hdu 5755 Gambler Bo 高斯消元

题目链接

给n*m的方格， 每个格子有值{0, 1, 2}。 然后可以对格子进行操作， 如果选择了一个格子， 那么这个格子的值+2， 这个格子上下左右的格子+1， 并且模3。

问你将所有格子变成0的操作方法。

其实就是一个模3的方程组， 高斯消元就可以了。 不知道为什么昨天比赛就是想不到......

#include <iostream>
#include <vector>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <complex>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <queue>
#include <stack>
#include <bitset>
using namespace std;
#define pb(x) push_back(x)
#define ll long long
#define mk(x, y) make_pair(x, y)
#define lson l, m, rt<<1
#define mem(a) memset(a, 0, sizeof(a))
#define rson m+1, r, rt<<1|1
#define mem1(a) memset(a, -1, sizeof(a))
#define mem2(a) memset(a, 0x3f, sizeof(a))
#define rep(i, n, a) for(int i = a; i<n; i++)
#define fi first
#define se second
typedef complex <double> cmx;
typedef pair<int, int> pll;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int mod = 3;
const int inf = 1061109567;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
int a[901][902];
int n, b[902], x[902], m;
int gcd(int a, int b)
{
    return b?gcd(b, a%b):a;
}
int lcm(int a, int b)
{
    return a/gcd(a, b)*b;
}
ll inv(ll a, ll m)
{
    if(a == 1)
        return 1;
    return inv(m%a, m)*(m-m/a)%m;
}
int gauss(int equ, int var)
{
    int max_r, col, k;
    for(k = 0, col = 0; k < equ && col < var; k++, col++) {
        max_r = k;
        for(int i = k + 1; i < equ; i ++) {
            if(abs(a[i][col]) > abs(a[max_r][col]))
                max_r = i;
        }
        if(a[max_r][col] == 0) {
            k--;
            continue;
        }
        if(max_r != k) {
            for(int j = col; j < var+1; j++) {
                swap(a[k][j], a[max_r][j]);
            }
        }
        for(int i = k + 1; i < equ; i++) {
            if(a[i][col]) {
                int LCM = lcm(abs(a[i][col]), abs(a[k][col]));
                int ta = LCM/abs(a[i][col]);
                int tb = LCM/abs(a[k][col]);
                if(a[i][col] * a[k][col] < 0)
                    tb = -tb;
                for(int j = col; j < var+1; j++) {
                    a[i][j] = ((a[i][j]*ta - a[k][j]*tb)%mod+mod)%mod;
                }
            }
        }
        for(int i = var-1; i >= 0; i--) {
            int tmp = a[i][var];
            for(int j = i+1; j < var; j++) {
                if(a[i][j]) {
                    tmp -= a[i][j]*x[j];
                    tmp = (tmp%mod+mod)%mod;
                }
            }
            x[i] = a[i][i]*tmp%mod;
        }
    }
}
int check(int x)
{
    return x >=0 && x < n*m;
}
int main()
{
    int t;
    cin>>t;
    while(t--) {
        scanf("%d%d", &n, &m);
        for(int i = 0; i < n*m; i++) {
            scanf("%d", &b[i]);
        }
        mem(x);
        mem(a);
        for(int i = 0; i < n*m; i++) {
            int u = i-m, d = i+m;
            int l = i-1, r = i+1;
            if(check(u)) {
                a[i][u] = 1;
            }
            if(check(d)) {
                a[i][d] = 1;
            }
            if(check(l)&&i%m!=0) {
                a[i][l] = 1;
            }
            if(check(r)&&(i+1)%m!=0) {
                a[i][r] = 1;
            }
            a[i][i] = 2;
            a[i][n*m] = (3-b[i])%3;
        }
        n *= m;
        gauss(n, n);
        int cnt = 0;
        for(int i = 0; i < n; i++) {
            if(x[i])
                cnt += x[i];
        }
        printf("%d
", cnt);
        for(int i = 0; i < n; i++) {
            while(x[i]) {
                printf("%d %d
", i/m+1, i%m+1);
                x[i]--;
            }
        }
    }
    return 0;
}