嘟嘟嘟

遇到這種（看似）構造的題，我好像一般都做不出來……

然而這題正解是高斯消元解異或方程組……
首先我們容易列出式子a[i][j] ^ a[i - 1][j] ^ a[i + 1][j] ^ a[i][j - 1] ^ a[i][j + 1] = 0。于是我們列出所有像這樣的\(n * m\)個式子，然后\(O((nm) ^ 3)\)高斯消元加bitset優化就過了。

講真我還不會高斯消元解異或方程組，就現學了一下。其實就是把運算改成了異或，然后bitset可以把一個一個消改成一行和一行消。

#include<cstdio> #include<iostream> #include<cmath> #include<algorithm> #include<cstring> #include<cstdlib> #include<cctype> #include<vector> #include<stack> #include<queue> #include<ctime> #include<bitset> using namespace std; #define enter puts("") #define space putchar(' ') #define Mem(a, x) memset(a, x, sizeof(a)) #define In inline typedef long long ll; typedef double db; const int INF = 0x3f3f3f3f; const db eps = 1e-8; const int maxn = 1605; inline ll read() {ll ans = 0;char ch = getchar(), last = ' ';while(!isdigit(ch)) last = ch, ch = getchar();while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();if(last == '-') ans = -ans;return ans; } inline void write(ll x) {if(x < 0) x = -x, putchar('-');if(x >= 10) write(x / 10);putchar(x % 10 + '0'); }int n, m; const int dx[] = {-1, 0, 1, 0, 0}, dy[] = {0, 1, 0, -1, 0}; bitset<maxn> f[maxn];In int num(int x, int y) {return (x - 1) * m + y; }int ans[maxn]; In void Gauss(int n) {for(int i = 1; i <= n; ++i){int pos = i;while(pos <= n && !f[pos][i]) ++pos;if(pos == n + 1) continue;swap(f[i], f[pos]);for(int j = i + 1; j <= n; ++j) if(f[j][i]) f[j] ^= f[i];}for(int i = n; i; --i)if(!f[i][i]) ans[i] = 1;else for(int j = i + 1; j <= n; ++j) if(f[i][j]) ans[i] ^= ans[j]; }int main() {n = read(), m = read();for(int i = 1; i <= n; ++i)for(int j = 1; j <= m; ++j)for(int k = 0; k <= 4; ++k){int x = i + dx[k], y = j + dy[k];if(x < 1 || x > n || y < 1 || y > m) continue;f[num(i, j)][num(x, y)] = 1;}// for(int i = 1; i <= n; ++i) cout << f[i].to_string() << endl;Gauss(n * m);for(int i = 1; i <= n; ++i){for(int j = 1; j <= m; ++j) write(ans[num(i, j)]), space;enter;}return 0; }

轉載于:https://www.cnblogs.com/mrclr/p/10461668.html

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