HDU 5755 Gambler Bo（高斯消元）2016 Multi-University Training Contest 3

Gambler Bo

Time Limit: 8000/4000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1058 Accepted Submission(s): 437
Special Judge

Problem Description
Gambler Bo is very proficient in a matrix game.

You have a N×M matrix, every cell has a value in {0,1,2}.

In this game, you can choose a cell in the matrix, plus 2 to this cell, and plus 1 to all the adjacent cells.

for example, you choose the cell (x,y), the value of (x,y) will be plused 2, and the value of (x−1,y)(x+1,y)(x,y−1)(x,y+1) will be plused 1.

if you choose the cell (1,2), the cell (1,2) will be plused 2, and the cell (2,2)(1,1)(1,3) will be plused 1, the cell (0,2) won’t be changed because it’s out of the matrix.

If the values of some cells is exceed 2, then these values will be modulo 3.

Gambler Bo gives you such a matrix, your task is making all value of this matrix to 0 by doing above operations no more than 2NM times.

Input
First line, an integer T. There are T test cases.

In each test, first line is two integers N,M, and following N lines describe the matrix of this test case.

T≤10,1≤N,M≤30, the matrix is random and guarantee that there is at least one operation solution.

Output
For each test, first line contains an integer num(0≤num≤2NM) describing the operation times.

Following num lines, each line contains two integers x,y(1≤x≤N,1≤y≤M) describing the operation cell.

The answer may not be unique, you can output any one.

