POJ 2947 Widget Factory

Widget Factory

Test instructions: There are n pieces of decorations, there is m group information. (1 <= N, m<=) Each group of information has the beginning of the week and the end of the week (is within the MoD 7 range) and also includes the type of num decorations (1~n), of which each ornament takes 3 <= x[i] <= 9; The input of the kind can be repeated;

Ideas:

1. Based on the input to establish the augmented matrix a[][], but in the process of establishing and solving because it is the mod sense , so the number of inputs and the final time to use the MoD 7; (analysis shows that when the number is 7 of the same class, the beginning of the same week is the same as the end of the week)

2. In front of a few Gaussian elimination, I use Free_var to determine whether there is a free variable, which is the number of input equations and the solution of the number of elements equal to the case is established. In the case of sample 1 will find that the number of equations originally more than the variable 1, at this time calculated Free_var = 1, but it is not going to have a dimension of free variable. It depends on the number of useful equations, the relationship between Row and Var;

3. To get the upper triangular array to solve the variable x[i], we need to solve a modal linear equation,a[i][i]*x[i] + 7*y = RET (mod 7);ret for A[i][col] with known x[j] Removal of a[i][i] obtained;

At this point the call EXGCD () can be solved, and finally the solution should be within the 3~9 range;

PS: Time performance is not very good, unexpectedly used 1782ms ... The shortest is 297ms. Gap Ah!!!

#include <iostream>#include<cstdio>#include<cstring>#include<string.h>#include<algorithm>#include<map>#include<queue>#include<vector>#include<cmath>#include<stdlib.h>#include<time.h>using namespacestd;#defineRep0 (I,L,R) for (int i = (l); i < (R); i++)#defineREP1 (I,L,R) for (int i = (l); I <= (r); i++)#defineRep_0 (i,r,l) for (int i = (r); i > (l); i--)#defineRep_1 (i,r,l) for (int i = (r); I >= (l); i--)#defineMS0 (a) memset (A,0,sizeof (a))#defineMS1 (a) memset (A,-1,sizeof (a))inta[305][305];intEquvar;intx[305];Const intMOD =7;voidDebug () {Puts ("********"); inti,j; Rep0 (i,0, Equ) {Rep1 (J,0,var) cout<<a[i][j]<<" "; cout<<Endl; }puts ("********");}int__GCD (intAintb) {    returnB?__GCD (b,a%b): A;}intLCM (intAintb) {    returnA/__GCD (A, b) *b;}voidEXGCD (intAintBint& D,int& X,int&y) {    if(!B) {d = a;x =1; y =0;} Else{EXGCD (b,a%b,d,y,x); Y-= x* (A/b); }}intGuass () {intI,j,k,free_var =0, Row,col;  for(row =0, col =0; row < equ && Col <var; row++,col++){        intMX =Row; Rep0 (J,row+1, equ)if(ABS (A[j][col]) > abs (A[mx][col])) MX =J; if(A[mx][col] = =0) {row--;//The number of free arguments cannot be recorded here, only the useless col can be recorded .            Continue; }        if(MX! =row) Rep1 (K,col,var) Swap (a[row][k],a[mx][k]); Rep0 (J,row+1, Equ) {            if(A[j][col]) {intLCM =LCM (ABS (A[row][col]), ABS (A[j][col)); intRation_row = Lcm/a[row][col],ration_j = lcm/A[j][col]; if(A[row][col]*a[j][col] <0) Ration_row =-ration_row;//The symbolic opposite becomes addition;REP1 (K,col,var) A[j][k]= ((a[j][k]*ration_j-a[row][k]*ration_row)%7+7)%7; }        }    }    //debug ();rep0 (i,row,equ)if(a[i][var] !=0)return-1;//No solution    if(Row <var)return var-row;//row represents a useful equation for the number of equations, but it is necessary to judge the premise of a solution to say that there are many groups of solutions;Rep_1 (I,var-1,0){//* * * If the only solution, in fact, is the Var matrix        intRET = a[i][var];  for(j = i+1; J <var; j + +)//using the obtained variable to eliminate the element behind Col in row rows, get Yi Yuanfang;RET-= x[j]*A[i][j]; RET= ((ret%7)+7)%7; intD,x1,y; //constructs a[i][i]*x[i] + 7*y = RET (mod 7), and gcd (a[row][col],7) = 1) because A[row][col]! = 0EXGCD (A[i][i],7, d,x1,y);//then take the RET to get 3~9 range;X[i] = ((ret*x1)%7+7)%7; if(X[i] <3) X[i] + =7; }    return 0;}Const Charstr[7][5] = {{"MON"},{"TUE"},{"WED"},{"THU"},{"FRI"},{"SAT"},{"SUN"}};intDATE_ID (Char*c) {     for(inti =0; I <7; i++)        if(strcmp (str[i],c) = =0)returni;}intMain () {inti,j,n,m; Chars[5],t[5];  while(SCANF ("%d%d", &n,&m) = =2&& n +m)        {MS0 (a); Equ= m;var=N; intKind,num; Rep0 (i,0, M) {scanf ("%d%s%s",&num,s,t); a[i][var] = date_id (t)-date_id (s) +1; if(a[i][var] <0) a[i][var] +=7; Rep0 (J,0, num) {scanf ("%d",&kind); a[i][--kind]++; } rep1 (J,0,var) A[i][j]%=7; }        //debug ();        intRET =Guass (); if(ret = =-1) puts ("inconsistent data."); Else if(Ret >0) puts ("multiple solutions."); Else{rep0 (i,0,var) printf ("%d%c", X[i],i = =var-1?'\ n':' '); }    }    return 0;}

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POJ 2947 Widget Factory