poj 1753 Flip Game（高斯消元 开关问题）

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Flip Game

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 33227 Accepted: 14532

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

4*4的网格中有黑色和白色 现在需要翻转  翻一个格子 它的上下左右也翻转 

现在需要通过翻转求出使用最少的次数使得格子中都是黑色或者白色 

因为不知道目标状态是黑色还是白色 所以需要两次高斯消元 求出较小的值

#include <cstdio>#include <iostream>#include <cstring>#include <cmath>#include <algorithm>#include <string.h>#include <string>#include <vector>#include <queue>#define MEM(a,x) memset(a,x,sizeof a)#define eps 1e-8#define MOD 10009#define MAXN 30#define MAXM 100010#define INF 99999999#define ll __int64#define bug cout<<"here"<<endl#define fread freopen("ceshi.txt","r",stdin)#define fwrite freopen("out.txt","w",stdout)using namespace std;int Read(){    char c = getchar();    while (c < '0' || c > '9') c = getchar();    int x = 0;    while (c >= '0' && c <= '9') {        x = x * 10 + c - '0';        c = getchar();    }    return x;}void Print(int a){     if(a>9)         Print(a/10);     putchar(a%10+'0');}int equ,var;//equ个方程 var个变元 增广矩阵行数为equ 列数为var+1 分别为0到varint a[MAXN][MAXN];//增广矩阵int x[MAXN];//解集int free_x[MAXN];//存储自由变元（多解枚举自由变元可以使用）int free_num;//自由变元的个数int n;void init(){    MEM(a,0); MEM(x,0);    equ=n*n; var=n*n;    for(int i=0;i<n;i++)        for(int j=0;j<n;j++)        {            int t=i*n+j;            a[t][t]=1;            if(i>0) a[(i-1)*n+j][t]=1;            if(i<n-1) a[(i+1)*n+j][t]=1;            if(j>0) a[i*n+j-1][t]=1;            if(j<n-1) a[i*n+j+1][t]=1;        }}//返回值为-1表示无解 为0是唯一解 否则返回自由变元个数int Gauss(){    int max_r,col,k;    free_num=0;    for(k=0,col=0;k<equ&&col<var;k++,col++)    {        max_r=k;        for(int i=k+1;i<equ;i++)        {            if(abs(a[i][col])>abs(a[max_r][col]))                max_r=i;        }        if(a[max_r][col]==0)        {            k--;            free_x[free_num++]=col;//这个是自由变元            continue;        }        if(max_r!=k)        {            for(int j=col;j<var+1;j++)                swap(a[k][j],a[max_r][j]);        }        for(int i=k+1;i<equ;i++)        {            if(a[i][col]!=0)            {                for(int j=col;j<var+1;j++)                    a[i][j]^=a[k][j];            }        }    }    for(int i=k;i<equ;i++)        if(a[i][col]!=0)            return -1;//无解    if(k<var) return var-k;//自由变元个数    //唯一解  回带    for(int i=var-1;i>=0;i--)    {        x[i]=a[i][var];        for(int j=i+1;j<var;j++)            x[i]^=(a[i][j]&&x[j]);    }    return 0;}int solve(){    int t=Gauss();    if(t==-1)    {        return INF;    }    else if(t==0)    {        int ans=0;        for(int i=0;i<n*n;i++)            ans+=x[i];       return ans;    }    else    {        int ans=INF;        int tot=(1<<t);        for(int i=0;i<tot;i++)        {            int cnt=0;            for(int j=0;j<t;j++)            {                if(i&(1<<j))                {                    x[free_x[j]]=1;                    cnt++;                }                else x[free_x[j]]=0;            }            for(int j=var-t-1;j>=0;j--)            {                int idx;                for(idx=j;idx<var;idx++)                    if(a[j][idx])                        break;                x[idx]=a[j][var];                for(int l=idx+1;l<var;l++)                    if(a[j][l])                        x[idx]^=x[l];                cnt+=x[idx];            }            ans=min(cnt,ans);        }        return ans;    }}char ch[5][5];int main(){//    fread;    while(scanf("%s",ch[0])!=EOF)    {        for(int i=1;i<4;i++)            scanf("%s",ch[i]);        n=4;        init();        for(int i=0;i<n;i++)        {            for(int j=0;j<n;j++)            {                if(ch[i][j]=='b')                    a[i*n+j][n*n]=0;                else a[i*n+j][n*n]=1;            }        }        int ans1=solve();        init();        for(int i=0;i<n;i++)        {            for(int j=0;j<n;j++)            {                if(ch[i][j]=='b')                    a[i*n+j][n*n]=1;                else a[i*n+j][n*n]=0;            }        }        int ans2=solve();        if(ans1==INF&&ans2==INF)            puts("Impossible");        else        {            int res=min(ans1,ans2);            printf("%d\n",res);        }    }    return 0;}