int n,m,sx,sy,ex,ey;
while (scanf("%d%d",&n,&m) && n)
{
priority_queue<Node>que;
for (int i = 0;i<n;i++)
scanf("%s",map[i]);
for (int i = 0;i<n;i++)
{
for (int j = 0;j<m;j++)
{
if (map[i][j]=='S')
{
map[i][j] = '.';
sx = i;sy =j;
}
else if (map[i][j]=='D')
{
map[i][j]='.';
ex=i;ey=j;
}
}
}
memset(vis,0,sizeof(vis));
node.x=sx;
node.y=sy;
for (int i = 0;i<n;i++)
for (int j = 0;j<m;j++)
node.mapp[i][j]=map[i][j];
node.step=0;
node.num=0;
vis[sx][sy][0]++;
que.push(node);
while (!que.empty())
{
node = que.top();
if (node.x==ex&&node.y==ey)
{
printf("%d\n",node.step);
break;
}
for (int i = 0;i<4;i++)
{
node.step++;
node.x+=net[i][0];
node.y+=net[i][1];
if (node.x>=0&&node.y>=0&&node.y<m&&node.x<n&&vis[node.x][node.y][node.num]<20)
{
if (node.mapp[node.x][node.y]=='.')
{
vis[node.x][node.y][node.num]++;
que.push(node);
}
else if (node.mapp[node.x][node.y]=='X' && node.num>0)
{
vis[node.x][node.y][node.num]++;
node.step++;
node.num--;
node.mapp[node.x][node.y]='.';
vis[node.x][node.y][node.num]++;
que.push(node);
}
else if (node.mapp[node.x][node.y]>='1' && node.mapp[node.x][node.y]<='9')
{
vis[node.x][node.y][node.num]++;
node.num+=node.mapp[node.x][node.y]-'0';
node.mapp[node.x][node.y]='.';
vis[node.x][node.y][node.num]++;
que.push(node);
}
}
node=que.top();
}
que.pop();
}
if (que.empty())
printf("-1\n");
}
return 0;
}
