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 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 <p>On the first and only line are the numbers <b>A</b>, <b>B</b>, and <b>C</b>. These are all integers in the range from 1 to 100 and <b>C</b>≤max(<b>A</b>,<b>B</b>).</p> Output <p>The first line of the output must contain the length of the sequence of operations <b>K</b>. The following <b>K</b> 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 ‘<b>impossible</b>’.</p> 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<string.h> #include<queue> using namespace std;
int a,b,c; int logal[1000][1000]={0}; struct dalao { int mla; int mlb; int step; string lalala[2000]; }; dalao t1,t2;
int bfs() { memset(logal,0,sizeof(logal));
queue<dalao>k;
t1.mla=0; t1.mlb=0; t1.step=0; //t1.lalala[0]=""; k.push(t1); logal[0][0]=1; //int y=0; while(k.empty()==0) { // cout<<y++<<endl;
//dalao j; t1=k.front(); //cout<<t1.mla<<" "<<t1.mlb<<" "<<" step"<<t1.step<<endl;
k.pop(); if(t1.mla==c||t1.mlb==c) { cout<<t1.step<<endl; for(int i=1;i<=t1.step;i++) { cout<<t1.lalala[i]<<endl; } return 1; } if(t1.mla==0) { t2=t1; t2.mla=a; t2.step++; t2.lalala[t2.step]="FILL(1)"; if(logal[t2.mla][t2.mlb]==0) { logal[t2.mla][t2.mlb]=1; k.push(t2); } } if(t1.mla>0&&t1.mla<=a) { t2=t1; t2.mla=0; t2.step++; t2.lalala[t2.step]="DROP(1)"; if(logal[t2.mla][t2.mlb]==0) { logal[t2.mla][t2.mlb]=1; k.push(t2); } }
if(t1.mlb<b&&t1.mla!=0) { t2=t1; t2.step++; if(t2.mla+t2.mlb<=b) { t2.mlb+=t2.mla; t2.mla=0; } else { t2.mla=t2.mla+t2.mlb-b; t2.mlb=b; } t2.lalala[t2.step]="POUR(1,2)"; if(logal[t2.mla][t2.mlb]==0) { logal[t2.mla][t2.mlb]=1; k.push(t2); } }
if(t1.mlb==0) { //cout<<"****"<<endl; t2=t1; t2.mlb=b; t2.step++; t2.lalala[t2.step]="FILL(2)"; if(logal[t2.mla][t2.mlb]==0) { logal[t2.mla][t2.mlb]=1; k.push(t2); }
} if(t1.mlb>0&&t1.mlb<=b)// { t2=t1; t2.mlb=0; t2.step++; t2.lalala[t2.step]="DROP(2)"; if(logal[t2.mla][t2.mlb]==0) { logal[t2.mla][t2.mlb]=1; k.push(t2); }
}
if(t1.mla<a&&t1.mlb!=0)//这里要特别注意,不要写错顺序 {
t2=t1; t2.step++; if(t2.mla+t2.mlb<=a) { t2.mla+=t2.mlb; t2.mlb=0; } else {
t2.mlb=t2.mla+t2.mlb-a; t2.mla=a;
}
t2.lalala[t2.step]="POUR(2,1)"; if(logal[t2.mla][t2.mlb]==0) { logal[t2.mla][t2.mlb]=1; k.push(t2); } }
} return 0; } int main() { //int a,b,c; while(cin>>a>>b>>c) { //memset(logal,0,sizeof(logal)); int ans=bfs(); if(ans==0) cout<<"impossible"<<endl;
} return 0;
}
