POJ 3414 Pots【BFS】

 PotsTime Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%lld& %lluSubmit Status Practice POJ3414DescriptionYou 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 itscontents 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.InputOn 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).OutputThe 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 thedesired 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)

#include<iostream>#include<cstdio>#include<cmath>#include<cstring>#include<climits>#include<string>#include<queue>#include<stack>#include<set>#include<map>#include<algorithm>using namespace std;#define rep(i,j,k)for(i=j;i<k;i++)#define per(i,j,k)for(i=j;i>k;i--)#define MS(x,y)memset(x,y,sizeof(x))typedef long long LL;const int INF=0x7ffffff;const int M=1e4+1;int A,B,C;int i,j,k,n,m;int vis[110][110];struct Node{int a,b;int step;string s[110];};bool BFS(){Node u,next;u.a=0;u.b=0;u.step=0;queue<Node>Q;Q.push(u);vis[0][0]=1;while(!Q.empty()){u=Q.front();Q.pop();if(u.a==C||u.b==C){printf("%d\n",u.step);for(int i=1;i<=u.step;i++){cout<<u.s[i]<<endl;}return true;}if(u.a==0){next=u;next.a=A;next.step++;next.s[next.step]="FILL(1)";if(!vis[next.a][next.b]){vis[next.a][next.b]=1;Q.push(next);}}if(u.a<=A&&u.a){next=u;next.a=0;next.step++;next.s[next.step]="DROP(1)";if(!vis[next.a][next.b]){vis[next.a][next.b]=1;Q.push(next);}}if(u.b<B&&u.a){next=u;next.step++;if(next.a+next.b<=B){next.b+=next.a;next.a=0;}else{next.a=next.a+next.b-B;next.b=B;}next.s[next.step]="POUR(1,2)";if(!vis[next.a][next.b]){vis[next.a][next.b]=1;Q.push(next);}}if(u.b==0){next=u;next.b=B;next.step++;next.s[next.step]="FILL(2)";if(!vis[next.a][next.b]){vis[next.a][next.b]=1;Q.push(next);}}if(u.b<=B&&u.b){next=u;next.b=0;next.step++;next.s[next.step]="DROP(2)";if(!vis[next.a][next.b]){vis[next.a][next.b]=true;Q.push(next);}}if(u.a<A&&u.b){next=u;next.step++;if(next.a+next.b<=A){next.a+=next.b;next.b=0;}else{next.b=next.a+next.b-A;next.a=A;}next.s[next.step]="POUR(2,1)";if(!vis[next.a][next.b]){vis[next.a][next.b]=true;Q.push(next);}}}return false;}int main(){while(~scanf("%d%d%d",&A,&B,&C)){MS(vis,0);if(!BFS())printf("impossible\n");}return 0;}