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POJ 3414 Pots （dfs，这个代码好长啊QAQ）

Description

You are given two pots, having the volume of A and B liters respectively. The following operations can be performed:

FILL(i) fill the pot i (1 ≤ i ≤ 2) from the tap;
DROP(i) empty the pot i to the drain;
POUR(i,j) pour from pot i to pot j; after this operation either the pot j is full (and there may be some water left in the pot i), or the pot i is empty (and all its contents have been moved to the pot j).

Write a program to find the shortest possible sequence of these operations that will yield exactly C liters of water in one of the pots.

Input

On the first and only line are the numbers A, B, and C. These are all integers in the range from 1 to 100 and C≤max(A,B).

Output

The first line of the output must contain the length of the sequence of operations K. The following K lines must each describe one operation. If there are several sequences of minimal length, output any one of them. If the desired result can’t be achieved, the first and only line of the file must contain the word ‘impossible’.

Sample Input

3 5 4

Sample Output

6
FILL(2)
POUR(2,1)
DROP(1)
POUR(2,1)
FILL(2)
POUR(2,1)

给你两个壶，容量a,b,让你三种操作 1.接满 2.一个壶倒到另一个壶（这个壶倒空了或者那个壶满了 才停止） 3.把一个壶倒空 问你几步两个壶中能有一个壶容量是能c。
这个题显然bfs，就是node结构体里面在加上一个queue和string保存下之前的状态。没啥了，代码好长啊QAQ。
代码如下：

  1 #include <iostream>
  2 #include <algorithm>
  3 #include <string>
  4 #include <cstring>
  5 #include <cmath>
  6 #include <cstdio>
  7 #include <queue>
  8 #include <vector>
  9 using namespace std;
 10 struct node
 11 {
 12     int na,nb,step;
 13     vector<int> num;
 14     string ns;
 15 };
 16 bool vis [110][110];
 17 int a,b,c;
 18 bool iffind;
 19 void init ()
 20 {
 21     memset(vis,false,sizeof vis);
 22     iffind=false;
 23 }
 24 void bfs(node now)
 25 {
 26     queue<node>q;
 27     q.push(now);
 28     while (!q.empty())
 29     {
 30         node temp=q.front();
 31         q.pop();
 32         if (vis[temp.na][temp.nb])
 33             continue ;
 34         vis[temp.na][temp.nb]=true;
 35         if (temp.na==c||temp.nb==c)//success
 36         {
 37             iffind=true;
 38             printf("%d
",temp.step);
 39             //printf("=== %d   %d
",sizeof(temp.ns),temp.num.size());
 40             for (int i=0;i<temp.num.size();++i)
 41             {
 42                 if (temp.ns[i]=='F')
 43                 {
 44                     printf("FILL");
 45                     if (temp.num[i]==1)
 46                     printf("(1)
");
 47                     else
 48                     printf("(2)
");
 49                 }
 50                 else if (temp.ns[i]=='P')
 51                 {
 52                     printf("POUR");
 53                     if (temp.num[i]==1)
 54                     printf("(1,2)
");
 55                     else
 56                     printf("(2,1)
");
 57                 }
 58                 else
 59                 {
 60                     printf("DROP");
 61                     if (temp.num[i]==1)
 62                     printf("(1)
");
 63                     else
 64                     printf("(2)
");
 65                 }
 66             }
 67             return ;
 68         }
 69         if (temp.na!=a)//fill a
 70         {
 71             node now=temp;
 72             now.step++;
 73             now.na=a;
 74             now.ns+="F";
 75             now.num.push_back(1);
 76             q.push(now);
 77         }
 78         if (temp.nb!=b)//fill b
 79         {
 80             node now=temp;
 81             now.step++;
 82             now.nb=b;
 83             now.ns+="F";
 84             now.num.push_back(2);
 85             q.push(now);
 86         }
 87         if (temp.na>0&&temp.nb<b)//pour a to b
 88         {
 89             node now=temp;
 90             now.step++;
 91             now.na-=min(temp.na,b-temp.nb);
 92             now.nb+=min(temp.na,b-temp.nb);
 93             now.ns+="P";
 94             now.num.push_back(1);
 95             q.push(now);
 96         }
 97         if (temp.nb>0&&temp.na<a)//pour b to a
 98         {
 99             node now=temp;
100             now.step++;
101             now.nb-=min(temp.nb,a-temp.na);
102             now.na+=min(temp.nb,a-temp.na);
103             now.ns+="P";
104             now.num.push_back(2);
105             q.push(now);
106         }
107         if(temp.na!=0)// drop a
108         {
109             node now=temp;
110             now.step++;
111             now.na=0;
112             now.ns+='D';
113             now.num.push_back(1);
114             q.push(now);
115         }
116         if(temp.nb!=0)//drop b
117         {
118             node now=temp;
119             now.step++;
120             now.nb=0;
121             now.ns+='D';
122             now.num.push_back(2);
123             q.push(now);
124         }
125     }
126     return ;
127 }
128 int main()
129 {
130     while (~scanf("%d%d%d",&a,&b,&c))
131     {
132         init();
133         node x;
134         x.na=0,x.nb=0,x.ns="",x.step=0,x.num.clear();
135         bfs(x);
136         if (iffind==false){
137             printf("impossible
");
138         }
139     }
140     return 0;
141 }

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原文地址：https://www.cnblogs.com/agenthtb/p/6063385.html