POJ 1753 Flip Game



Flip Game

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 34437 Accepted: 15058

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

bwwbbbwbbwwbbwww

Sample Output

4

Source

Northeastern Europe 2000

题意：

在一个4x4的棋盘上放满了棋子，棋子有黑白两种颜色。你可以对棋子进行翻转操作，白色可以变成黑色，黑色变成白色。

具体翻转规则：

1  任取16个棋子的一个

2 你在改变该棋子的时候，它的上下左右的棋子都要改变。

做法：

直接BFS，犯了一个错误，就是，BFS时候没有判断 输入情况是否是目标情况。

#include <iostream>#include <cstdio>#include <cmath>#include <vector>#include <cstring>#include <algorithm>#include <string>#include <set>#include <functional>#include <numeric>#include <sstream>#include <stack>#include <map>#include <queue>using namespace std;string g[100];long long ans;long long gao[1000000];struct zhengling{long long x;//状态 long long rounds;//第几轮 long long k;//翻动第K棋子 zhengling(long long x1,long long r1,long long k1){x = x1;rounds = r1;k = k1;}};zhengling fan(zhengling tmpp,long long p)//翻转函数 {long long i = p/4;long long j = p%4;long long tmp = tmpp.x;//zhengling c;tmp^=(0x1<<p);if(i-1>=0)tmp^=(0x1<<(p-4));if(j-1>=0)tmp^=(0x1<<(p-1));if(i+1<4)tmp^=(0x1<<(p+4));if(j+1<4)tmp^=(0x1<<(p+1));zhengling c = zhengling(tmp,tmpp.rounds+1,p);return c;}void bfs(){queue<zhengling> q;zhengling fff = zhengling(ans,0,-1);;//fff = zhengling(ans,0,-1);q.push(fff);gao[ans] = 1;if((fff.x==0)||(fff.x == (0xFFFF))){cout<<fff.rounds<<endl;return;}while(!q.empty()){zhengling tmp = q.front();q.pop();for(int i=0;i<16;i++){if(tmp.k == i)continue;fff = fan(tmp,i);if(!gao[fff.x]){q.push(fff);gao[fff.x]=1;if((fff.x==0)||(fff.x == (0xFFFF)))break;}} if((fff.x==0)||(fff.x == (0xFFFF)))break;}if(q.empty())cout<<"Impossible\n";elsecout<<fff.rounds<<endl;}void solve(){memset(gao,0,sizeof(gao));bfs();}int main(){ans = 0;for(int i=0;i<4;i++){cin >> g[i];for(int j = 0;j<4;j++){if(g[i][j]=='b')ans|=(0x1<<(i*4+j));}}//init();solve();return 0;}