pku 2947 Widget Factory 高斯消元

http://poj.org/problem?id=2947

题意：

工人生产不同的部件需要不同的时间，最少需要3天最多需要9天。

给出n种部件和m条记录，每条记录包括该工人生产部件的总数k和他开始生产时间和结束时间（只给出是周几，不给出具体时间），之后给出这K个部件跟的所属的种类。最后求出生产这n种部件分别所需要的时间。

思路：

高斯消元：

例子来讲

首先我们能够列出方程

(x + y)%7 = 4;

(2*x + y)%7 = 5;

(x + 2*y)%7 = 0;

这里直接套高斯消元模板就行，关键是处理%7的情况，我们先按[0,6]计算结果最后再处理到[3,9];

View Code

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

#include <queue>

#include <stack>

#include <set>

#include <map>

#include <string>



#define CL(a,num) memset((a),(num),sizeof(a))

#define iabs(x)  ((x) > 0 ? (x) : -(x))

#define Min(a,b) (a) > (b)? (b):(a)

#define Max(a,b) (a) > (b)? (a):(b)



#define ll long long

#define inf 0x7f7f7f7f

#define MOD 100000007

#define lc l,m,rt<<1

#define rc m + 1,r,rt<<1|1

#define pi acos(-1.0)

#define test puts("<------------------->")

#define maxn 100007

#define M 100007

#define N 307

using namespace std;

//freopen("din.txt","r",stdin);



int a[N][N],X[N];

int equ,var;



int find(char *s){

    if (!strcmp(s,"MON")) return 1;

    if (!strcmp(s,"TUE")) return 2;

    if (!strcmp(s,"WED")) return 3;

    if (!strcmp(s,"THU")) return 4;

    if (!strcmp(s,"FRI")) return 5;

    if (!strcmp(s,"SAT")) return 6;

     return 7;

}

int cha(char *s1,char *s2){

    return (find(s2) - find(s1) + 1 + 7)%7;

}

int GCD(int x,int y){

    if (y == 0) return x;

    return GCD(y,x%y);

}

int LCM(int x,int y){

    return x/GCD(x,y)*y;

}

void d()

{

    int i,j;

    for (i = 0; i < equ; ++i){

        for (j = 0; j < var + 1; ++j)

        printf("%d ",a[i][j]);

        puts("");

    }

}

int Guass(){

    int i,j,k,col;

    CL(X,0);

    for (k = 0,col = 0; k < equ && col < var; ++k,++col){

        int max_r = k;

        for (i = k + 1; i < equ; ++i){

            if (iabs(a[i][col]) > iabs(a[max_r][col])) max_r = i;

        }

        if (max_r != k){

            for (i = k; i < var + 1; ++i) swap(a[k][i],a[max_r][i]);

        }

        if (a[k][col] == 0){

            k--; continue;

        }

        for (i = k + 1; i < equ; ++i){

            //注意处理不等于0 的

            if (a[i][col] != 0){

                int lcm = LCM(iabs(a[k][col]),iabs(a[i][col]));//注意这里取绝对值

                int ta = lcm/iabs(a[i][col]); int tb = lcm/iabs(a[k][col]);

                if (a[i][col]*a[k][col] < 0) tb = -tb;

                for (j = col; j < var + 1; ++j){

                    a[i][j] = ((ta*a[i][j])%7 - (tb*a[k][j])%7 + 7)%7;

                }

            }

        }

    }

    for (i = k; i < equ; ++i){

        if (a[i][col]%7 != 0) return -1;

    }

    if (k < var) return var - k;

    else {

        for (i = var - 1; i >= 0; --i){

            int tmp = a[i][var];

            for (j = i + 1; j < var; ++j){

                if (a[i][j] != 0) tmp -= a[i][j]*X[j];

            }

            while (tmp%a[i][i] != 0) tmp += 7;

            X[i] = tmp/a[i][i];

            if (X[i] < 3){

                while (X[i] < 3) X[i] += 7;

            }

            else if (X[i] > 9){

                while (X[i] > 9) X[i] -= 7;

            }



        }

    }

    return 0;

}



int main(){

    //freopen("din.txt","r",stdin);

    int i,k;

    char s1[10],s2[10];

    while (~scanf("%d%d",&var,&equ))

    {

        if (!var && !equ) break;

        CL(a,0);

        //构造增广矩阵

        for (i = 0; i < equ; ++i){

            scanf("%d%s%s",&k,s1,s2);

            a[i][var] = cha(s1,s2);

            int type;

            while (k--){

                scanf("%d",&type); type--;

                a[i][type]++;

                a[i][type] %= 7;

            }

        }

        //d();

        int free_num = Guass();



        if (free_num == -1) printf("Inconsistent data.\n");

        else if (free_num > 0) printf("Multiple solutions.\n");

        else{

            for (i = 0; i < var - 1; ++i) printf("%d ",X[i]);

            printf("%d\n",X[i]);

        }

    }

    return 0;

}