POJ-3414c(Pots)

最新推荐文章于 2022-10-27 18:26:38 发布

YancyKahn

最新推荐文章于 2022-10-27 18:26:38 发布

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CC 4.0 BY-SA版权

分类专栏：搜索

本文链接：https://blog.youkuaiyun.com/qq_37753409/article/details/79440552

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本文介绍了一个经典的水壶问题，并通过广度优先搜索算法（BFS）解决该问题。文章详细展示了如何使用C++实现算法，包括状态表示、状态转移等关键步骤。

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POJ-3414

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

string strpath[] = {
    "FILL(1)", "FILL(2)", "DROP(1)", "DROP(2)", "POUR(1,2)", "POUR(2,1)"
};

struct _Node{
    int val, depth;
    _Node(){
    };
    _Node(int _val, int _depth){
        val = _val; depth = _depth; 
    }
};
struct _Path{
    int from, idx;
}path[1000005];
int a, b, at, bt, ct;
stack<int> s;
int depth;

int hash(int a, int b)
{
    return a*1000 + b;
}

void reverse_hash(int val)
{
    a = val/1000;
    b = val%1000;
}
int visit[1000005];

bool judge(int end)
{
    if(a == end || b == end) return 1;
    else return 0;
}

void bfs(int start, int end)
{
    queue<_Node> q;
    q.push(_Node(start, 0));
    visit[start] = 1;
    path[start].from = path[start].idx = -1;
    while(!q.empty())
    {
        _Node x = q.front(); q.pop();
        int u = x.val;
        reverse_hash(u);
        if(judge(end)){
            while(path[u].from != -1)
            {
                if(a == 0 && b == 0) break;
                s.push(path[u].idx);
                u = path[u].from;
            }
            cout << x.depth << endl;
            while(!s.empty())
            {
                cout << strpath[s.top()] << endl;
                s.pop();
            }
            return ;
        }


        //FIll;
        int t = hash(at, b);
        if(a != at && !visit[t])
        {   

            q.push(_Node(t, x.depth + 1));
            visit[t] = 1;
            path[t].from = u;
            path[t].idx = 0;
        }
        t = hash(a, bt);
        if(b != bt && !visit[t])
        {       
            q.push(_Node(t, x.depth + 1));
            visit[t] = 1;
            path[t].from = u;
            path[t].idx = 1;
        }

        //DROP;
        t = hash(0, b);
        if(a != 0 && !visit[t])
        {   
            q.push(_Node(t, x.depth + 1));
            visit[t] = 1;
            path[t].from = u;
            path[t].idx = 2;
        }

        t = hash(a, 0);
        if(b != 0 && !visit[t])
        {   
            q.push(_Node(t, x.depth + 1));
            visit[t] = 1;
            path[t].from = u;
            path[t].idx = 3;
        }

        //POUR;
        //a -> b
        if(a > 0 && b < bt)
        {
            int need = bt - b;


            if(a > need && !visit[hash(a - need, bt)])      //bFILL
            {
                t = hash(a - need, bt);
                q.push(_Node(t, x.depth + 1));
                visit[t] = 1;
                path[t].from = u;
                path[t].idx = 4;
            }else if(a <= need && !visit[hash(0, a + b)]){  //aout
                t = hash(0, a+b);
                q.push(_Node(t, x.depth + 1));
                visit[t] = 1;
                path[t].from = u;
                path[t].idx = 4;
            }
        }
        //b -> a
        if(b > 0 && a < at)
        {
            int need = at - a;
            if(b > need &&  !visit[hash(at, b - need)]) //aFIll
            {
                t = hash(at, b - need);
                q.push(_Node(t, x.depth + 1));
                visit[t] = 1;
                path[t].from = u;
                path[t].idx = 5;
            }else if(b <= need && !visit[hash(a + b, 0)]){  //bout
                t = hash(a + b, 0);
                q.push(_Node(t, x.depth + 1));
                visit[t] = 1;
                path[t].from = u;
                path[t].idx = 5;
            }
        }
    }   
    cout << "impossible" << endl;
    return;
}
int main()
{
    while(cin >> at >> bt >> ct)
    {
        memset(visit, 0, sizeof(visit));
        memset(path, 0, sizeof(path));
        bfs(hash(0, 0), ct);
     } 
}