POJ 2947 Widget Factory （高斯消元 判多解 无解 和解集 模7情况）

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <cstdlib>
  5 #include <cmath>
  6 #include <algorithm>
  7 #define LL __int64
  8 const int maxn = 300+10;
  9 const int INF = 1<<28;
 10 using namespace std;
 11 int equ, var, fn;
 12 int a[maxn][maxn], x[maxn];
 13 bool free_x[maxn];
 14 
 15 int gcd(int a, int b)
 16 {
 17     return b==0?a:gcd(b, a%b);
 18 }
 19 int lcm(int a, int b)
 20 {
 21     return a*b/gcd(a, b);
 22 }
 23 int Gauss()
 24 {
 25     int x_mo;
 26     x_mo = 7;
 27     int i, j, k, max_r, col;
 28     int ta, tb, LCM, tmp, fx_num;
 29     int free_index;
 30     col = 0;
 31 
 32     for(k = 0; k<equ && col<var; k++, col++)
 33     {
 34         max_r = k;
 35         for(i = k+1; i < equ; i++)
 36             if(abs(a[i][col])>abs(a[max_r][col]))
 37                 max_r = i;
 38 
 39         if(max_r != k)
 40             for(j = k; j < var+1; j++)
 41                 swap(a[k][j], a[max_r][j]);
 42 
 43         if(a[k][col]==0)
 44         {
 45             k--;
 46             continue;
 47         }
 48         for(i = k+1; i < equ; i++)
 49         {
 50             if(a[i][col] != 0)
 51             {
 52                 LCM = lcm(abs(a[i][col]), abs(a[k][col]));
 53                 ta = LCM/abs(a[i][col]);
 54                 tb= LCM/abs(a[k][col]);
 55                 if(a[i][col]*a[k][col] < 0) tb = -tb;
 56 
 57                 for(j = col; j < var+1; j++)
 58                     a[i][j] = ((a[i][j]*ta - a[k][j]*tb)%x_mo+x_mo)%x_mo;
 59             }
 60         }
 61     }
 62     for(i = k; i < equ; i++)
 63         if(a[i][col] != 0)
 64             return -1;
 65 
 66     if(k < var)
 67         return var-k;
 68 
 69     for(i = var-1; i >= 0; i--)
 70     {
 71         tmp = a[i][var];
 72         for(j = i+1; j < var; j++)
 73             if(a[i][j] != 0)
 74                 tmp = ((tmp-a[i][j]*x[j])%x_mo+x_mo)%x_mo;
 75 
 76         if(a[i][i]==0)
 77             x[i] = 0;
 78         else
 79         {
 80             //if(tmp%a[i][i] != 0) return -2;
 81             while(tmp%a[i][i]!=0) tmp += x_mo;
 82             x[i] = (tmp/a[i][i])%x_mo;
 83         }
 84     }
 85     return 0;
 86 }
 87 int check(char s[])
 88 {
 89     if(strcmp(s,"MON")==0) return 1;
 90     else if(strcmp(s,"TUE")==0) return 2;
 91     else if(strcmp(s,"WED")==0) return 3;
 92     else if(strcmp(s,"THU")==0) return 4;
 93     else if(strcmp(s,"FRI")==0) return 5;
 94     else if(strcmp(s,"SAT")==0) return 6;
 95     else  return 7;
 96 }
 97 int main()
 98 {
 99     int n, m, i, k;
100     char s1[20], s2[20];
101     while(~scanf("%d%d", &n, &m))
102     {
103         equ = m; var = n;  //m个方程，n个未知数
104         if(n==0&&m==0) break;
105         memset(a, 0, sizeof(a));
106         memset(x, 0, sizeof(x));
107         memset(free_x, -1, sizeof(free_x));
108         for(i = 0; i < m; i++)
109         {
110             scanf("%d", &k);
111             getchar();
112             scanf("%s %s", s1, s2);
113             a[i][n] = ((check(s2)-check(s1)+1)%7+7)%7; //s1s2顺序不能乱，而且因为有负的所以需要这样处理
114             while(k--)
115             {
116                 int tmp;
117                 scanf("%d", &tmp);
118                 tmp --;  //因为a里面是从0开始的
119                 a[i][tmp] ++;
120                 a[i][tmp] %= 7;   //一定要对7取余，不然会wa,大概因为上面a[i][n]取余了，如果这里不取余会使方程无解吧。
121             }
122         }
123         fn = Gauss();
124         if(fn<0)
125         printf("Inconsistent data.\n");
126         if(fn>0)
127         printf("Multiple solutions.\n");
128         else if(fn==0)
129         {
130             for(i = 0; i < n; i++)
131             {
132                 if(x[i]<=2) x[i] += 7; //注意题目说3——9天
133                 if(i==n-1)
134                 printf("%d\n", x[i]);
135                 else
136                 printf("%d ", x[i]);
137             }
138         }
139     }
140     return 0;
141 }