【高斯消元】 HDOJ 5257 翻转游戏

来源：互联网发布：ubuntu kernel 编辑：程序博客网时间：2024/06/12 19:08

如果第一行的状态确定了，那么矩阵的所有状态也会随之确定。。。那么我们就将第一行写成变量，这样能够推出矩阵的m个方程。。。然后对于k，可以写出k个限制方程。。因此我们总共列出m+k个方程，高斯消元，bitset优化即可。。。

#include <iostream>#include <queue>#include <stack>#include <map>#include <set>#include <bitset>#include <cstdio>#include <algorithm>#include <cstring>#include <climits>#include <cstdlib>#include <cmath>#include <time.h>#define maxn 305#define maxm 100005#define eps 1e-7#define mod 1000000007#define INF 0x3f3f3f3f#define PI (acos(-1.0))#define lowbit(x) (x&(-x))#define mp make_pair#define ls o<<1#define rs o<<1 | 1#define lson o<<1, L, mid #define rson o<<1 | 1, mid+1, R#define pii pair<int, int>#pragma comment(linker, "/STACK:16777216")typedef long long LL;typedef unsigned long long ULL;//typedef int LL;using namespace std;LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}// headbitset<maxn> bs[maxn][maxn], equ[maxn];char G[maxn][maxn];bool g[maxn][maxn];void init(int n, int m){for(int i = 0; i <= n+1; i++)for(int j = 0; j <= m+1; j++) bs[i][j] = 0;}bool gauss(int n, int m){for(int i = 1, r = 1; i <= m; i++, r++) {int t = -1;for(int j = r; j <= n; j++) if(equ[j][i]) t = j;if(t == -1) continue;swap(equ[r], equ[t]);for(int j = r+1; j <= n; j++) if(equ[j][i]) equ[j] ^= equ[r];}for(int i = 1; i <= n; i++) {int flag = 0;for(int j = 1; j <= m; j++) flag |= equ[i][j];if(!flag && equ[i][m+1]) return false;}return true;}void debug(int n, int m){for(int i = 1; i <= n; i++) {for(int j = 1; j <= m + 1; j++) {int t = equ[i][j];printf("%d ", t);}printf("\n");}}void work(){int n, m, kk, x, y;scanf("%d%d%d", &n, &m, &kk);init(n, m);for(int i = 1; i <= n; i++) scanf("%s", G[i] + 1);for(int i = 1; i <= n; i++)for(int j = 1; j <= m; j++)g[i][j] = G[i][j] == 'B';for(int i = 1; i <= m; i++) bs[1][i][i] = 1;for(int i = 2; i <= n+1; i++)for(int j = 1; j <= m; j++) {bs[i][j] ^= bs[i-1][j] ^ bs[i-2][j] ^ bs[i-1][j-1] ^ bs[i-1][j+1];bs[i][j][m+1] = bs[i][j][m+1] ^ g[i-1][j];}for(int i = 1; i <= m; i++) equ[i] = bs[n + 1][i];for(int i = m+1; i <= m+kk; i++) {scanf("%d%d", &x, &y);equ[i] = bs[x-1][y] ^ bs[x+1][y] ^ bs[x][y-1] ^ bs[x][y+1];equ[i][m+1] = equ[i][m+1] ^ g[x][y];}if(gauss(m+kk, m)) printf("YES\n");else printf("NO\n");}int main(){int _;scanf("%d", &_);for(int i = 1; i <= _; i++) {printf("Case #%d:\n", i);work();}return 0;}

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