POJ 3414 Pots

BFS入门，经典倒水问题，想当年在UVa上做的第一道BFS，今天在POJ上拿来复习一下。

拓展队列时注意去掉一些不必要的情况和细节：

1.空杯子不能向外倒水

2.水从A倒入B后，B水满或不满分情况考虑。据说可以一个式子解决……我是没一个式子解决掉，就分开写了。

3.水满的杯子不能再向里倒水

 

  1 #include <cstdio>

  2 #include <cstring>

  3 #include <cstdlib>

  4 #include <queue>

  5 

  6 using namespace std;

  7 

  8 const int MAXN = 30000;

  9 

 10 bool vis[MAXN];

 11 int A, B, C;

 12 int head[MAXN];

 13 int step[MAXN];

 14 int cnt[MAXN];

 15 

 16 int BFS()

 17 {

 18     memset( vis, false, sizeof(vis) );

 19     queue<int> Q;

 20 

 21     int st = 0;

 22     Q.push(st);

 23 

 24     vis[st] = true;

 25     head[st] = -1;

 26     cnt[st] = 0;

 27 

 28     while ( !Q.empty() )

 29     {

 30         int cur = Q.front();

 31         Q.pop();

 32 

 33         int AA = cur / 100;

 34         int BB = cur % 100;

 35 

 36         if ( AA == C || BB == C ) return cur;

 37         int temp;

 38 

 39         if ( AA != A ) // fill A

 40         {

 41             temp = A * 100 + BB ;

 42 

 43             if ( !vis[temp] )

 44             {

 45                 vis[temp] = true;

 46                 head[temp] = cur;

 47                 step[temp] = 1;

 48                 cnt[temp] = cnt[cur] + 1;

 49                 Q.push( temp );

 50             }

 51         }

 52 

 53         if ( AA != 0 )

 54         {

 55             if ( BB != B )  // pour A to B

 56             {

 57                 if ( AA + BB <= B ) temp = AA + BB;

 58                 else temp = ( AA - ( B - BB ) ) * 100 + B;

 59 

 60                 if ( !vis[temp] )

 61                 {

 62                     vis[temp] = true;

 63                     head[temp] = cur;

 64                     step[temp] = 2;

 65                     cnt[temp] = cnt[cur] + 1;

 66                     Q.push( temp );

 67                 }

 68             }

 69 

 70             temp = BB;  //drop A

 71 

 72             if ( !vis[temp] )

 73             {

 74                 vis[temp] = true;

 75                 head[temp] = cur;

 76                 step[temp] = 3;

 77                 cnt[temp] = cnt[cur] + 1;

 78                 Q.push( temp );

 79             }

 80         }

 81 

 82         if ( BB != B )  // fill B

 83         {

 84             temp = AA * 100 + B ;

 85    

 86             if ( !vis[temp] )

 87             {

 88                 vis[temp] = true;

 89                 head[temp] = cur;

 90                 step[temp] = 4;

 91                 cnt[temp] = cnt[cur] + 1;

 92                 Q.push( temp );

 93             }

 94         }

 95 

 96         if ( BB != 0 )

 97         {

 98             if ( AA != A )   // pour B to A

 99             {

100                 if ( AA + BB <= A ) temp = ( AA + BB ) * 100;

101                 else temp = A * 100 + BB - ( A - AA );

102 

103                 if ( !vis[temp] )

104                 {

105                     vis[temp] = true;

106                     head[temp] = cur;

107                     step[temp] = 5;

108                     cnt[temp] = cnt[cur] + 1;

109                     Q.push(temp);

110                 }

111             }

112 

113             temp = AA * 100; // drop B

114             if ( !vis[temp] )

115             {

116                 vis[temp] = true;

117                 head[temp] = cur;

118                 step[temp] = 6;

119                 cnt[temp] = cnt[cur] + 1;

120                 Q.push( temp );

121             }

122         }

123     }

124 

125     return -1;

126 }

127 

128 void DFS( int cur )

129 {

130     if ( cur == -1 ) return;

131     DFS( head[cur] );

132     switch( step[cur] )

133     {

134         case 1: puts("FILL(1)"); break;

135         case 2: puts("POUR(1,2)"); break;

136         case 3: puts("DROP(1)"); break;

137         case 4: puts("FILL(2)"); break;

138         case 5: puts("POUR(2,1)"); break;

139         case 6: puts("DROP(2)"); break;

140     }

141     return;

142 }

143 

144 int main()

145 {

146     while ( ~scanf( "%d%d%d", &A, &B, &C ) )

147     {

148         int ans = BFS();

149         if ( ans != -1 )

150         {

151             printf( "%d\n", cnt[ans] );

152             DFS( ans );

153         }

154         else puts("impossible");

155     }

156     return 0;

157 }