J - EXTENDED LIGHTS OUT

Posted 2020-12-18 zzhzzh123

POJ - 1222

与这道题一样，都是高斯消元求异或方程组。

一共(30)盏灯，每盏灯影响上下左右的灯，基本上就是矩阵改一下。

最后求解方程，自由元随你定。

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

int n;

int a[35],b[35];
int g[35][35];

long long ksm(long long x,int y){
    long long z = 1;
    while(y){
        if(y & 1) z = z * x;
        y >>= 1;
        x = x * x;
    }
    return z;
}

int ans[35];

int main(){
    int T; scanf("%d",&T);
    for(int cas = 1; cas <= T; ++ cas){
        memset(g,0,sizeof(g));
        int n = 30;
        for(int i = 0; i < 5; ++ i)
        for(int j = 1; j <= 6; ++ j){
            scanf("%d",&g[i * 6 + j][31]);
        }
        for(int i = 0; i < 5; ++ i)
        for(int j = 1; j <= 6; ++ j){
            int id, ps = i * 6 + j; 
            if(i > 0) id = (i - 1) * 6 + j, g[id][ps] = 1;
            if(i < 5) id = (i + 1) * 6 + j, g[id][ps] = 1;
            if(j > 1) id = i * 6 + j - 1, g[id][ps] = 1;
            if(j < 6) id = i * 6 + j + 1, g[id][ps] = 1;
        }
        for(int i = 1; i <= n; ++ i) g[i][i] = 1;
        
        int now = 1;
        for(int i = 1; i <= n; ++ i){
            bool flag = 0; int pos = -1;
            for(int j = now; j <= n; ++ j){
                if(g[j][i] == 1) { flag = 1; pos = j; break; }
            }
            if(!flag) continue;
            for(int j = 1; j <= n + 1; ++ j) swap(g[now][j],g[pos][j]);
            
            for(int j = now + 1; j <= n; ++ j){
                if(g[j][i] == 0) continue;
                for(int k = 1; k <= n + 1; ++ k) 
                g[j][k] ^= g[now][k];
            }
            
            ++ now;
        }
        
        printf("PUZZLE #%d
",cas);
        
        bool flag = 1;
        for(int i = now; i <= n; ++ i){
            if(g[i][n + 1] != 0) { flag = 0; break; }
        }
        if(!flag) { puts("Oh,it‘s impossible~!!"); continue; }
        
        memset(ans,0,sizeof(ans));
        for(int i = now - 1; i >= 1; -- i){
            int pos = -1;
            for(int j = 1; j <= n; ++ j) if(g[i][j] != 0) { pos = j; break; }
            ans[pos] = g[i][n + 1];
            for(int j = n; j > pos; -- j) ans[pos] ^= (ans[j] & g[i][j]);
        }
        
        for(int i = 0; i < 5; ++ i){
            for(int j = 1; j <= 6; ++ j)
            printf("%d ",ans[i * 6 + j]);
            puts("");
        }
    }
    return 0;
}