Pots
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 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.
Sample Input
3 5 4
Sample Output
6
FILL(2)
POUR(2,1)
DROP(1)
POUR(2,1)
FILL(2)
POUR(2,1)
Source
Northeastern Europe 2002, Western Subregion
题意：倒水 6种操作。各自是1：向1杯装满水。2：向2杯装满水。3：1杯水倒入2杯。4：2杯水倒入1杯。5：1杯水倒出；6：2杯水倒出
思路：简单BFS+回溯 广搜时借用节点下标。记录父亲节点，同一时候记录操作相应的数字，以及终点节点 从终点节点回溯，打印路径
#include <cstdio>
#include <queue>
#include <cstring>
#include <cstdlib>
#include <vector>
using namespace std ;
int a , b , c ;
int num , flag ; //num 记录搜索步数 ， flag 记录终点搜索步数
int ans[101][101] ; //记录步数
struct next {
int op ; //操作
int par ; //父亲
} nn[10001] ;
void dfs( int flag ) { //深搜打印路径
if( flag > 0 ) { dfs(nn[flag].par) ;}
if(nn[flag].op == 1 ) cout << "FILL(1)" << endl;
else if(nn[flag].op == 2 ) cout << "FILL(2)" << endl;
else if(nn[flag].op == 3 ) cout << "POUR(1,2)" << endl;
else if(nn[flag].op == 4 ) cout << "POUR(2,1)" << endl;
else if(nn[flag].op == 5 ) cout << "DROP(1)" << endl;
else if(nn[flag].op == 6 ) cout << "DROP(2)" << endl;
}
bool bfs() {
memset(ans, 0 ,sizeof(ans)) ;
ans[0][0] = 1;
queue<int> A ;
queue<int> B ;
while(!A.empty()||!B.empty() ) {
A.pop() ;B.pop() ;
}
A.push(0) ;
B.push(0) ;
int time = 0 , tempa , tempb ; //time 当前节点数
while(!A.empty()) {
int inita = A.front () ;
int initb = B.front() ;
A.pop() ;B.pop() ;
//FILL(1)
tempa = a ;
tempb = initb ;
if ( !ans[tempa][tempb] ) { //保证未訪问过
A.push(tempa) ;
B.push(tempb) ;
ans[tempa][tempb] = ans[inita][initb] +1 ;
nn[num].op = 1 ;
nn[num++].par = time ; //记录父亲节点
if ( tempa==c || tempb==c ) {
flag = num-1 ; //记录终点节点
break;
}
}
//FILL(2)
tempa = inita ;
tempb = b ;
if( !ans[tempa][tempb] ) {
A.push(tempa) ;
B.push(tempb) ;
ans[tempa][tempb] = ans[inita][initb] +1;
nn[num].op = 2 ;
nn[num++].par = time ;
if(tempa==c||tempb==c) {
flag = num -1;
break ;
}
}
//POUR(1,2)
if(inita <= b-initb) tempa = 0 ; //倒完
else tempa = inita - ( b - initb) ;//倒部分
tempb = (initb + inita) ;
if(tempb> b) tempb = b ; //小心溢出
if( !ans[tempa][tempb] ) {
A.push(tempa) ;
B.push(tempb) ;
ans[tempa][tempb] = ans[inita][initb] +1;
nn[num].op = 3 ;
nn[num++].par = time ;
if(tempa==c||tempb==c) {
flag = num-1 ;
break ;
}
}
//POUR(2,1)
if(initb <= a-inita) tempb = 0 ;
else tempb = initb - ( a - inita) ;
tempa = (initb + inita) ;
if(tempa> a) tempa = a ;
if( !ans[tempa][tempb] ) {
A.push(tempa) ;
B.push(tempb) ;
ans[tempa][tempb] = ans[inita][initb] +1;
nn[num].op = 4 ;
nn[num++].par = time ;
if(tempa==c||tempb==c) {
flag = num-1 ;
break ;
}
}
//DROP(1)
tempa = 0 ;
tempb = initb ;
if( !ans[tempa][tempb] ) {
A.push(tempa) ;
B.push(tempb) ;
ans[tempa][tempb] = ans[inita][initb] +1;
nn[num].op =5 ;
nn[num++].par = time ;
if(tempa==c||tempb==c) {
flag = num-1 ;
break;
}
}
//DROP(2)
tempb = 0 ;
tempa = inita ;
if( !ans[tempa][tempb] ) {
A.push(tempa) ;
B.push(tempb) ;
ans[tempa][tempb] = ans[inita][initb] +1;
nn[num].op = 6 ;
nn[num++].par = time ;
if(tempa==c||tempb==c) {
flag = num-1 ;
break;
}
}
++time ; //记录节点数
}
if( flag ) {
printf("%d
",max(ans[tempa][tempb],ans[tempb][tempa])-1) ;//ans[0][0] = 1 , 故减之
dfs(flag) ;
return true ;
}
return false ;
}
int main( ) {
while(~scanf("%d %d %d",&a,&b,&c) ) {
num = 1 ;
flag = 0;
if(!bfs()) cout << "impossible" << endl ;
}
return 0 ;
}