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)
雖然不能打acm了，但是還是要補一下之前的坑。
直接按著狀態廣搜，標記這個狀態出沒出現過，出現過就不要向下擴展了。沒出現就向下擴展。
代碼如下：

#include<iostream> #include<cstdio> #include<algorithm> #include<queue> #include<stack> #include<map> #define ll long long using namespace std;char ss[6][20]={"FILL(2)","FILL(1)","POUR(2,1)","POUR(1,2)","DROP(1)","DROP(2)"}; struct node{int a;int b;string s;node(int x,int y,string z){a=x,b=y,s=z;} }; int a,b,c;inline void bfs() {queue<node> q;map<pair<int,int>,bool> mp;q.push(node(0,0,""));mp[make_pair(0,0)]=0;while(q.size()){node bb=q.front();q.pop();if(bb.a==c||bb.b==c){printf("%d\n",bb.s.length());for(int i=0;i<bb.s.length();i++) {int x=bb.s[i]-'0';printf("%s\n",ss[x]);}return ;}if(mp[make_pair(bb.a,bb.b)]) continue;else mp[make_pair(bb.a,bb.b)]=1;if(bb.a<=a) {q.push(node(a,bb.b,bb.s+"1"));int x=a-bb.a;if(bb.b>0&&x){ if(x>=bb.b) q.push(node(bb.a+bb.b,0,bb.s+"2"));else q.push(node(a,bb.b-x,bb.s+"2"));}}if(bb.b<=b) {q.push(node(bb.a,b,bb.s+"0"));int x=b-bb.b;if(bb.a>0&&x){if(x>=bb.a) q.push(node(0,bb.b+bb.a,bb.s+"3"));else q.push(node(bb.a-x,b,bb.s+"3"));}}if(bb.a) q.push(node(0,bb.b,bb.s+"4"));if(bb.b) q.push(node(bb.a,0,bb.s+"5"));}printf("impossible\n"); } int main() {while(~scanf("%d%d%d",&a,&b,&c)){bfs();}return 0; }

喜歡算法，不一定非要打比賽，當個樂趣消遣消遣也是不錯的。
努力加油a啊，(^o)/~

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