Sample Input
2
2 3
2 1 2
0 2 0
3 3
1 0 1
0 1 0
1 0 1

Sample Output
1
1 2
5
1 1
1 3
2 2
3 1
3 3

Author
绍兴一中

Source
2016 Multi-University Training Contest 3

高斯消元裸题，用了kuangbin的模板，但是里面有一些小改动，还有一些需要奇技淫巧的地方。

#include "cstring"#include "cstdio"#include "iostream"#include "string.h"#include "vector"using namespace std;const int MAXN=1000;vector<int> ans;int a[MAXN][MAXN];//增广矩阵int x[MAXN];//解集bool free_x[MAXN];//标记是否是不确定的变元/* void Debug(void) { int i, j; for (i = 0; i < equ; i++) { for (j = 0; j < var + 1; j++) { cout << a[i][j] << " "; } cout << endl; } cout << endl; } */inline int gcd(int a,int b){    int t;    while(b!=0)    {        t=b;        b=a%b;        a=t;    }    return a;}inline int lcm(int a,int b){    return a/gcd(a,b)*b;//先除后乘防溢出}// 高斯消元法解方程组(Gauss-Jordan elimination).(-2表示有浮点数解，但无整数解，//-1表示无解，0表示唯一解，大于0表示无穷解，并返回自由变元的个数)//有equ个方程，var个变元。增广矩阵行数为equ,分别为0到equ-1,列数为var+1,分别为0到var.int Gauss(int equ,int var){    int i,j,k;    int max_r;// 当前这列绝对值最大的行.    int col;//当前处理的列    int ta,tb;    int LCM;    int temp;    int free_x_num;    int free_index;    for(int i=0;i<=var;i++)    {        x[i]=0;        free_x[i]=true;    }    //转换为阶梯阵.    col=0; // 当前处理的列    for(k = 0;k < equ && col < var;k++,col++)    {// 枚举当前处理的行.        // 找到该col列元素绝对值最大的那行与第k行交换.(为了在除法时减小误差)        max_r=k;        for(i=k+1;i<equ;i++)        {            if(abs(a[i][col])>abs(a[max_r][col])) max_r=i;        }        if(max_r!=k)        {// 与第k行交换.            for(j=k;j<var+1;j++) swap(a[k][j],a[max_r][j]);        }        if(a[k][col]==0)        {// 说明该col列第k行以下全是0了，则处理当前行的下一列.            k--;            continue;        }        for(i=k+1;i<equ;i++)        {// 枚举要删去的行.            if(a[i][col]!=0)            {                LCM = lcm(abs(a[i][col]),abs(a[k][col]));                ta = LCM/abs(a[i][col]);                tb = LCM/abs(a[k][col]);                if(a[i][col]*a[k][col]<0)tb=-tb;//异号的情况是相加                for(j=col;j<var+1;j++)                {                    a[i][j] = (a[i][j]*ta-a[k][j]*tb);                    a[i][j]=(a[i][j]%3+3)%3;                    //a[i][j]%=3;                }            }        }    }    /*     //  Debug();     // 1. 无解的情况: 化简的增广阵中存在(0, 0, ..., a)这样的行(a != 0).     for (i = k; i < equ; i++)     { // 对于无穷解来说，如果要判断哪些是自由变元，那么初等行变换中的交换就会影响，则要记录交换.     if (a[i][col] != 0) return -1;     }     // 2. 无穷解的情况: 在var * (var + 1)的增广阵中出现(0, 0, ..., 0)这样的行，即说明没有形成严格的上三角阵.     // 且出现的行数即为自由变元的个数.     if (k < var)     {     // 首先，自由变元有var - k个，即不确定的变元至少有var - k个.     for (i = k - 1; i >= 0; i--)     {     // 第i行一定不会是(0, 0, ..., 0)的情况，因为这样的行是在第k行到第equ行.     // 同样，第i行一定不会是(0, 0, ..., a), a != 0的情况，这样的无解的.     free_x_num = 0; // 用于判断该行中的不确定的变元的个数，如果超过1个，则无法求解，它们仍然为不确定的变元.     for (j = 0; j < var; j++)     {     if (a[i][j] != 0 && free_x[j]) free_x_num++, free_index = j;     }     if (free_x_num > 1) continue; // 无法求解出确定的变元.     // 说明就只有一个不确定的变元free_index，那么可以求解出该变元，且该变元是确定的.     temp = a[i][var];     for (j = 0; j < var; j++)     {     if (a[i][j] != 0 && j != free_index) temp -= a[i][j] * x[j];     }     x[free_index] = temp / a[i][free_index]; // 求出该变元.     free_x[free_index] = 0; // 该变元是确定的.     }     return var - k; // 自由变元有var - k个.     }*/    // 3. 唯一解的情况: 在var * (var + 1)的增广阵中形成严格的上三角阵.    // 计算出Xn-1, Xn-2 ... X0.    for (i = var - 1; i >= 0; i--)    {        temp = a[i][var];        for (j = i + 1; j < var; j++)        {            if (a[i][j] != 0)            {                temp -= a[i][j] * x[j];                temp=(temp%3+3)%3;                //temp%=3;            }        }        /*         if (temp % a[i][i] != 0) return -2; // 说明有浮点数解，但无整数解.         x[i] = temp / a[i][i];*/        x[i] = temp*a[i][i]; //用逆元把除法变成乘法 化简之后刚好是这个形式        x[i] %= 3;    }    return 0;}int main(){    int t;    scanf("%d",&t);    while(t--)    {        ans.clear();        int m,n;        scanf("%d%d",&m,&n);        memset(a,0,sizeof(a));        for(int i=0;i<m*n;i++)        {            int c=i%n,r=i/n;            a[i][i]=2;            if(c>0)                a[i][i-1]=1;            if(c<n-1)                a[i][i+1]=1;            if(r>0)                a[i][i-n]=1;            if(r<m-1)                a[i][i+n]=1;        }        for(int i=0;i<m;i++)        {            for(int j=0;j<n;j++)            {                int temp;                scanf("%d",&temp);                //???                a[i*n+j][n*m]=((3-temp)%3);//要加几才能为0            }        }        int s=Gauss(n*m, n*m);        for(int i=0;i<m*n;i++)        {            while(x[i])            {                x[i]--;                ans.push_back(i);            }        }        printf("%d\n",ans.size());        for(int i=0;i<ans.size();i++)        {            int c=ans[i]%n,r=ans[i]/n;            printf("%d %d\n",r+1,c+1);        }    }    return 0;}