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100A留个纪念/* * Created: 2016年04月05日 22时28分26秒 星期二 * Author: Akrusher * */ #include <cstdio> #include <cstdlib> #include <cstring> #include <cmath> #include <ctime> #include <iostream> #include <algorithm> #include <string> #include <vector> #include <deque> #include <list> #include <set> #include <map> #include <stack> #include <queue> #include <numeric> #include <iomanip> #include <bitset> #include <sstream> #include <fstream> using namespace std; #define rep(i,a,n) for (int i=a;i<n;i++) #define per(i,a,n) for (int i=n-1;i>=a;i--) #define in(n) scanf("%d",&(n)) #define in2(x1,x2) scanf("%d%d",&(x1),&(x2)) #define in3(x1,x2,x3) scanf("%d%d%d",&(x1),&(x2),&(x3)) #define inll(n) scanf("%I64d",&(n)) #define inll2(x1,x2) scanf("%I64d%I64d",&(x1),&(x2)) #define inlld(n) scanf("%lld",&(n)) #define inlld2(x1,x2) scanf("%lld%lld",&(x1),&(x2)) #define inf(n) scanf("%f",&(n)) #define inf2(x1,x2) scanf("%f%f",&(x1),&(x2)) #define inlf(n) scanf("%lf",&(n)) #define inlf2(x1,x2) scanf("%lf%lf",&(x1),&(x2)) #define inc(str) scanf("%c",&(str)) #define ins(str) scanf("%s",(str)) #define out(x) printf("%d\n",(x)) #define out2(x1,x2) printf("%d %d\n",(x1),(x2)) #define outf(x) printf("%f\n",(x)) #define outlf(x) printf("%lf\n",(x)) #define outlf2(x1,x2) printf("%lf %lf\n",(x1),(x2)); #define outll(x) printf("%I64d\n",(x)) #define outlld(x) printf("%lld\n",(x)) #define outc(str) printf("%c\n",(str)) #define pb push_back #define mp make_pair #define fi first #define se second #define SZ(x) ((int)(x).size()) #define mem(X,Y) memset(X,Y,sizeof(X)); typedef vector<int> vec; typedef long long ll; typedef pair<int,int> P; const int dx[5]={-1,0,0,0,1},dy[5]={0,-1,0,1,0}; const int INF=0x3f3f3f3f; const ll mod=1e9+7; ll powmod(ll a,ll b) {ll res=1;a%=mod;for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;} const bool AC=true; const int M=5; const int N=6; int tile[5][6]; int opt[5][6]; int flip[5][6]; int k; int get(int x,int y){ int c=tile[x][y]; rep(i,0,5){ int nx=x+dx[i],ny=y+dy[i]; if(nx>=0&&ny>=0&&nx<M&&ny<N){ c+=flip[nx][ny]; } } return c%2; } int calc(){ rep(i,1,M){ rep(j,0,N){ if(get(i-1,j)) flip[i][j]=1; } } rep(j,0,N){ if(get(M-1,j)){ return -1; } } int res=0; rep(i,0,M) rep(j,0,N) if(flip[i][j]) res++; return res; } void solve(){ int res=-1; rep(i,0,1<<N){ mem(flip,0); rep(j,0,N){ flip[0][N-j-1]=(i>>j)&1; } int num=calc(); if(num>=0&&(res<0||res>num)){ //此处可能为0 res=num; memcpy(opt,flip,sizeof(flip)); } } printf("PUZZLE #%d\n",k); rep(i,0,M){ rep(j,0,N){ printf("%d ",opt[i][j]); } printf("\n"); } } int main() { int t; in(t); k=0; while(t--){ k++; rep(i,0,M) rep(j,0,N){ in(tile[i][j]); } solve(); } return 0; } Followed by: Post your reply here: |
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