HDU 1495 非常可乐

题目链接 http://acm.hdu.edu.cn/showproblem.php?pid=1495

有三个瓶子 A B C，每次都有6个状态

A->B 　　A->C 　　B->A 　　B->C　　C->A　　C->B

用BFS跑一遍就行了

代码：

 1 #include <cstdio>
 2 #include <cstdlib>
 3 #include <cstring>
 4 #include <queue>
 5 
 6 using namespace std;
 7 
 8 const int inf = 0x3f;
 9 const int INF = 0x3f3f3f3f;
10 const int maxn = 500;
11 struct Coke
12 {
13     int a, b, c;
14     int step;
15 }coke[maxn];
16 
17 int a, b, c, ok;
18 int Hash[maxn][maxn], dir[3];
19 queue<Coke> q;
20 Coke now, Ne;
21 
22 bool judge(int x, int y, int z)
23 {
24     //printf("In %d %d %d\n", x, y, z);
25     if(x < 0 || x > a) return false;
26     if(y < 0 || y > b) return false;
27     if(z < 0 || z > c) return false;
28     if(Hash[x][y]) return false;
29     Ne.a = x; Ne.b = y; Ne.c = z; Ne.step = now.step+1;
30     //printf("out %d %d %d\n", x, y, z);
31     Hash[x][y] = 1;
32     q.push(Ne);
33     if(x==a/2 && y==a/2) ok=1;
34     if(x==a/2 && z==a/2) ok=1;
35     if(z==a/2 && y==a/2) ok=1;
36     return true;
37 }
38 int BFS()
39 {
40     while(!q.empty()) q.pop();
41 
42     int x, y, z, mi, ma;
43     Coke s;
44     s.a = a; s.b = 0; s.c = 0; s.step = 0;
45     q.push(s);
46     Hash[a][0] = 1;
47     while(!q.empty())
48     {
49         now = q.front(); q.pop();
50         Ne = now;
51         //a->b
52         mi = min(now.a, (b-now.b));
53         x = now.a - mi; y = now.b + mi; z = now.c;
54         judge(x, y, z);
55         //a->c
56         mi = min(now.a, (c-now.c));
57         x = now.a - mi; z = now.c + mi; y = now.b;
58         judge(x, y, z);
59         //b->a
60         mi = min(now.b, (a-now.a));
61         y = now.b - mi; x = now.a + mi; z = now.c;
62         judge(x, y, z);
63         //b->c
64         mi = min(now.b, (c-now.c));
65         y = now.b - mi; z = now.c + mi; x = now.a;
66         judge(x, y, z);
67         //c->a
68         mi = min(now.c, (a-now.a));
69         z = now.c - mi; x = now.a + mi; y = now.b;
70         judge(x, y, z);
71         //c->b
72         mi = min(now.c, (b-now.b));
73         z = now.c - mi; y = now.b + mi; x = now.a;
74         judge(x, y, z);
75         
76         if(ok == 1) return now.step+1;
77     }
78     return 0;
79 }
80 
81 int main()
82 {
83     while(scanf("%d %d %d", &a, &b, &c) && (a+b+c))
84     {
85         if(a%2)
86         {
87             puts("NO"); continue;
88         }
89         ok = 0;
90         memset(Hash, 0, sizeof(Hash));
91         int ans = 0;
92         if(ans = BFS()) printf("%d\n", ans);
93         else puts("NO");
94     }
95     return 0;
96 }
View Code

 

posted @ 2015-05-10 16:59  豪气干云  阅读(165)  评论(0)  编辑  收藏  举报