http://poj.org/problem?id=2947
各种逗啊。。还好1a了。。
题意我就不说了，百度一大把。
转换为mod的方程组，即
(x[1,1]*a[1])+(x[1,2]*a[2])+...+(x[1,n]*a[n])=x[1, n+1] (mod m)
(x[2,1]*a[1])+(x[2,2]*a[2])+...+(x[2,n]*a[n])=x[2, n+1] (mod m)
...
(x[n,1]*a[1])+(x[n,2]*a[2])+...+(x[n,n]*a[n])=x[n, n+1] (mod m)
没有mod就是裸的高斯消元。。。
我们来考虑怎么消元。
显然如果有方程1和方程2，他们都有相同系数不为0的元素y，那么我们消元只需要将他们的系数调成一样即可，即调成公倍数，那么因为mod意义下满足a=b(mod m), a*c=b*c(mod m)，所以将两个方程左式和右式都乘上这个公倍数即可。
而回代麻烦一些，减完当前方程其它元素的对应值后，此时假设当前元素的系数为A[i][i]，而值是A[i][n+1]，那么因为有x*A[i][i]=A[i][n+1](mod m)，我们可以用拓展欧几里得或者一直加m，知道出现成立（此时因为一定有解，所以一定找得到一个x使得等式成立）
那么得到的x就是答案。
做的时候注意n和m是方程数还是未知量的数就行了。。
#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
#define rep(i, n) for(int i=0; i<(n); ++i)
#define for1(i,a,n) for(int i=(a);i<=(n);++i)
#define for2(i,a,n) for(int i=(a);i<(n);++i)
#define for3(i,a,n) for(int i=(a);i>=(n);--i)
#define for4(i,a,n) for(int i=(a);i>(n);--i)
#define CC(i,a) memset(i,a,sizeof(i))
#define read(a) a=getint()
#define print(a) printf("%d", a)
#define dbg(x) cout << (#x) << " = " << (x) << endl
#define printarr2(a, b, c) for1(_, 1, b) { for1(__, 1, c) cout << a[_][__]; cout << endl; }
#define printarr1(a, b) for1(_, 1, b) cout << a[_] << ‘\t‘; cout << endl
inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<‘0‘||c>‘9‘; c=getchar()) if(c==‘-‘) k=-1; for(; c>=‘0‘&&c<=‘9‘; c=getchar()) r=r*10+c-‘0‘; return k*r; }
inline const int max(const int &a, const int &b) { return a>b?a:b; }
inline const int min(const int &a, const int &b) { return a<b?a:b; }
const int N=305;
typedef int mtx[N][N];
string Dt[7]={"MON", "TUE", "WED", "THU", "FRI", "SAT", "SUN"};
int gauss(mtx A, int n, int m, int MD) {
int x=1, y=1, pos;
while(x<=n && y<=m) {
pos=x;
while(!A[pos][y] && pos<=n) ++pos;
if(A[pos][y]) {
for1(i, 1, m+1) swap(A[pos][i], A[x][i]);
for1(i, x+1, n) if(A[i][y]) {
int l=A[x][y], r=A[i][y];
for1(j, y, m+1) A[i][j]=((A[i][j]*l-A[x][j]*r)%MD+MD)%MD;
}
++x;
}
++y;
}
for1(i, x, n) if(A[i][m+1]) return -1;
if(x<=m) return m-x+1;
for3(i, m, 1) {
for1(j, i+1, m) if(A[i][j]) A[i][m+1]=((A[i][m+1]-(A[j][m+1]*A[i][j]))%MD+MD)%MD;
while(A[i][m+1]%A[i][i]!=0) A[i][m+1]+=MD; //这里可以用拓欧搞掉。。
A[i][m+1]=(A[i][m+1]/A[i][i])%MD;
}
return 0;
}
inline int get(string s) { return find(Dt, Dt+7, s)-Dt+1; }
int main() {
int n=getint(), m=getint(), t;
char s[2][5];
mtx a;
while(n|m) {
CC(a, 0);
for1(i, 1, m) {
read(t); scanf("%s%s", s[0], s[1]);
a[i][n+1]=(get(s[1])-get(s[0])+1+7)%7;
for1(j, 1, t) ++a[i][getint()];
for1(j, 1, n) a[i][j]%=7;
}
int ans=gauss(a, m, n, 7);
if(ans==-1) puts("Inconsistent data.");
else if(ans) puts("Multiple solutions.");
else {
for1(i, 1, n) if(a[i][n+1]<3) a[i][n+1]+=7;
for1(i, 1, n-1) printf("%d ", a[i][n+1]); printf("%d\n", a[n][n+1]);
}
n=getint(), m=getint();
}
return 0;
}
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.
Source