POJ 2947 Widget Factory(高斯消元)

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题目连接～～～
　　本题是对同余方程组消元，方程比较好列，本题可以列m个方程，每个方程可以列成k1*x1 + k2*x2……kn*xn = y1 (mod 7)
y1是加工这一批零件所需要的时间。然后解方程就可以了。

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <cmath>#include <ctime>#include <algorithm>using namespace std ;#define INF 0x3f3f3f3fconst double esp = 0.00000001 ;const int MX = 300 + 10 ;const int MY = 100 + 10 ;int g[MX][MX] ,ans[MX] ;int Solve(char s[]){    int f_num = 0 ;    if(strcmp(s ,"MON") == 0) f_num = 1 ;    else if(strcmp(s ,"TUE") == 0) f_num = 2 ;    else if(strcmp(s ,"WED") == 0) f_num = 3 ;    else if(strcmp(s ,"THU") == 0) f_num = 4 ;    else if(strcmp(s ,"FRI") == 0) f_num = 5 ;    else if(strcmp(s ,"SAT") == 0) f_num = 6 ;    else if(strcmp(s ,"SUN") == 0) f_num = 7 ;    return f_num ;}int Gcd(int a ,int b){    int tmp ;    while(b){        tmp = a ;        a = b ;        b = tmp % b ;    }    return a ;}int Lcm(int a ,int b){    return a/Gcd(a ,b)*b ;}int Gauss(int n ,int m ,int g[][MX] ,int mod){//n行 m列    int max_r ;    int k = 0 ,col = 0 ;    for( ;k < n && col < m ; ++k ,++col){        max_r = k ;        for(int i = k+1 ;i < n ; ++i)            if(abs(g[max_r][col])  < abs(g[i][col]))                max_r = i ;        if(!g[max_r][col]){            k-- ; continue ;        }        if(max_r != k){            for(int i = col ;i <= m ; ++i)                swap(g[k][i] ,g[max_r][i]) ;        }        for(int i = k + 1 ;i < n ; ++i){            if(g[i][col]){                int LCM = Lcm(abs(g[k][col]) ,abs(g[i][col])) ;                int ta = LCM/abs(g[i][col]) ;                int tb = LCM/abs(g[k][col]) ;                if(g[i][col] * g[k][col] < 0) tb = -tb ;                for(int j = col ;j <= m ; ++j)                    g[i][j] = ((g[i][j]*ta - g[k][j]*tb)%mod +mod)%mod ;            }        }    }    for(int i = k ;i < n ; ++i)        if(g[i][col]) return -1 ;    if(k < m) return m-k ;    for(int i = n-1 ;i >= 0 ; --i){        int tmp = g[i][m] ;        for(int j = i+1 ;j < m ; ++j){            tmp -= g[i][j]*ans[j] ;            tmp = (tmp%mod + mod)%mod ;        }        int sum ;        for(int t = 3 ;t <= 9 ; ++t){            sum = ((g[i][i]*t)%mod + mod)%mod ;            if(sum == tmp){                ans[i] = t ;                break ;            }        }    }    return 0 ;}int main(){    //freopen("input.txt" ,"r" ,stdin) ;    char s1[MX] ,s2[MX] ;    int n ,m ,mod = 7 ,num ,tmp ;    while(~scanf("%d%d" ,&n ,&m)){        if(!n && !m) break ;        memset(g ,0 ,sizeof(g)) ;        for(int i = 0 ;i < m ; ++i){            scanf("%d%s%s" ,&num ,s1 ,s2) ;            int d1 = Solve(s1) ;            int d2 = Solve(s2) ;            g[i][n] = ((d2 - d1 + mod)%mod + 1)%mod ;            for(int j = 0 ;j < num ; ++j){                scanf("%d" ,&tmp) ;                g[i][tmp-1] = (g[i][tmp-1]+1)%mod ;            }        }        memset(ans ,0 ,sizeof(ans)) ;        int f_num = Gauss(m ,n ,g ,mod) ;        if(!f_num) {            for(int i = 0 ;i < n ; ++i)                if(!i) printf("%d" ,ans[i]) ;                else printf(" %d" ,ans[i]) ;            puts("") ;        }        else if(f_num == -1) puts("Inconsistent data.") ;        else puts("Multiple solutions.") ;    }    return 0 ;}

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