POJ 2947：Widget Factory 求同余方程

Widget Factory

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 5173		Accepted: 1790

Description

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. 

题意是给出了n件商品加工时间，m个同余方程。每个方程右边的值自然是时间差，左边就是第i种商品有多少个在这个期间加工。

感觉之前模板里面的写得有一些小问题。对于同余方程，还是最后枚举下来更正确，套用之前的模板对于POJ1166输出有问题。

代码：

#pragma warning(disable:4996)
#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
#include <cstdlib>
#include <cstdio>
using namespace std;

int n, m;
int x[305];//解集
int val[305][305];//增广矩阵

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
}

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;
	}
	//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.(为了在除法时减少误差)
		max_r = k;
		for (i = k + 1; i < equ; i++)
		{
			if (abs(val[i][col])>abs(val[max_r][col]))
				max_r = i;
		}
		if (max_r != k)
		{//与第k行交换
			for (j = k; j < var + 1; j++)
				swap(val[k][j], val[max_r][j]);
		}
		if (val[k][col] == 0)
		{
			k--;
			continue;
		}
		for (i = k + 1; i < equ; i++)
		{//枚举要删去的行
			if (val[i][col] != 0)
			{
				LCM = lcm(abs(val[i][col]), abs(val[k][col]));
				ta = LCM / abs(val[i][col]);
				tb = LCM / abs(val[k][col]);
				if (val[i][col] * val[k][col] < 0)
					tb = -tb;
				for (j = col; j < var + 1; j++)
				{
					val[i][j] = ((val[i][j] * ta - val[k][j] * tb) % 7 + 7) % 7;
				}
			}
		}
	}

	//无解的情况
	for (i = k; i < equ; i++)
	{
		if (val[i][col] != 0)
			return -1;
	}

	if (k < var)
	{
		return var - k;
	}
	for (i = var - 1; i >= 0; i--)
	{
		temp = val[i][var];
		for (j = i + 1; j < var; j++)
		{
			if (val[i][j] != 0)
			{
				temp = temp - val[i][j] * x[j];
				temp = (temp % 7 + 7) % 7;
			}
		}
		for (x[i] = 0; x[i] < 7; x[i]++)
		{
			if ((x[i] * val[i][i] + 7) % 7 == temp)
			{
				break;
			}
		}
	}
	return 0;
}

int tran(char *s)
{
	if (strcmp(s, "MON") == 0)
		return 1;
	else if (strcmp(s, "TUE") == 0)//"MON","TUE", "WED","THU", "FRI","SAT","SUN"
		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 if (strcmp(s, "SUN") == 0)
		return 7;
}

int main()
{
	//freopen("i.txt", "r", stdin);
	//freopen("o.txt", "w", stdout);

	int i, j, num, temp_num, ans;
	char temp1[20], temp2[20];
	while (scanf("%d%d", &n, &m) != EOF)
	{
		if (n == 0 && m == 0)
			break;
		memset(val, 0, sizeof(val));
		for (i = 0; i < m; i++)
		{
			scanf("%d%s%s", &num, temp1, temp2);

			val[i][n] = ((tran(temp2) - tran(temp1) + 1) % 7 + 7) % 7;
			for (j = 0; j < num; j++)
			{
				scanf("%d", &temp_num);
				val[i][temp_num - 1]++;
				val[i][temp_num - 1] = val[i][temp_num - 1] % 7;
			}
		}
		ans = Gauss(m, n);
		if (ans == 0)
		{
			for (int h = 0; h < n; h++)
			{
				if (x[h] < 3)x[h] = x[h] + 7;
				if (h == 0)
					printf("%d", x[h]);
				else
				{
					printf(" %d", x[h]);
				}
			}
			printf("\n");
		}
		else if (ans == -1)
		{
			printf("Inconsistent data.\n");
		}
		else
		{
			printf("Multiple solutions.\n");
		}
	}
	//system("pause");
	return 0;
}