POJ 2947 Widget Factory（高斯消元）

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).

 

题目大意：有n个装饰品，每个装饰品要生产3~9天。给出m种作业，每个作业生产k种装饰品，从星期X生产到星期Y（未必是同一个星期，一天只能生产一个产品），然后给出这k种装饰品分别是什么。问是否能求出n个装饰品分别须要多少天来生产，若有多组解输出Multiple solutions.，无解输出Inconsistent data.。

思路：可以列出m个方程组成方程组。对于每一个作业，设ki为生产装饰品 i 多少次（用输入数据统计一下就好），xi为生产装饰品 i 需要多少天，那么可以列出方程

k1 * x1 + k2 * x2 + …… + kn * xn = Y - X + 1(mod 7)。

得出方程组后，用高斯消元解，其中需要保证每个系数的范围都在0~6中，除法可以对7求逆元。

消掉下三角之后，若存在k1, k2, …… , kn = 0而右边系数不为0的方程，则无解。

若矩阵的秩rank < n则方程有多组解。

否则算出每一个xi，若xi < 3则 xi += 7

PS：白书训练指南上的模板居然有BUG！

 

代码（2360MS）：

  1 #include <iostream>

  2 #include <cstdio>

  3 #include <algorithm>

  4 #include <cstring>

  5 #include <string>

  6 #include <map>

  7 using namespace std;

  8 

  9 int inv[7];

 10 

 11 struct MOD7 {

 12     int val;

 13     MOD7() {}

 14     MOD7(int val): val((val + 7) % 7) {}

 15     MOD7 operator + (const MOD7 &rhs) const {

 16         return MOD7(val + rhs.val);

 17     }

 18     MOD7 operator - (const MOD7 &rhs) const {

 19         return MOD7(val - rhs.val);

 20     }

 21     MOD7 operator * (const MOD7 &rhs) const {

 22         return MOD7(val * rhs.val);

 23     }

 24     MOD7 operator / (const MOD7 &rhs) const {

 25         return MOD7(val * inv[rhs.val]);

 26     }

 27     void operator -= (const MOD7 &rhs) {

 28         val = (val - rhs.val + 7) % 7;

 29     }

 30     bool isZero() {

 31         return val == 0;

 32     }

 33     void inc() {

 34         val = (val + 1) % 7;

 35     }

 36     void print() {

 37         if(val < 3) printf("%d", val + 7);

 38         else printf("%d", val);

 39     }

 40 };

 41 

 42 map<string, int> mymap;

 43 

 44 void init() {

 45     for(int i = 1; i < 7; ++i) {

 46         inv[i] = 1;

 47         for(int j = 1; j < 6; ++j) inv[i] = (inv[i] * i) % 7;

 48     }

 49     mymap["MON"] = 1;

 50     mymap["TUE"] = 2;

 51     mymap["WED"] = 3;

 52     mymap["THU"] = 4;

 53     mymap["FRI"] = 5;

 54     mymap["SAT"] = 6;

 55     mymap["SUN"] = 0;

 56 }

 57 

 58 const int MAXN = 310;

 59 

 60 MOD7 mat[MAXN][MAXN];

 61 int n, m;

 62 

 63 int guess_eliminatioin() {

 64     int rank = 0;

 65     for(int i = 0, t = 0; i < m && t < n; ++i, ++t) {

 66         int r = i;

 67         for(int j = i + 1; j < m; ++j)

 68             if(mat[r][t].val < mat[j][t].val) r = j;

 69         if(mat[r][t].isZero()) { --i; continue;}

 70         else ++rank;

 71         if(r != i) for(int j = 0; j <= n; ++j) swap(mat[i][j], mat[r][j]);

 72         for(int j = n; j >= t; --j)

 73             for(int k = i + 1; k < m; ++k) mat[k][j] -= mat[i][j] * mat[k][t] / mat[i][t];

 74     }

 75     for(int i = rank; i < m; ++i)

 76         if(!mat[i][n].isZero()) return -1;

 77     if(rank < n) return 0;

 78     for(int i = n - 1; i >= 0; --i) {

 79         for(int j = i + 1; j < n; ++j)

 80             mat[i][n] -= mat[j][n] * mat[i][j];

 81         mat[i][n] = mat[i][n] / mat[i][i];

 82     }

 83     return 1;

 84 }

 85 

 86 int main() {

 87     init();

 88     while(scanf("%d%d", &n, &m) != EOF) {

 89         if(n == 0 && m == 0) break;

 90         memset(mat, 0, sizeof(mat));

 91         int t, p;

 92         string s1, s2;

 93         for(int i = 0; i < m; ++i) {

 94             cin>>t>>s1>>s2;

 95             while(t--) {

 96                 scanf("%d", &p);

 97                 mat[i][p - 1].inc();

 98             }

 99             mat[i][n] = MOD7(mymap[s2] - mymap[s1] + 8);

100         }

101         int result = guess_eliminatioin();

102         if(result == -1) puts("Inconsistent data.");

103         else if(result == 0) puts("Multiple solutions.");

104         else {

105             for(int i = 0; i < n - 1; ++i) {

106                 mat[i][n].print();

107                 putchar(' ');

108             }

109             mat[n - 1][n].print();

110             puts("");

111         }

112     }

113 }

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