"CodeVS1226" pour water problem

Title Description Description

There are two water kettles with no tick marks, each of which can be fitted with an X-liter and a Y-liter (an integer with x, Y and no more than 100). There is another water tank which can be used for watering the kettle or pouring it out from the kettle, and the water can also be dumped between the two kettles. The X-liter pot is known as an empty pot, and the Y-liter pot is an empty pot. Ask how to use the water or irrigation operation, with a minimum number of steps can be in the X or y liters of the pot to measure the Z (z≤100) liters.

Enter a description Input Description

One row, three data, representing X, Y and z, respectively;

Output description Output Description

One row, outputs the minimum number of steps, and outputs "impossible" if the target cannot be reached

Sample input Sample Input

3 22 1

Sample output Sample Output

14

Exercises

BFS, a total of 6 possible:

1, a barrel empty 2, b barrels empty

3, a barrel full 4, B barrels full

5, a barrel into the B-barrel

1) Overflow

2) No overflow

6, b barrels poured into a barrel

1) Overflow

2) No overflow

It was wrong to write it.

    // 5 x->y    inch ;     = P.x-min (Y-p.y,p.x)    ; = p.y + min (Y-p.y,p.x);    Q.push (p);     // 6 y->x    inch ;     = p.x + min (x-p.x,p.y)    ; = P.y-min (x-p.x,p.y);    Q.push (p);

#include <iostream>#include<queue>#include<cstdlib>#include<cstdio>#defineN 1000using namespacestd;structcup{intX,y,step;} First,now;intX,y,z;queue<cup>Q;BOOLF[n][n];voidExtend (Cupinch){    if(f[inch. x][inch. Y])return; f[inch. x][inch. y] =1; inch. step++; //1Cup p =inch; P.x=0;    Q.push (P); //2p =inch; P.Y=0;    Q.push (P); //3p =inch; P.x=x;    Q.push (P); //4p =inch; P.Y=y;    Q.push (P); //5 X->yp =inch; P.x=inch. X-min (Yinch. Y,inch. x); P.Y=inch. Y + min (yinch. Y,inch. x);    Q.push (P); //6 y->xp =inch; P.x=inch. x + min (x-inch. x,inch. Y); P.Y=inch. y-min (Xinch. x,inch. Y); Q.push (P);}voidBFs () {Q.push (first);  while(!Q.empty ()) { Now=Q.front ();        Q.pop (); if(now.x = = Z | | now.y = =z) {printf ("%d", Now.step); Exit (0);    } extend (now); } printf ("Impossible");}intMain () {scanf ("%d%d%d",&x,&y,&z); First.step= First.x = First.y =0; BFS ();}

"CodeVS1226" pour water problem