Hdu4771(杭州賽區)
Stealing Harry Potter's Precious
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1875 Accepted Submission(s): 878
Problem Description
Harry Potter has some precious. For example, his invisible robe, his wand and his owl. When Hogwarts school is in holiday, Harry Potter has to go back to uncle Vernon's home. But he can't bring his precious with him. As you know, uncle Vernon never allows
such magic things in his house. So Harry has to deposit his precious in the Gringotts Wizarding Bank which is owned by some goblins. The bank can be considered as a N × M grid consisting of N × M rooms. Each room has a coordinate. The coordinates of the upper-left
room is (1,1) , the down-right room is (N,M) and the room below the upper-left room is (2,1)..... A 3×4 bank grid is shown below:
Some rooms are indestructible and some rooms are vulnerable. Goblins always care more about their own safety than their customers" properties, so they live in the indestructible rooms and put customers' properties in vulnerable rooms. Harry Potter's precious
are also put in some vulnerable rooms. Dudely wants to steal Harry's things this holiday. He gets the most advanced drilling machine from his father, uncle Vernon, and drills into the bank. But he can only pass though the vulnerable rooms. He can't access
the indestructible rooms. He starts from a certain vulnerable room, and then moves in four directions: north, east, south and west. Dudely knows where Harry's precious are. He wants to collect all Harry's precious by as less steps as possible. Moving from
one room to another adjacent room is called a 'step'. Dudely doesn't want to get out of the bank before he collects all Harry's things. Dudely is stupid.He pay you $1,000,000 to figure out at least how many steps he must take to get all Harry's precious.
Input
There are several test cases.
In each test cases:
The first line are two integers N and M, meaning that the bank is a N × M grid(0
Then a N×M matrix follows. Each element is a letter standing for a room. '#' means a indestructible room, '.' means a vulnerable room, and the only '@' means the vulnerable room from which Dudely starts to move.
The next line is an integer K ( 0 < K <= 4), indicating there are K Harry Potter's precious in the bank.
In next K lines, each line describes the position of a Harry Potter's precious by two integers X and Y, meaning that there is a precious in room (X,Y).
The input ends with N = 0 and M = 0
Output
For each test case, print the minimum number of steps Dudely must take. If Dudely can't get all Harry's things, print -1.
Sample Input
2 3
##@
#.#
1
2 2
4 4
#@##
....
####
....
2
2 1
2 4
0 0
Sample Output
-1
5
Source
2013 Asia Hangzhou Regional Contest
這題有些人用什麼壓縮dp寫的，俺不會，後來發現一種超級巧妙的方法
你看啊k最多總共只有4個點加入a1,a2,a3,a4，起點是a0，那麼從a0一直遍歷所有點不就是a0->a1->a2-》a3->a4的a1,a2,a3,a4的全排列嗎，最多4！直接爆力，每次用next_permutaion（）更新排列即可，當天如果k比較大這種方法不行
還有注意next_permuation(a,a+n)如果你是從下標1開始的就是（a+1,a+n+1)不然會一直WA！
#include
#include
#include
#include
#include
using namespace std;
int n , m,k;
int visit[110][110];
int p[5];
char g[110][110];
int sx,sy;
int dx[] = {-1,1,0,0};
int dy[] = {0,0,-1,1};
struct node
{
int x,y,step;
node(int a,int b, int c): x(a),y(b),step(c) {}
node(){}
}ss[6];
int bfs()
{
queue q;
q.push(node(sx,sy,0));
memset(visit,0,sizeof(visit));
visit[sx][sy] = 1;
for(int i = 0; !q.empty(); )
{
node temp = q.front();
q.pop();
for(int j = 0; j < 4; j++)
{
int xx = temp.x + dx[j];
int yy = temp.y + dy[j];
int step = temp.step + 1;
if(xx < 0 || yy < 0 || xx >= n || yy >= m || g[xx][yy] == '#' || visit[xx][yy]) continue;
int flag = xx == ss[p[i]].x && yy == ss[p[i]].y;
if(flag)
{
while(!q.empty()) q.pop();
memset(visit,0,sizeof(visit));
if(++i == k) return step;
}
q.push(node(xx,yy,step));
visit[xx][yy] = 1;
if(flag) break;
}
}
return -1;
}
int main()
{
#ifdef xxz
freopen("in.txt","r",stdin);
#endif
while(scanf("%d%d",&n,&m)!=EOF && n != 0)
{
for(int i = 0; i < n; i++)
{
scanf("%s",g[i]);
for(int j = 0; j < m; j++)
{
if(g[i][j] == '@')
{
sx = i;
sy = j;
}
}
}
scanf("%d",&k);
int Case = 1;
for(int i = 0; i < k; i++)
{
scanf("%d%d",&ss[i].x,&ss[i].y);
ss[i].x--;
ss[i].y--;
p[i] = i;
Case *= i+1;
}
int ans = -1;
while(Case--)
{
int temp = bfs();
// cout< -1 && (temp < ans || ans == -1)) ans = temp;
next_permutation(p,p+k);
}
printf("%d\n",ans);
}
return 0;
}
