标签：des style blog color java os io strong
Nightmare
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7083 Accepted Submission(s): 3409
Problem Description
Ignatius had a nightmare last night. He found himself in a labyrinth with a time bomb on him. The labyrinth has an exit, Ignatius should get out of the labyrinth before the bomb explodes. The initial exploding time of the bomb is set to 6 minutes. To prevent the bomb from exploding by shake, Ignatius had to move slowly, that is to move from one area to the nearest area(that is, if Ignatius stands on (x,y) now, he could only on (x+1,y), (x-1,y), (x,y+1), or (x,y-1) in the next minute) takes him 1 minute. Some area in the labyrinth contains a Bomb-Reset-Equipment. They could reset the exploding time to 6 minutes.
Given the layout of the labyrinth and Ignatius‘ start position, please tell Ignatius whether he could get out of the labyrinth, if he could, output the minimum time that he has to use to find the exit of the labyrinth, else output -1.
Here are some rules:
1. We can assume the labyrinth is a 2 array.
2. Each minute, Ignatius could only get to one of the nearest area, and he should not walk out of the border, of course he could not walk on a wall, too.
3. If Ignatius get to the exit when the exploding time turns to 0, he can‘t get out of the labyrinth.
4. If Ignatius get to the area which contains Bomb-Rest-Equipment when the exploding time turns to 0, he can‘t use the equipment to reset the bomb.
5. A Bomb-Reset-Equipment can be used as many times as you wish, if it is needed, Ignatius can get to any areas in the labyrinth as many times as you wish.
6. The time to reset the exploding time can be ignore, in other words, if Ignatius get to an area which contain Bomb-Rest-Equipment, and the exploding time is larger than 0, the exploding time would be reset to 6.
Input
The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.
Each test case starts with two integers N and M(1<=N,Mm=8) which indicate the size of the labyrinth. Then N lines follow, each line contains M integers. The array indicates the layout of the labyrinth.
There are five integers which indicate the different type of area in the labyrinth:
0: The area is a wall, Ignatius should not walk on it.
1: The area contains nothing, Ignatius can walk on it.
2: Ignatius‘ start position, Ignatius starts his escape from this position.
3: The exit of the labyrinth, Ignatius‘ target position.
4: The area contains a Bomb-Reset-Equipment, Ignatius can delay the exploding time by walking to these areas.
Output
For each test case, if Ignatius can get out of the labyrinth, you should output the minimum time he needs, else you should just output -1.
Sample Input
3
3 3
2 1 1
1 1 0
1 1 3
4 8
2 1 1 0 1 1 1 0
1 0 4 1 1 0 4 1
1 0 0 0 0 0 0 1
1 1 1 4 1 1 1 3
5 8
1 2 1 1 1 1 1 4
1 0 0 0 1 0 0 1
1 4 1 0 1 1 0 1
1 0 0 0 0 3 0 1
1 1 4 1 1 1 1 1
Sample Output
4
-1
13
题目意思：
给出一个n*m的图，初始时一个人在map[i][j]=2处，只能向上、下、左、右移动1格，每走一步时间减1，初始化时间为6，当时间为1时还没找到出口3位置处，就输出1，否则输出最少走的步数，0是墙，1是空地，2是初始所在地方，3是目标，4可以把时间重置为6。
思路：
最优走迷宫问题一般就是bfs，模拟一下即可，判断下一步是否可走，若可走且走一步后时间不为0，就走这一步，然后判断此处是否是目标，是的话返回步数，是否是4，是的话更新时间。在我的代码中加了个visited[][]数组进行优化，因为每个位置最多走4次即可，再多的话进行无用功或者一直在两个地方之间徘徊跳不出bfs函数。
代码：
1 #include <cstdio>
2 #include <cstring>
3 #include <algorithm>
4 #include <vector>
5 #include <queue>
6 #include <iostream>
7 using namespace std;
8
9 struct node {
10 int x, y, w, step;
11 };
12
13 int n, m;
14 int map[10][10];
15 int visited[10][10];
16 int xx[]={1,-1,0,0};
17 int yy[]={0,0,1,-1};
18
19 int judge(int x,int y){
20 if(map[x][y]==0||x<0||x>=n||y<0||y>=m) return 0;
21 return 1;
22 }
23
24 int bfs(int x,int y){
25 int i, j;
26 node p, q;
27 p.x=x;p.y=y;p.w=6;p.step=0;
28 queue<node>Q;
29 Q.push(p);
30 while(!Q.empty()){
31 p=Q.front();
32 Q.pop();
33 for(i=0;i<4;i++){
34 q.x=p.x+xx[i];
35 q.y=p.y+yy[i];
36 q.w=p.w-1;
37 q.step=p.step+1;
38 if(judge(q.x,q.y)){
39 if(q.w>0&&visited[q.x][q.y]<4&&map[q.x][q.y]!=2){ //因为从2出发，再回到2已经多余了
40 if(map[q.x][q.y]==3) return q.step;
41 else if(map[q.x][q.y]==4){
42 q.w=6;
43 }
44 visited[q.x][q.y]++;
45 Q.push(q);
46 }
47 }
48 }
49 }
50 return 0;
51 }
52 main()
53 {
54 int i, j, t, x, y;
55 cin>>t;
56 while(t--){
57 memset(visited,0,sizeof(visited));
58 scanf("%d %d",&n,&m);
59 for(i=0;i<n;i++){
60 for(j=0;j<m;j++){
61 scanf("%d",&map[i][j]);
62 if(map[i][j]==2){
63 x=i;y=j;
64 }
65 }
66 }
67 int ans=bfs(x,y);
68 if(!ans) printf("-1\n");
69 else printf("%d\n",ans);
70 }
71 }
HDU 1072 bfs,布布扣,bubuko.com
HDU 1072 bfs
标签：des style blog color java os io strong
原文地址：http://www.cnblogs.com/qq1012662902/p/3893491.html
