Widget Factory （高斯消元解线性方程组）

Posted 2020-12-30 mrh-acmer

The widget factory produces several different kinds of widgets. Each widget is carefully built by a skilled widgeteer. The time required to build a widget depends on its type: the simple widgets need only 3 days, but the most complex ones may need as many as 9 days.

The factory is currently in a state of complete chaos: recently, the factory has been bought by a new owner, and the new director has fired almost everyone. The new staff know almost nothing about building widgets, and it seems that no one remembers how many days are required to build each diofferent type of widget. This is very embarrassing when a client orders widgets and the factory cannot tell the client how many days are needed to produce the required goods. Fortunately, there are records that say for each widgeteer the date when he started working at the factory, the date when he was fired and what types of widgets he built. The problem is that the record does not say the exact date of starting and leaving the job, only the day of the week. Nevertheless, even this information might be helpful in certain cases: for example, if a widgeteer started working on a Tuesday, built a Type 41 widget, and was fired on a Friday,then we know that it takes 4 days to build a Type 41 widget. Your task is to figure out from these records (if possible) the number of days that are required to build the different types of widgets.

Input

The input contains several blocks of test cases. Each case begins with a line containing two integers: the number 1 ≤ n ≤ 300 of the different types, and the number 1 ≤ m ≤ 300 of the records. This line is followed by a description of the m records. Each record is described by two lines. The first line contains the total number 1 ≤ k ≤ 10000 of widgets built by this widgeteer, followed by the day of week when he/she started working and the day of the week he/she was fired. The days of the week are given bythe strings `MON‘, `TUE‘, `WED‘, `THU‘, `FRI‘, `SAT‘ and `SUN‘. The second line contains k integers separated by spaces. These numbers are between 1 and n , and they describe the diofferent types of widgets that the widgeteer built. For example, the following two lines mean that the widgeteer started working on a Wednesday, built a Type 13 widget, a Type 18 widget, a Type 1 widget, again a Type 13 widget,and was fired on a Sunday.

4 WED SUN
13 18 1 13

Note that the widgeteers work 7 days a week, and they were working on every day between their first and last day at the factory (if you like weekends and holidays, then do not become a widgeteer!).

The input is terminated by a test case with n = m = 0 .

Output

For each test case, you have to output a single line containing n integers separated by spaces: the number of days required to build the different types of widgets. There should be no space before the first number or after the last number, and there should be exactly one space between two numbers. If there is more than one possible solution for the problem, then write `Multiple solutions.‘ (without the quotes). If you are sure that there is no solution consistent with the input, then write `Inconsistent data.‘(without the quotes).

Sample Input

2 3
2 MON THU
1 2
3 MON FRI
1 1 2
3 MON SUN
1 2 2
10 2
1 MON TUE 
3
1 MON WED
3
0 0

Sample Output

8 3
Inconsistent data.

Hint

Huge input file, ‘scanf‘ recommended to avoid TLE.

 

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <cmath>
using namespace std;
const int maxn = 310;
int equ, var; // 有equ个方程，var个变元。增广阵行数为equ, 分别为0到equ - 1，列数为var + 1，分别为0到var.
int a[maxn][maxn];//增广矩阵
int x[maxn]; // 解集.
int free_num;


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*b/gcd(a,b);
}
// 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)
int change(char s[])
{
    if(strcmp(s,"MON")==0) return 1;
    else if(strcmp(s,"TUE")==0) return 2;
    else if(strcmp(s,"WED")==0) return 3;
    else if(strcmp(s,"THU")==0) return 4;
    else if(strcmp(s,"FRI")==0) return 5;
    else if(strcmp(s,"SAT")==0) return 6;
    else  return 7;
}
int Gauss(void)
{
    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;
    // 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; // 异号的情况是两个数相加.
                for (j = col; j < var + 1; j++)
                {
                    a[i][j] =(((a[i][j] * ta - a[k][j] * tb)%7+7)%7);
                }
            }
        }
    }
    //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)
        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];//等式右边的数
        for (j = i + 1; j < var; j++)
        {
            if (a[i][j] != 0) temp -= a[i][j] * x[j];//把已知的解带入，减去，只剩下，一个未知的解
            temp=(temp%7+7)%7;
        }
        while(temp%a[i][i]!=0)//外层每次循环都是为了求 a[i][i],因为它是每个方程中唯一一个未知的变量（求该方程时）
            temp+=7;//因为天数不确定，而a[i][i]必须得为整数才可以，周期为7
        x[i]=(temp/a[i][i])%7;
    }
    return 0;
}

int main(void)
{
    int n,m,k,num;
    char s[5],e[5];
    while(scanf("%d%d",&n,&m)!=EOF&&(n||m))
    {
        memset(a,0,sizeof(a));
        for(int i=0;i<m;i++)
        {
            scanf("%d",&k);
            scanf("%s%s",s,e);
            a[i][n]=((change(e)-change(s)+1)%7+7)%7;
            for(int j=1;j<=k;j++)//k是他打造的数量
            {
                scanf("%d",&num);//可能是相同的数
                num--;
                a[i][num]++;//系数++
                a[i][num]%=7;//有重复的。
            }
        }
        equ=m;//有m个方程
        var=n;//有多少个变量
        free_num = Gauss();
        if(free_num==0)
        {
            for(int i=0;i<n;i++)//根据题意，每个零件的加工时间在3-9天.
                if(x[i]<=2)
                x[i]+=7;
            for(int i=0;i<n-1;i++)
             cout<<x[i]<<" ";
            cout<<x[n-1]<<endl;
        }
        else if(free_num==-1)
            cout<<"Inconsistent data."<<endl;
        else
            cout<<"Multiple solutions."<<endl;
    }
    return 0;
}