poj3414Pots（bfs）

题目链接：http://poj.org/problem?id=3414

很有趣的题目。

每次六种决策，下面注释中已说明。

用string记录路径。

用二维vis数组记录两个杯子的状态，避免重复。

  1 #include<cstdio>
  2 #include<algorithm>
  3 #include<cstring>
  4 #include<iostream>
  5 #include<queue>
  6 #include<string>
  7 using namespace std;
  8 char *st[6]={"FILL(1)","FILL(2)","DROP(1)","DROP(2)","POUR(1,2)","POUR(2,1)"};
  9 int vis[110][110];  //用二维数组记录两个杯子的状态
 10 string s;
 11 struct node
 12 {
 13     int a,b;
 14     string s;  //记录路径
 15
 16 }now,nex;
 17
 18 int ca,cb,cc;
 19
 20 string bfs()
 21 {
 22     queue<node> q;
 23     now.a=0;
 24     now.b=0;
 25     now.s="";
 26     q.push(now);
 27     vis[0][0]=1;
 28     while(!q.empty())
 29     {
 30         now=q.front();
 31         q.pop();
 32         if(now.a==cc||now.b==cc) return now.s;
 33         if(now.a<ca&&!vis[ca][now.b])  //a加满
 34         {
 35             vis[ca][now.b]=1;
 36             nex.a=ca;
 37             nex.b=now.b;
 38             nex.s="";
 39             nex.s=now.s+‘0‘;
 40             q.push(nex);
 41
 42         }
 43           if(now.b<cb&&!vis[now.a][cb])  //b加满
 44         {
 45             vis[now.a][cb]=1;
 46             nex.a=now.a;
 47             nex.b=cb;
 48             nex.s="";
 49             nex.s=now.s+‘1‘;
 50             q.push(nex);
 51         }
 52         if(now.a>0&&!vis[0][now.b])  //a倒掉
 53         {
 54             vis[0][now.b]=1;
 55             nex.a=0;
 56             nex.b=now.b;
 57             nex.s="";
 58             nex.s=now.s+‘2‘;
 59             q.push(nex);
 60         }
 61         if(now.b>0&&!vis[now.a][0])  //b倒掉
 62         {
 63             vis[now.a][0]=1;
 64             nex.a=now.a;
 65             nex.b=0;
 66             nex.s="";
 67             nex.s=now.s+‘3‘;
 68             q.push(nex);
 69         }
 70         if(now.a>0&&now.b<cb)  //a倒给b
 71         {
 72             int v=min(now.a,cb-now.b);
 73             nex.a=now.a-v;
 74             nex.b=now.b+v;
 75             if(!vis[nex.a][nex.b])
 76             {
 77                 vis[nex.a][nex.b]=1;
 78                 nex.s="";
 79                 nex.s=now.s+‘4‘;
 80                 q.push(nex);
 81             }
 82         }
 83         if(now.b>0&&now.a<ca)  //b倒给a
 84         {
 85             int v=min(now.b,ca-now.a);
 86             nex.a=now.a+v;
 87             nex.b=now.b-v;
 88             if(!vis[nex.a][nex.b])
 89             {
 90                 vis[nex.a][nex.b]=1;
 91                 nex.s="";
 92                 nex.s=now.s+‘5‘;
 93                 q.push(nex);
 94             }
 95         }
 96
 97     }
 98     return "";
 99
100 }
101
102 int main()
103 {
104     while(scanf("%d%d%d",&ca,&cb,&cc)!=EOF)
105     {
106         memset(vis,0,sizeof(vis));
107        string s1 =bfs();
108        if(s1.length())
109         {
110             printf("%d\n",s1.length());
111             int len=s1.length();
112             for(int i=0;i<len;i++)
113             printf("%s\n",st[s1[i]-‘0‘]);
114         }
115         else puts("impossible");
116     }
117 }