POJ3414——Pots

Pots

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 10021 Accepted: 4209 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 potj is full (and there may be some water left in the pot i), or the poti 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 exactlyC liters of water in one of the pots.

Input

On the first and only line are the numbers A, B, andC. 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 operationsK. 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

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

Source

Northeastern Europe 2002, Western Subregion


啊哈，和杭电的一题差不多，每个状态拿过来最多可以新产生6个状态，做好剪枝和标记，就可以AC了

#include<queue>#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;bool vis[105][105]; struct node{    int A, B;    int num;    int code;    int from, to;    int cnt;};node pre[1001000];void change(node &temp2, node &temp1, int t){temp2.A = temp1.A;temp2.B = temp1.B;temp2.cnt = temp1.cnt + 1;temp2.num = temp1.num + t;pre[temp2.num] = temp1;}node bfs(int A, int B, int C){queue< node >qu;memset( vis, 0, sizeof(vis) );while( !qu.empty() )qu.pop();    int cnt = 1;node temp1, temp2, ans;bool flag = false;temp1.A = 0;temp1.B = 0;temp1.num = 0;//????temp1.cnt = 0;temp1.code = -1;//????,//pre[temp1.num] = temp1;qu.push(temp1);while( !qu.empty() ){temp1 = qu.front();qu.pop();if( temp1.A == C || temp1.B == C){flag = true;ans = temp1;break;}if(temp1.A != 0)//??A???{    change(temp2, temp1, cnt ++);temp2.A = 0;temp2.code = 2;temp2.from = 1;if( !vis[temp2.A][temp2.B] ){vis[temp2.A][temp2.B] = 1;qu.push(temp2);}}if(temp1.B != 0){    change(temp2, temp1, cnt ++);temp2.B = 0;temp2.code = 2;temp2.from = 2;if( !vis[temp2.A][temp2.B] ){vis[temp2.A][temp2.B] = 1;qu.push(temp2);}}        if(temp1.A != A)//???,????{    change(temp2, temp1, cnt ++);temp2.A = A;temp2.code = 1;temp2.from = 1;if( !vis[temp2.A][temp2.B] ){vis[temp2.A][temp2.B] = 1;qu.push(temp2);}}if(temp1.B != B)//???,????{    change(temp2, temp1, cnt ++);temp2.B = B;temp2.code = 1;temp2.from = 2;if( !vis[temp2.A][temp2.B] ){vis[temp2.A][temp2.B] = 1;qu.push(temp2);}}if( temp1.A != 0 && temp1.B != B)//A??,B??,?A?B{    change(temp2, temp1, cnt ++);temp2.code = 3;temp2.from = 1;temp2.to = 2;if(temp2.A + temp2.B <= B){temp2.B += temp2.A;temp2.A = 0;if( !vis[temp2.A][temp2.B] ){vis[temp2.A][temp2.B] = 1;qu.push(temp2);}}else{temp2.A -= (B - temp2.B);temp2.B = B;if( !vis[temp2.A][temp2.B] ){vis[temp2.A][temp2.B] = 1;qu.push(temp2);}}}if( temp1.B != 0 && temp1.A != A)//B??,A??,?B?A{    change(temp2, temp1, cnt ++);temp2.code = 3;temp2.from = 2;temp2.to = 1;if(temp2.A + temp2.B <= A){temp2.A += temp2.B;temp2.B = 0;if( !vis[temp2.A][temp2.B] ){vis[temp2.A][temp2.B] = 1;qu.push(temp2);}}else{temp2.B -= (A - temp2.A);temp2.A = A;if( !vis[temp2.A][temp2.B] ){vis[temp2.A][temp2.B] = 1;qu.push(temp2);}}}}if(flag)return ans;ans.num = -1;return ans;}void print(node temp){if(temp.num != 0)print( pre[temp.num] );if(temp.code == 1){printf("FILL(%d)\n", temp.from);}else if(temp.code == 2){printf("DROP(%d)\n", temp.from);}else if(temp.code == 3){printf("POUR(%d,%d)\n", temp.from, temp.to);}}int main(){int A, B, C;while(~scanf("%d%d%d", &A, &B, &C)){node ans = bfs(A, B, C);if(ans.num == -1){printf("impossible\n");continue;}printf("%d\n", ans.cnt);print(ans);}return 0;}