poj3414 Pots--BFS

题目来源: poj 3414

问题描述
有两个空壶，最终要将其中一个壶装水量达到给定值。对壶有3种操作方式，一是开水龙头装满，二是倒光，三是将其中一个壶的水倒到另一个壶里，且水倒光或者另一个壶装满。若能实现目标则输出最小步骤数且按顺序输出操作，若不能实现，则输出“impossible”

未AC…之后再看

#include <iostream>
#include <queue>
#include <string>
#include <vector>
using namespace std;
struct node
{
    int num[2];
    int step;
    vector<string> v;
} start;
int volume[2];
int k; //要量的值
bool has = false;
void BFS()
{
    bool vis[201][201] = {false};
    queue<node> Q;
    start.num[0] = start.num[1] = 0;
    start.step = 0;
    vis[0][0] = true;
    Q.push(start);
    while (!Q.empty())
    {
        node top = Q.front();
        Q.pop();
        if (top.num[0] == k || top.num[1] == k)
        {
            cout << top.step << endl;
            for (int i = 0; i < top.step; i++)
            {
                cout << top.v[i] << endl;
                has = true;
            }
            return;
        }
        if (top.num[0] < volume[0]) //fill A
        {
            node temp = top;
            temp.num[0] = volume[0];
            temp.step = top.step + 1;
            string ss = "FILL (1)";
            if (vis[temp.num[0]][temp.num[1]] == false)
            {
                vis[temp.num[0]][temp.num[1]] = true;
                temp.v.push_back(ss);
                Q.push(temp);
            }
        }
        if (top.num[1] < volume[1]) //fill B
        {
            node temp = top;
            temp.num[1] = volume[1];
            temp.step = top.step + 1;
            string ss = "FILL (2)";
            if (vis[temp.num[0]][temp.num[1]] == false)
            {
                vis[temp.num[0]][temp.num[1]] = true;
                temp.v.push_back(ss);
                Q.push(temp);
            }
        }
        if (top.num[0] > 0) //DROP A
        {
            node temp = top;
            temp.num[0] = 0;
            temp.step = top.step + 1;
            string ss = "DROP (1)";
            if (vis[temp.num[0]][temp.num[1]] == false)
            {
                vis[temp.num[0]][temp.num[1]] = true;
                temp.v.push_back(ss);
                Q.push(temp);
            }
        }
        if (top.num[1] > 0) //DROP B
        {
            node temp = top;
            temp.num[1] = 0;
            temp.step = top.step + 1;
            string ss = "DROP (2)";
            if (vis[temp.num[0]][temp.num[1]] == false)
            {
                vis[temp.num[0]][temp.num[1]] = true;
                temp.v.push_back(ss);
                Q.push(temp);
            }
        }
        if (top.num[0] >= top.num[1]) //POUR (1,2)
        {
            node temp = top;
            int rest = volume[1] - temp.num[1]; //2中还可以装多少水
            if (temp.num[0] > rest)             //2可以装满
            {
                temp.num[0] -= rest;
                temp.num[1] = volume[1];
            }
            else
            {
                temp.num[0] = 0;
                temp.num[1] += temp.num[0];
            }
            temp.step = top.step + 1;
            string ss = "POUR (1,2)";
            temp.v.push_back(ss);
            if (vis[temp.num[0]][temp.num[1]] == false)
            {
                vis[temp.num[0]][temp.num[1]] = true;
                Q.push(temp);
            }
        }
        if (top.num[0] <= top.num[1]) //POUR (2,1)
        {
            node temp = top;
            int rest = volume[0] - temp.num[0]; //2中还可以装多少水
            if (temp.num[1] > rest)             //2可以装满
            {
                temp.num[1] -= rest;
                temp.num[0] = volume[0];
            }
            else
            {
                temp.num[1] = 0;
                temp.num[0] += temp.num[1];
            }
            temp.step = top.step + 1;
            string ss = "POUR (2,1)";
            temp.v.push_back(ss);
            if (vis[temp.num[0]][temp.num[1]] == false)
            {
                vis[temp.num[0]][temp.num[1]] = true;
                Q.push(temp);
            }
        }
    }
}

int main()
{

    while (cin >> volume[0] >> volume[1] >> k)
    {
        BFS();
        if (!has)
        {
            cout << "impossible" << endl;
        }
        has = false;
    }
    return 0;
}