Stealing Harry Potter‘s Precious
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1295 Accepted Submission(s): 618
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<N,M <= 100).
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
题目意思:
给一个n*m的迷宫，给一个k，下面k行为x, y。 一个人从@处开始走，只能上下左右走，走一步时间+1，#为墙，问人能否走遍给出的k个x,y。
若能输出最小步数，否则输出-1.
思路：
该题难点在于怎么判断一个人是否遍历k所有的x,y。 而且不同的路径已经经过的x,y也不同。若在每个路径中存放一个地图，那么很容易超时，由于k最大为4，所有对k个坐标状压即可。
代码：
1 #include <cstdio>
2 #include <cstring>
3 #include <iostream>
4 #include <algorithm>
5 #include <vector>
6 #include <queue>
7 using namespace std;
8
9 #define N 105
10
11 char map[N][N];
12 int visited[N][N][1<<4];
13 int n, m, k;
14 int xx[]={1,-1,0,0};
15 int yy[]={0,0,1,-1};
16
17 struct node{
18 int x, y, pr, t;
19 };
20
21 int bfs(int stx,int sty){
22 memset(visited,0,sizeof(visited));
23 node p, q;
24 int i, j;
25 queue<node>Q;
26 p.x=stx;p.y=sty;p.pr=0;p.t=0;
27 visited[p.x][p.y][p.pr]=1;
28 Q.push(p);
29 while(!Q.empty()){
30 p=Q.front();
31 Q.pop();
32 if(p.pr==(1<<k)-1) return p.t; //当遍历完k个坐标就return
33
34 for(i=0;i<4;i++){
35 p.x+=xx[i];
36 p.y+=yy[i];
37 if(p.x>=0&&p.x<n&&p.y>=0&&p.y<m&&map[p.x][p.y]!=‘#‘&&!visited[p.x][p.y][p.pr]){
38 visited[p.x][p.y][p.pr]=1;
39 if(map[p.x][p.y]!=‘.‘){ //当走到给的k个坐标处
40 if(p.pr&(1<<(map[p.x][p.y]-‘0‘))){ //已经走过该坐标
41 p.t+=1;
42 Q.push(p);
43 p.t-=1;
44 }
45 else{ //没走过该坐标
46 int u=p.pr;
47 p.pr=p.pr|(1<<(map[p.x][p.y]-‘0‘));
48 p.t+=1;
49 visited[p.x][p.y][p.pr]=1;
50 Q.push(p);
51 p.t-=1;
52 p.pr=u;
53 }
54 }
55 else{
56 p.t+=1;
57 Q.push(p);
58 p.t-=1;
59 }
60 }
61 p.x-=xx[i];
62 p.y-=yy[i];
63 }
64
65 }
66 return -1;
67 }
68
69 main()
70 {
71 int i, j, x, y;
72 while(scanf("%d %d",&n,&m)==2){
73 if(!n&&!m) break;
74 for(i=0;i<n;i++) scanf("%s",map[i]);
75 int stx, sty, f=0;
76 for(i=0;i<n;i++){
77 for(j=0;j<m;j++){
78 if(map[i][j]==‘@‘){
79 stx=i;sty=j;f=1;break;
80 }
81 }if(f) break;
82 }
83 scanf("%d",&k);
84 f=0;int cnt=0, ka=k;
85 while(ka--){
86 scanf("%d %d",&x,&y);
87 x--;y--;
88 if(f) continue;
89 if(map[x][y]==‘#‘){ //若给的坐标处为墙，肯定没法遍历完k个坐标
90 f=1;
91 }
92 else{
93 map[x][y]=cnt+‘0‘;
94 cnt++;
95 }
96 }
97 if(f) {
98 printf("-1\n");continue;
99 }
100 int ans=bfs(stx,sty);
101 if(ans!=-1) printf("%d\n",ans);
102 else printf("-1\n");
103 }
104 }