有一个瓶子A和一个瓶子B,可以有三种操作倒满,倒空,或者把瓶子A倒向瓶子B(或者把瓶子B倒向瓶子A),可以扩展出6种操作,没什么简单的写法,只能一种一种的写.....
当然是使用广搜.......................直到一个瓶子里面有C升水,或者倒不出来这种结果,还需要输出到得步骤, 可以递归输出路径......
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<queue>
using namespace std;
#define maxn 110
int A, B, C;
char a[10][10] ={ "FILL(1)","FILL(2)","DROP(1)","DROP(2)","POUR(1,2)","POUR(2,1)"};
struct node
{
int fromX, fromY, step, op;
}v[maxn][maxn];
struct point
{
int x, y;
};//当前瓶子A,B的水量
//输出路径
void DFS(int x, int y)
{
if(x == 0 && y == 0)
return ;
DFS(v[x][y].fromX, v[x][y].fromY);
printf("%s\n", a[ v[x][y].op ]);
}
void BFS()
{
queue<point> Q;
point s, q;
s.x = s.y = 0;
v[0][0].step = 1;
Q.push(s);
while(Q.size())
{
s = Q.front();Q.pop();
if(s.x == C || s.y == C)
{
printf("%d\n", v[s.x][s.y].step-1);
DFS(s.x, s.y);
return ;
}
for(int i=0; i<6; i++)
{
q = s;
if(i == 0)
q.x = A;
else if(i == 1)
q.y = B;
else if(i == 2)
q.x = 0;
else if(i == 3)
q.y = 0;
else if(i == 4)
{
if(q.x+q.y <= B)
q.y += q.x, q.x = 0;//把A里面的水全倒B里面
else
q.x -= (B-q.y), q.y = B;//把B倒满
}
else
{
if(q.x+q.y <= A)
q.x += q.y, q.y = 0;
else
q.y -= (A-q.x), q.x = A;
}
if(v[q.x][q.y].step == 0)
{
v[q.x][q.y].step = v[s.x][s.y].step+1;
v[q.x][q.y].fromX = s.x;
v[q.x][q.y].fromY = s.y;
v[q.x][q.y].op = i;
Q.push(q);
}
}
}
printf("impossible\n");
}
int main()
{
while(scanf("%d%d%d", &A, &B, &C) != EOF)
{
memset(v, 0, sizeof(v));
BFS();
}
return 0;