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[N][N];
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[N];
    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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