poj 1753 Flip Game（高斯消元）

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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

Output

Sample Input

bwwb
bbwb
bwwb
bwww

Sample Output

4

高斯消元简单题：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
using namespace std;

int a[35][35],x[35];
int equ,var;//方程数和变元数

void Debug()
{
    for(int i=0;i<equ;i++)
    {
        for(int j=0;j<=var;j++)
            printf("%d ",a[i][j]);
        puts("");
    }
    puts("");
}

int gauss()
{
    int max_r,col=0;
    for(int i=0;i<equ&&col<var;i++,col++)
    {
        max_r=i;
        for(int j=i+1;j<equ;j++)
        {
            if(a[j][col]>a[max_r][col]) max_r=j;
        }
        if(max_r!=i)
        {
            for(int k=0;k<=var;k++)
            {
                swap(a[i][k],a[max_r][k]);
            }
        }
        if(a[i][col]==0)
        {
            i--;
            continue;
        }

        for(int j=i+1;j<equ;j++)
        if(i!=j&&a[j][col])
        {
            for(int k=col;k<=var;k++)
            a[j][k]=a[i][k]^a[j][k];
        }
    }

    //Debug();
    for(int i=equ-4;i<equ;i++)
    if(a[i][var]) return -1;

    int ans=100,sum=0;
    for(int k=0;k<=15;k++)
    {
       x[equ-1]=(k&1)>0?1:0;x[equ-2]=(k&2)>0?1:0;x[equ-3]=(k&4)>0?1:0;x[equ-4]=(k&8)>0?1:0;
       //printf("%d%d%d%d\n",x[equ-1],x[equ-2],x[equ-3],x[equ-4]);
       sum=0;
       for(int i=equ-5;i>=0;i--)
       {
        int temp=a[i][var];
        for(int j=i+1;j<var;j++)
        {
            temp-=a[i][j]*x[j];
        }
        x[i]=temp/a[i][i];
       }
       for(int i=0;i<=15;i++)
       if(x[i]%2!=0)
       {
           sum++;
       }

       //printf("  sum:%d\n",sum);
       ans=min(ans,sum);
    }
    return ans;
}


void init()
{
    memset(a,0,sizeof(a));
    memset(x,0,sizeof(x));
    for(int i=0;i<4;i++)
    for(int j=0;j<4;j++)
    {
        if(i!=0) a[i*4+j][(i-1)*4+j]=1;
        if(i!=3) a[i*4+j][(i+1)*4+j]=1;
        if(j!=0) a[i*4+j][i*4+j-1]=1;
        if(j!=3) a[i*4+j][i*4+j+1]=1;
        a[i*4+j][i*4+j]=1;
    }
    //Debug();
}
int main()
{
    equ=var=16;
    char s[20],t[7];
    memset(s,0,sizeof(s));
    memset(t,0,sizeof(t));
    while(scanf("%s",s)!=EOF)
    {
        for(int i=0;i<3;i++)
        {
            scanf("%s",t);
            strcat(s,t);
        }
        init();
        for(int i=0;i<16;i++)
        {
            if(s[i]=='b') a[i][var]=1;
            else a[i][var]=0;
        }
        int ans1=gauss();

        init();
        for(int i=0;i<16;i++)
        {
            if(s[i]=='w') a[i][var]=1;
            else a[i][var]=0;
        }
        int ans2=gauss();

        if(ans1!=-1&&ans2!=-1)
            printf("%d\n",min(ans1,ans2));
        else if(ans1!=-1) printf("%d\n",ans1);
        else if(ans2!=-1) printf("%d\n",ans2);
        else printf("Impossible\n");
    }
    return 0;
}

poj 1753 Flip Game（高斯消元）

标签：des   style   http   color   os   io   strong   for   ar   

原文地址：http://blog.csdn.net/knight_kaka/article/details/38797121