POJ

题目链接

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

题意
给出两个杯子 容量分别为 A B 然后给出C 是目标容量
有三种操作
1 将一个杯子装满
2.将一个杯子全都倒掉
3.将一个杯子的水倒到另一个杯子里面

如果某个杯子里面的水 能够达到 目标容量 那么就输出步骤

思路
BFS 并且要存储步骤

每一步一共有六步操作 记得标记

AC代码

#include <cstdio>
#include <cstring>
#include <ctype.h>
#include <cstdlib>
#include <cmath>
#include <climits>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <map>
#include <stack>
#include <set>
#include <numeric>
#include <sstream>
#include <iomanip>
#include <limits>

#define CLR(a) memset(a, 0, sizeof(a))
#define pb push_back

using namespace std;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;
typedef pair<string, int> psi;
typedef pair<string, string> pss;

const double PI = acos(-1.0);
const double E = exp(1.0);
const double eps = 1e-8;

const int INF = 0x3f3f3f3f;
const int maxn = 1e2 + 5;
const int MOD = 1e9 + 7;

/*
    0 FILL 1
    1 FILL 2
    2 DROP 1
    3 DROP 2
    4 POUR 1 2
    5 POUR 2 1
*/

int visit[maxn][maxn];

vector <int> ans;

int flag;

int a, b, c;

struct node
{
    int a, b;
    vector <int> v;
};

void bfs()
{
    queue <node> q;
    node tmp;
    tmp.a = 0;
    tmp.b = 0;
    tmp.v.clear();
    q.push(tmp);
    visit[0][0] = 1;
    while (!q.empty())
    {
        node u = q.front(), v;
        q.pop();
        if (u.a == c || u.b == c)
        {
            flag = 1;
            ans = u.v;
            return;
        }
        if (u.a < a)
        {
            v.a = a;
            v.b = u.b;
            if (visit[v.a][v.b] == 0)
            {
                v.v = u.v;
                v.v.pb(0);
                q.push(v);
                visit[v.a][v.b] = 1;
            }
        }
        if (u.b < b)
        {
            v.a = u.a;
            v.b = b;
            if (visit[v.a][v.b] == 0)
            {
                v.v = u.v;
                v.v.pb(1);
                q.push(v);
                visit[v.a][v.b] = 1;
            }
        }
        if (u.a > 0)
        {
            v.a = 0;
            v.b = u.b;
            if (visit[v.a][v.b] == 0)
            {
                v.v = u.v;
                v.v.pb(2);
                q.push(v);
                visit[v.a][v.b] = 1;
            }
        }
        if (u.b > 0)
        {
            v.a = u.a;
            v.b = 0;
            if (visit[v.a][v.b] == 0)
            {
                v.v = u.v;
                v.v.pb(3);
                q.push(v);
                visit[v.a][v.b] = 1;
            }
        }
        if (u.a < a)
        {
            int c = a - u.a;
            if (u.b >= c)
            {
                v.b = u.b - c;
                v.a = a;
            }
            else
            {
                v.b = 0;
                v.a = u.a + u.b;
            }
            if (visit[v.a][v.b] == 0)
            {
                v.v = u.v;
                v.v.pb(5);
                q.push(v);
                visit[v.a][v.b] = 1;
            }
        }
        if (u.b < b)
        {
            int c = b - u.b;
            if (u.a >= c)
            {
                v.a = u.a - c;
                v.b = b;
            }
            else
            {
                v.a = 0;
                v.b = u.a + u.b;
            }
            if (visit[v.a][v.b] == 0)
            {
                v.v = u.v;
                v.v.pb(4);
                q.push(v);
                visit[v.a][v.b] = 1;
            }
        }
    }
}

int main()
{
    map <int, string> M;
    M[0] = "FILL(1)";
    M[1] = "FILL(2)";
    M[2] = "DROP(1)";
    M[3] = "DROP(2)";
    M[4] = "POUR(1,2)";
    M[5] = "POUR(2,1)";
    CLR(visit, 0);
    scanf("%d%d%d", &a, &b, &c);
    flag = 0;
    bfs();
    if (flag == 0)
        printf("impossible
");
    else
    {
        int len = ans.size();
        cout << len << endl;
        for (int i = 0; i < len; i++)
            cout << M[ans[i]] << endl;
    }
}