坦克大战
时间限制：1000 ms | 内存限制：65535 KB
难度：3
- 描述
-
Many of us had played the game "Battle city" in
our childhood, and some people (like me) even often play it on computer
now.
What we are discussing is a simple edition of this game. Given a
map that consists of empty spaces, rivers, steel walls and brick walls only.
Your task is to get a bonus as soon as possible suppose that no enemies will
disturb you (See the following picture).
Your
tank can‘t move through rivers or walls, but it can destroy brick walls by
shooting. A brick wall will be turned into empty spaces when you hit it,
however, if your shot hit a steel wall, there will be no damage to the wall.
In each of your turns, you can choose to move to a neighboring (4 directions,
not 8) empty space, or shoot in one of the four directions without a move. The
shot will go ahead in that direction, until it go out of the map or hit a
wall. If the shot hits a brick wall, the wall will disappear (i.e., in this
turn). Well, given the description of a map, the positions of your tank and
the target, how many turns will you take at least to arrive
there?
- 输入
- The input consists of several test cases. The first line of each test case
contains two integers M and N (2 <= M, N <= 300). Each of the following
M lines contains N uppercase letters, each of which is one of ‘Y‘ (you), ‘T‘
(target), ‘S‘ (steel wall), ‘B‘ (brick wall), ‘R‘ (river) and ‘E‘ (empty
space). Both ‘Y‘ and ‘T‘ appear only once. A test case of M = N = 0 indicates
the end of input, and should not be processed. - 输出
- For each test case, please output the turns you take at least in a
separate line. If you can‘t arrive at the target, output "-1" instead. - 样例输入
3 4
YBEB
EERE
SSTE
0 0- 样例输出
8
- 来源
- POJ
- 上传者
- sadsad
较为水的一题。只需要转变为权值，然后就搜索即可，当然这道题，若果采用盲深搜索的话，会tie
贴一下tle代码，记录做题的思路吧！
代码：
1 #include<iostream>
2 #include<cstring>
3 #include<cstdio>
4 #include<vector>
5 #include<cstdlib>
6 using namespace std;
7 const int maxn=302;
8 int map[maxn][maxn];
9 //先采用深度搜索
10 int dir[4][2]={{0,1},{-1,0},{1,0},{0,-1}};
11 int m,n,sx,sy,ex,ey,minc,res;
12 void dfs(int x,int y)
13 {
14 //剪枝
15 if(x>0&&y>0&&x<=m&&y<=n)
16 {
17 if(x==ex&&y==ey)
18 {
19 if(res>minc) res=minc;
20 return ;
21 }
22 for(int i=0;i<4;i++)
23 {
24 int temp_val=map[x+dir[i][0]][y+dir[i][1]];
25 if(temp_val>0)
26 {
27 minc+=temp_val;
28 map[x+dir[i][0]][y+dir[i][1]]=-0x3f3f3f3f;
29 dfs(x+dir[i][0],y+dir[i][1]);
30 map[x+dir[i][0]][y+dir[i][1]]=temp_val;
31 minc-=temp_val;
32 }
33 }
34 }
35 }
36
37 int main()
38 {
39 int i,j;
40 char ss;
41 //freopen("test.in","r",stdin);
42 while(scanf("%d%d",&m,&n)!=EOF&&n+m)
43 {
44 getchar();
45 memset(map,0,sizeof(map));
46 minc=0;
47 res=0x3f3f3f;
48 for(i=1;i<=m;i++)
49 {
50 for(j=1;j<=n;j++)
51 {
52
53 scanf("%c",&ss);
54 if(ss==‘E‘)
55 map[i][j]=1;
56 else if(ss==‘B‘)
57 map[i][j]=2;
58 else if(ss==‘Y‘)
59 {
60 sx=i;
61 sy=j;
62 }
63 else if(ss==‘T‘)
64 {
65 map[i][j]=1;
66 ex=i;
67 ey=j;
68 }
69 else
70 map[i][j]=-0x3f3f3f3f; //代表障碍物
71 }
72 getchar();
73 }
74 //algorithm functiom
75 dfs(sx,sy);
76 printf("%d\n",res);
77 }
78 return 0;
79 }
上面的思路是tle了的，下面的是ac的，加了一个贪心的思想.
代码：时间为 4ms
1 #include<iostream>
2 #include<cstring>
3 #include<cstdio>
4 #include<vector>
5 #include<cstdlib>
6 using namespace std;
7 const int maxn=302;
8 struct node
9 {
10 int val;
11 int minc;
12 };
13 node map[maxn][maxn];
14 //先采用深度搜索
15 int dir[4][2]=
16 {
17 {0,1}, //向右
18 {-1,0}, //向左
19 {1,0}, //向上
20 {0,-1} //向下
21 };
22 int m,n,sx,sy,ex,ey;
23 void dfs(int x,int y)
24 {
25 //剪枝
26 if(x>0&&y>0&&x<=m&&y<=n)
27 {
28 for(int i=0;i<4;i++)
29 {
30 if(map[x+dir[i][0]][y+dir[i][1]].val>0&&map[x+dir[i][0]][y+dir[i][1]].minc>(map[x][y].minc+map[x+dir[i][0]][y+dir[i][1]].val)) //非智能 shit
31 {
32 map[x+dir[i][0]][y+dir[i][1]].minc=(map[x][y].minc+map[x+dir[i][0]][y+dir[i][1]].val);
33 dfs(x+dir[i][0],y+dir[i][1]);
34 }
35 }
36 }
37 }
38
39 int main()
40 {
41 int i,j;
42 char ss;
43 // freopen("test.in","r",stdin);
44 while(scanf("%d%d",&m,&n)!=EOF&&n+m)
45 {
46 getchar();
47 memset(map,0,sizeof(map));
48 for(i=1;i<=m;i++)
49 {
50 for(j=1;j<=n;j++)
51 {
52 map[i][j].minc=0x3f3f3f3f;
53 scanf("%c",&ss);
54 if(ss==‘E‘)
55 map[i][j].val=1;
56 else if(ss==‘B‘)
57 map[i][j].val=2;
58 else if(ss==‘Y‘)
59 {
60 map[i][j].minc=0;
61 sx=i;
62 sy=j;
63 }
64 else if(ss==‘T‘)
65 {
66 map[i][j].val=1;
67 ex=i;
68 ey=j;
69 }
70 else
71 map[i][j].val=-1; //代表障碍物
72 }
73 getchar();
74 }
75 //algorithm functiom
76 dfs(sx,sy);
77 if(map[ex][ey].minc==0x3f3f3f3f)
78 printf("-1\n");
79 else
80 printf("%d\n",map[ex][ey].minc);
81 }
82 return 0;
83 }
当然还可以用bfs来做，时间会更快什么的........
nyoj-----284坦克大战（带权值的图搜索）,布布扣,bubuko.com