POJ 3414　Pots 均分水（bfs）

题目链接：

POJ 3414 Pots

题目大意：

有两个水杯，一开始都没有装水，给出两个水杯的容量，和要达到的目标容量

每个水被有三种操作：１．装满水，２．倒掉所有水，３，将水倒入另一个杯子中．

思路：

两个杯子，六种操作，，求最短路，并打印路径，６入口bfs，跟非常可乐一题类似

Description

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 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. 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 ‘impossible’.

Sample Input

＞3 5 4

Sample Output

6

FILL(2)

POUR(2,1)

DROP(1)

POUR(2,1)

FILL(2)

POUR(2,1)

代码：

/*************************************************************************
> File Name: poj_3414.cpp
> Author: dulun
> Mail: dulun@xiyoulinux.org
> Created Time: 2016年04月13日 星期三 12时59分54秒
************************************************************************/

#include<iostream>
#include<stdio.h>
#include<cstring>
#include<cstdlib>
#include<queue>
#include<algorithm>
#define LL long long
using namespace std;

const int N = 186;
struct Node
{
int a, b, step;
char str

;
};

bool v

;

void bfs(int a, int b, int c)
{
memset(v, 0, sizeof(v));
Node t, s;
queue<Node> q;
t = (Node){0, 0, 0};
q.push(t);
v[t.a][t.b] = true;

while(!q.empty())
{
t = q.front();
q.pop();
v[t.a][t.b] = true;
if(t.a == c || t.b == c)
{
printf("%d\n", t.step);
for(int i = 1; i <= t.step; i++)
{
printf("%s\n", t.str[i]);
}
return ;
}

if(t.a == 0)
{
s = t;
s.a = a;
s.step++;
strcpy(s.str[s.step], "FILL(1)");
if(!v[s.a][s.b])
{
v[s.a][s.b] = true;
q.push(s);
}
}
else if(t.a <= a)
{
s = t;
s.a = 0;
s.step++;
strcpy(s.str[s.step], "DROP(1)");
if(!v[s.a][s.b])
{
v[s.a][s.b] = true;
q.push(s);
}
if(t.b < b)
{
s = t;
if(t.a+t.b <= b)
{
s.a = 0;
s.b = t.a+t.b;
}
else{
s.a = t.a - (b-t.b);
s.b = b;
}
s.step++;
strcpy(s.str[s.step], "POUR(1,2)");
if(!v[s.a][s.b])
{
v[s.a][s.b] = true;
q.push(s);
}
}
}

if(t.b == 0)
{
s = t;
s.b = b;
s.step++;
strcpy(s.str[s.step], "FILL(2)");
if(!v[s.a][s.b])
{
v[s.a][s.b] = true;
q.push(s);
}
}
else if(t.b <= b)
{
s = t;
s.b = 0;
s.step++;
strcpy(s.str[s.step], "DROP(2)");
if(!v[s.a][s.b])
{
v[s.a][s.b] = true;
q.push(s);
}
if(t.a < a)
{
if(t.a+t.b < a)
{
s = t;
s.a = t.a+t.b;
s.b = 0;
}
else{
s = t;
s.a = a;
s.b = t.b - (a-t.a);
}
s.step++;
strcpy(s.str[s.step], "POUR(2,1)");
if(!v[s.a][s.b])
{
v[s.a][s.b] = true;
q.push(s);
}
}
}

}
printf("impossible\n");
return;

}

int main()
{
int a, b, c;
scanf("%d%d%d", &a, &b, &c);

bfs(a, b, c);
return 0;
}