Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 20199 | Accepted: 8752 |
Description
- Choose any one of the 16 pieces.
- 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
Source
#include<iostream> #include<cstring> using namespace std; const int mm=937; const int oo=1e9; int map[mm][mm+1],b[mm][mm+1]; int ans[mm]; int var,equ; int max_ans; void debug(); void swap(int&a,int &b) { int tmp;tmp=a;a=b;b=tmp; } /* int abs(int x) { if(x<0)return -x;return x; }*/ int gcd(int a,int b) { int t; while(b!=0) { t=b;b=a%b;a=t; }return a; } int lcm(int a,int b) { return a*b/gcd(a,b); } void dfs(int z,int dep,int k,int kans) { ans[var-dep]=z; if(dep==k) { for(int i=var-k-1;i>=0;i--) { int tmp; tmp=map[i][var]%2; for(int j=i+1;j<var;j++) if(map[i][j]) tmp=(tmp-(map[i][j]*ans[j])%2+2)%2; ans[i]=(tmp/map[i][i])%2;if(ans[i])++kans; } if(kans<max_ans)max_ans=kans; //cout<<"ans="<<kans<<endl; } else { dfs(1,dep+1,k,kans+1);dfs(0,dep+1,k,kans); } } int Guss() { int k,col;max_ans=0; for(k=0,col=0;k<equ&&col<var;k++,col++) {/**行变换消元*/ int max_i=k; for(int i=k+1;i<equ;i++) if(map[i][col]>map[max_i][col])max_i=i; if(max_i!=k) {//for(int i=k;i<var+1;i++)好像也OK for(int i=col;i<var+1;i++)//是K不是col因为k--; swap(map[k][i],map[max_i][i]); } /**除去对角线0,列变换*/ for(int i = 0; i <equ; ++i)//每一行主元素化为非零 if(!map[i][i]) { int j; for(j = i+1;j<var;++j) if(map[i][j]) break; if(j == var) break; for(int kk = 0;kk < equ; ++kk) swap(map[kk][i],map[kk][j]); } /**遇到对角线0,k退1*/ if(map[k][col]==0) { k--;continue; } /**消元*/ for(int i=k+1;i<equ;i++) if(map[i][col]!=0) { int LCM=lcm(map[k][col],map[i][col]); int ta,tb; ta=LCM/map[i][col];tb=LCM/map[k][col]; if(map[i][col]*map[k][col]<0)tb=-tb; for(int j=col;j<var+1;j++) { map[i][j]=((map[i][j]*ta)%2-(map[k][j]*tb)%2+2)%2; } } // cout<<"k="<<k<<endl; } //debug(); ///判断回代 for(int i=k;i<equ;i++) if(map[i][col])return -1; if(var==k) for(int i=k-1;i>=0;i--) { int tmp; tmp=map[i][var]%2; for(int j=i+1;j<var;j++) if(map[i][j]) tmp=(tmp-(map[i][j]*ans[j])%2+2)%2; ans[i]=(tmp/map[i][i])%2;if(ans[i])max_ans++; } else { max_ans=oo; dfs(1,1,var-k,1);dfs(0,1,var-k,0); } return 0; } void debug() { for(int i=0;i<var;i++) {for(int j=0;j<var+1;j++) cout<<map[i][j]; cout<<"\n"; } } int main() { int num; char s; int k_ans=oo; memset(map,0,sizeof(map)); memset(ans,0,sizeof(ans)); num=4; var=equ=num*num; for(int i=0;i<num;i++) { for(int j=0;j<num;j++) { map[i*num+j][i*num+j]=1; if(i!=0)map[i*num+j][(i-1)*num+j]=1; if(j!=0)map[i*num+j][i*num+j-1]=1; if(i!=num-1)map[i*num+j][(i+1)*num+j]=1; if(j!=num-1)map[i*num+j][i*num+j+1]=1; cin>>s; if(s=='b')map[i*num+j][var]=0; else map[i*num+j][var]=1; //cout<<map[i*num+j][var]; //cout<<"i="<<i<<"j="<<j<<endl; } } for(int i=0;i<var;i++) {for(int j=0;j<var;j++) b[i][j]=map[i][j]; b[i][var]=map[i][var]^1; } //debug(); /*for(int i=0;i<var;i++) {for(int j=0;j<var;j++) cout<<b[i][j]; cout<<b[i][var]<<"\n"; }*/ int flag=Guss(); if(flag!=-1&&max_ans<k_ans) k_ans=max_ans; for(int i=0;i<var;i++) {for(int j=0;j<var;j++) map[i][j]=b[i][j]; map[i][var]=b[i][var]; } flag=Guss(); if(flag!=-1&&max_ans<k_ans) k_ans=max_ans; if(k_ans==oo)cout<<"Impossible\n"; else { cout<<k_ans<<"\n"; } }