poj3414 Pots（BFS）

题目链接

http://poj.org/problem?id=3414

题意

有两个杯子，容量分别为A升，B升，可以向杯子里倒满水，将杯子里的水倒空，将一个杯子里的水倒到另一个杯子里，求怎样倒才能使其中的一个杯子里的水恰为C升，输出最少步数和操作；如果不能倒到C升，输出“impossible”。

思路

这题与poj1606基本相同，在poj1606的基础上添加了输出最少步数，修改了操作的表示，以及不可能达到目标时输出impossible。将poj1606的代码略作修改即可。

代码

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <stack>
using namespace std;

struct Node
{
int a, b;
int steps;
int flag;
Node* pre;
};

const int N = 1010;
int ca, cb, n;
int visit

;
stack<int> s;

void print()
{
while (!s.empty())
{
switch (s.top())
{
case 0:
cout << "FILL(1)" << endl;
break;
case 1:
cout << "FILL(2)" << endl;
break;
case 2:
cout << "DROP(1)" << endl;
break;
case 3:
cout << "DROP(2)" << endl;
break;
case 4:
cout << "POUR(1,2)" << endl;
break;
case 5:
cout << "POUR(2,1)" << endl;
break;
}
s.pop();
}
}

void bfs(int a, int b)
{
Node state
;
int cnt = -1;
memset(visit, 0, sizeof(visit));
Node node;
node.a = node.b = 0;
node.steps = 0;
node.pre = NULL;
queue<Node> q;
q.push(node);
visit[node.a][node.b] = 1;
while (!q.empty())
{
Node node = q.front();
q.pop();
state[++cnt] = node;
Node next = node;
for (int i = 0; i < 6; i++)
{
next = node;
int amount;
switch (i)
{
case 0:        //FILL(1)
next.a = ca;
next.flag = 0;
break;
case 1:        //FILL(2)
next.b = cb;
next.flag = 1;
break;
case 2:        // DROP(1)
next.a = 0;
next.flag = 2;
break;
case 3:        //DROP(2)
next.b = 0;
next.flag = 3;
break;
case 4:        //POUR(1,2)
amount = cb - node.b;
if (node.a > amount)
{
next.a -= amount;
next.b = cb;
}
else {
next.a = 0;
next.b = node.a + node.b;
}
next.flag = 4;
break;
case 5:        //POUR(2,1)
amount = ca - node.a;
if (node.b > amount)
{
next.a = ca;
next.b -= amount;
}
else {
next.a = node.a + node.b;
next.b = 0;
}
next.flag = 5;
break;
}

if (!visit[next.a][next.b])
{
visit[next.a][next.b] = 1;
next.pre = &state[cnt];
next.steps = node.steps + 1;
if (next.a == n || next.b == n)
{
cout << next.steps << endl;
while (next.pre)
{
s.push(next.flag);
next = *next.pre;
}
print();
return;
}
q.push(next);
}
}
}
cout << "impossible" << endl;
}

int main()
{
cin >> ca >> cb >> n;
bfs(0, 0);
return 0;
}

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