zoukankan      html  css  js  c++  java
  • POJ1753 Flip Game(bfs、枚举)

    链接:http://poj.org/problem?id=1753

    Flip Game

    Description

    Flip game is played on a rectangular 4x4 field with two-sided pieces placed on each of its 16 squares. One side of each piece is white and the other one is black and each piece is lying either it's black or white side up. Each round you flip 3 to 5 pieces, thus changing the color of their upper side from black to white and vice versa. The pieces to be flipped are chosen every round according to the following rules: 
    1. Choose any one of the 16 pieces. 
    2. Flip the chosen piece and also all adjacent pieces to the left, to the right, to the top, and to the bottom of the chosen piece (if there are any).

    Consider the following position as an example: 

    bwbw 
    wwww 
    bbwb 
    bwwb 
    Here "b" denotes pieces lying their black side up and "w" denotes pieces lying their white side up. If we choose to flip the 1st piece from the 3rd row (this choice is shown at the picture), then the field will become: 

    bwbw 
    bwww 
    wwwb 
    wwwb 
    The goal of the game is to flip either all pieces white side up or all pieces black side up. You are to write a program that will search for the minimum number of rounds needed to achieve this goal. 

    Input

    The input consists of 4 lines with 4 characters "w" or "b" each that denote game field position.

    Output

    Write to the output file a single integer number - the minimum number of rounds needed to achieve the goal of the game from the given position. If the goal is initially achieved, then write 0. If it's impossible to achieve the goal, then write the word "Impossible" (without quotes).

    Sample Input

    bwwb
    bbwb
    bwwb
    bwww

    Sample Output

    4


    题目里是4*4的格子,可以用二进制的0、1表示每个格子的黑白,第i个格子为黑即第i位的二进制数为1。这样16个格子需要16位二进制,即一共有2^16-1个状态。从初始状态开始枚举每一种状态改变16个格子的颜色所能导致的状态。出现16位全1或者全0就输出Impossible。
    代码:
    #include <iostream>
    #include <stdio.h>
    #include <queue>
    #include <string.h>
    using namespace std;
    #define MAXX (1<<16)
    bool flag[MAXX];
    int step[MAXX];
    queue<int> que;
    void input()
    {
        int state=0;
        char ch;
        int i;
        for(i=1;i<MAXX;i<<=1)
        {
            scanf("%c",&ch);
            if(ch=='
    ')
                scanf("%c",&ch);
            if(ch=='b')
                state+=i;
        }
        que.push(state);
        memset(flag,false,sizeof(flag));
        memset(step,0,sizeof(step));
        flag[state]=true;
        //printf("%d ",state);
    }
    void change(int n,int state)
    {
        int temp=state;
        if(n-4>=0)
            state^=(1<<(n-4));
        if(n+4<16)
            state^=(1<<(n+4));
        if(n!=0&&n!=4&&n!=8&&n!=12)
            state^=(1<<(n-1));
        if(n!=3&&n!=7&&n!=11&&n!=15)
            state^=(1<<(n+1));
        state^=(1<<n);
        if(!flag[state])
        {
            flag[state]=true;
            que.push(state);
            step[state]=step[temp]+1;
        }
    }
    void getall()
    {
        int state;
        int i;
        while(!que.empty())
        {
            state=que.front();
            if(state==0||state==MAXX-1)
            {
                printf("%d
    ",step[state]);
                return;
            }
            que.pop();
            for(i=0;i<16;i++)
            {
                change(i,state);
            }
        }
        printf("Impossible
    ");
    }
    int main()
    {
        input();
        getall();
        return 0;
    }
    View Code
  • 相关阅读:
    Java四种引用类型+ReferenceQueue+WeakHashMap
    浅谈怎样写好一篇(技术)博客?
    MySQL-5.7.14-WinX64安装配置详解
    网络编程梳理:Android网络基础知识复习
    Git时间:常用Git命令收集整理(持续更新)
    一些常见技术问题收集(二)持续更新
    开源库AndroidSwipeLayout分析(一),炫酷ItemView滑动呼出效果
    开源库AndroidSwipeLayout分析(二),SwipeLayout源码探究
    ES 基础操作
    pymongo
  • 原文地址:https://www.cnblogs.com/vwqv/p/5857716.html
Copyright © 2011-2022 走看看