POJ 3414 BFS 输出过程

Pots
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 17456   Accepted: 7407   Special Judge

Description

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

  1. FILL(i)        fill the pot i (1 ≤ i ≤ 2) from the tap;
  2. DROP(i)      empty the pot i to the drain;
  3. 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)

Source

Northeastern Europe 2002, Western Subregion
思路：和非常可乐很像，需要输出过程。
代码：
  1 #include "cstdio"
  2 #include "stdlib.h"
  3 #include "iostream"
  4 #include "algorithm"
  5 #include "string"
  6 #include "cstring"
  7 #include "queue"
  8 #include "cmath"
  9 #include "vector"
 10 #include "map"
 11 #include "set"
 12 #define mj
 13 #define db double
 14 #define ll long long
 15 using namespace std;
 16 const int N=1e2+5;
 17 int cas=1;
 18 int a,b,c;
 19 int ok=0;
 20 int v[N][N];
 21 char s[10][10]={"","FILL(1)","FILL(2)","DROP(1)","DROP(2)","POUR(1,2)","POUR(2,1)"};
 22 struct P
 23 {
 24     int a,b,t;
 25     char s[1000];//记录操作
 26 };
 27 
 28 void bfs()
 29 {
 30     queue<P>q;
 31     P p,u;
 32     p.a=0,p.b=0,p.t=0,p.s[0]='0';
 33     v[p.a][p.b]=1;
 34     q.push(p);
 35     while(!q.empty()){
 36         u=q.front();
 37         q.pop();
 38         if(u.a==c || u.b==c){
 39             ok=1;
 40             printf("%d\n",u.t);
 41             for(int i=0;i<u.t;i++){
 42                 printf("%s\n",s[u.s[i]-'0']);//输出操作
 43             }
 44             return ;
 45         }
 46         if(u.a<a){
 47             p.a=a;
 48             p.b=u.b;
 49             p.t=u.t+1;
 50             strcpy(p.s,u.s);//复制之前的操作
 51             p.s[u.t]=1+'0';
 52             if(!v[p.a][p.b]){
 53                 v[p.a][p.b]=1;
 54                 q.push(p);
 55             }
 56         }
 57         if(u.b<b){
 58             p.b=b;
 59             p.a=u.a;
 60             p.t=u.t+1;
 61             strcpy(p.s,u.s);
 62             p.s[u.t]=2+'0';
 63             if(!v[p.a][p.b]){
 64                 v[p.a][p.b]=1;
 65                 q.push(p);
 66             }
 67         }
 68         if(u.a!=0){
 69             p.a=0;
 70             p.b=u.b;
 71             p.t=u.t+1;
 72             strcpy(p.s,u.s);
 73             p.s[u.t]=3+'0';
 74             if(!v[p.a][p.b]){
 75                 v[p.a][p.b]=1;
 76                 q.push(p);
 77             }
 78         }
 79         if(u.b!=0){
 80             p.b=0;
 81             p.a=u.a;
 82             p.t=u.t+1;
 83             strcpy(p.s,u.s);
 84             p.s[u.t]=4+'0';
 85             if(!v[p.a][p.b]){
 86                 v[p.a][p.b]=1;
 87                 q.push(p);
 88             }
 89         }
 90         if(u.a!=0 && u.b<b){
 91             if(b-u.b>=u.a){
 92                 p.b=u.a+u.b;
 93                 p.a=0;
 94             }
 95             else{
 96                 p.a=u.a+u.b-b;
 97                 p.b=b;
 98             }
 99             p.t=u.t+1;
100             strcpy(p.s,u.s);
101             p.s[u.t]=5+'0';
102             if(!v[p.a][p.b]){
103                 v[p.a][p.b]=1;
104                 q.push(p);
105             }
106         }
107         if(u.b!=0 && u.a<a){
108             if(a-u.a>=u.b){
109                 p.a=u.a+u.b;
110                 p.b=0;
111             }
112             else{
113                 p.b=u.a+u.b-a;
114                 p.a=a;
115             }
116             p.t=u.t+1;
117             strcpy(p.s,u.s);
118             p.s[u.t]=6+'0';
119             if(!v[p.a][p.b]){
120                 v[p.a][p.b]=1;
121                 q.push(p);
122             }
123         }
124     }
125 }
126 int main()
127 {
128     scanf("%d %d %d",&a,&b,&c);
129     memset(v,0, sizeof(v));
130     bfs();
131     if(!ok) printf("impossible\n");
132     return 0;
133 }

 

 
posted @ 2017-07-18 15:08  thges  阅读(140)  评论(0)  编辑  收藏  举报