Poj3414 extensive search

<span style= "color: #330099;" >/*d-dtime limit:1000ms Memory limit:65536kb 64bit IO format:%i64d &%i64usubmit Status Practice POJ 3414 Descriptionyou is given the 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 PO t i to the drain; Pour (i,j) pour from pot I to pot J; After this operation either the pot J was full (and there could be some water left in the pot I), or the pot I was empty (and All its contents has been moved to the pot J). Write a program to find the shortest possible sequence of these operations that'll yield exactly C liters of water in on E of the pots. Inputon the first and only line is the numbers A, B, and C. These is all integers in the range from 1 to 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 is several sequences of minimal length, output any one of them. If the desired result can ' t is achieved, the first and only line of the file must contain the word ' impossible '. Sample Input3 5 4Sample output6fill (2) pour (2,1) DROP (1) pour (2,1) FILL (2) pour (2,1) by Grant yuan2014.7.14poj 3414 wide Search */# include<iostream> #include <cstdio> #include <cstring> #include <cstdlib>using namespace std;  BOOL Flag=0;int next[6]={0,1,2,3,4,5};int a,b,c;int aa,bb,cc;typedef struct{int A;  int b;  int F;  int sum; int ope;} Node;int res;node q[10000];bool mark[101][101];int top,base;int top1;int s[10000];bool can (int x1,int y1) {if (x1>=0&    amp;&x1<=aa&&y1>=0&&y1<=bb&&mark[x1][y1]==0) return 1; return 0;}    void Slove () {int a1,b1,f1,a2,b2;     while (top>=base) {//cout<< "Zhang" <<endl; if (q[base].a==cc| |           Q[BASE].B==CC) {flag=1;           Res=q[base].sum;          Break        } for (int i=0;i<6;i++) {  if (i==0) {a1=aa;               b1=q[base].b;                if (Can (A1,B1)) {q[++top].a=a1;                Q[TOP].B=B1;                Q[top].f=base;                q[top].sum=q[base].sum+1;                Q[top].ope=i;            Mark[a1][b1]=1;}               } else if (i==1) {a1=q[base].a;               B1=BB;                if (Can (A1,B1)) {q[++top].a=a1;                Q[TOP].B=B1;                Q[top].f=base;                q[top].sum=q[base].sum+1;                Q[top].ope=i;            Mark[a1][b1]=1;}               } else if (i==2)//1dao2 {int m,n;               M=Q[BASE].A;               n=bb-q[base].b;                   if (m>=n) {a1=m-n;                B1=BB;                if (Can (A1,B1)) {q[++top].a=a1;                Q[TOP].B=B1;                Q[top].f=base;                q[top].sum=q[base].sum+1;         Q[top].ope=i;       Mark[a1][b1]=1;}                  } else{a1=0;                  b1=m+q[base].b;                if (Can (A1,B1)) {q[++top].a=a1;                Q[TOP].B=B1;                Q[top].f=base;                q[top].sum=q[base].sum+1;                Q[top].ope=i;                  Mark[a1][b1]=1;}               }} else if (i==3)//1dao2 {int m,n;               M=AA-Q[BASE].A;               n=q[base].b;                   if (n>=m) {a1=aa;                B1=n-m;                if (Can (A1,B1)) {q[++top].a=a1;                Q[TOP].B=B1;                Q[top].f=base;                q[top].sum=q[base].sum+1;                Q[top].ope=i; Mark[a1][b1]=1;}                  } else{b1=0;                  A1=N+Q[BASE].A;                if (Can (A1,B1)) {q[++top].a=a1;                Q[TOP].B=B1;                Q[top].f=base;   q[top].sum=q[base].sum+1;             Q[top].ope=i;                  Mark[a1][b1]=1;}                   }} else if (i==4) {a1=0;                   b1=q[base].b;                    if (Can (A1,B1)) {q[++top].a=a1;                    Q[TOP].B=B1;                    Q[top].f=base;                     q[top].sum=q[base].sum+1;                     Q[top].ope=i;                    Mark[a1][b1]=1;                   }} else if (i==5) {b1=0;                   A1=Q[BASE].A;                    if (Can (A1,B1)) {q[++top].a=a1;                    Q[TOP].B=B1;                    Q[top].f=base;                    q[top].sum=q[base].sum+1;                     Q[top].ope=i;                    Mark[a1][b1]=1; }}} base++;}}    void print () {top1=-1;    int i=base,j;        while (1) {s[++top1]=q[i].ope;        J=Q[I].F;        I=j; if (i==0) BreAk        } for (j=top1;j>=0;j--) {if (s[j]==0) cout<< "FILL (1)" <<endl;        else if (s[j]==1) cout<< "FILL (2)" <<endl;        else if (s[j]==2) cout<< "pour" <<endl;        else if (s[j]==3) cout<< "pour (2,1)" <<endl;        else if (s[j]==4) cout<< "DROP (1)" <<endl;    else if (s[j]==5) cout<< "DROP (2)" <<endl;    }}int Main () {cin>>aa>>bb>>cc;    Top=-1;    memset (Mark,0,sizeof (Mark));    Mark[0][0]=1;    base=0;    q[++top].a=0;    q[top].b=0;    q[top].f=0;    q[top].sum=0;    q[top].ope=0;    Slove ();    if (flag==0) cout<< "Impossible" <<endl;    else{cout<<res<<endl;    Print ();} return 0;} </span>

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Poj3414 extensive search