POJ 3185 the water bowls

The water bowls

Test instructions: given 20 01 strings (The final state), each point changes to affect the left and right points, the question is finally 20 0 required minimum number of operations?

Water problem. Modify the augmented matrix directly; it seems that the optimal solution is not Gaussian elimination (if there is a Gauss 0ms a message ~ ~ ~), a lot of 0ms, I used 32ms;

#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))#defineINF 0x3f3f3f3fintdir[2][4] = {{0,1,0,-1},{1,0,-1,0}};inta[ at][ at];intEquvar;intx[ at],free_var[ -];voidDebug () {Puts ("********"); inti,j; Rep0 (i,0, Equ) {Rep1 (J,0,var) cout<<a[i][j]<<" "; cout<<Endl; }puts ("********");}intGauss () {inti,j,k,row,col,cnt =0;  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 .free_var[cnt++] = col;//recording the marking of free variable elements;            Continue; }        if(MX! =row) Rep1 (K,col,var) Swap (a[row][k],a[mx][k]); Rep0 (J,row+1, Equ) {            if(A[j][col]) {REP1 (K,col,var) A[j][k]^=A[row][k]; }        }    }    //debug (); //rep0 (i,row,equ)//if (A[i][var]! = 0) return-1; //No solution//The free variable is enumerated, row represents a useful equation number equation, but we can say that there are many groups of solutions under the premise of judging the solution;//if (Row < Var) return var-row; //returns the number of free arguments directly when no enumeration is required    intAns = Inf,tot =1<< (var-row); Rep0 (i,0, tot) {        intCNT =0, TMP =i; Rep0 (J,0,var-row) {X[free_var[j]]= (tmp&1); if(X[free_var[j]]) cnt++;//**TMP >>=1; } rep_1 (I,row-1,0) {X[i]= a[i][var];//now assigned to A[i][var], if the free variable is still equal to 0, will not be repeated calculation;Rep0 (j,i+1, Equ) {X[i]^= (A[i][j] && x[j]);//The J Lantern will affect the first lamp, and the first J lamp is also lit            }            if(X[i]) cnt++; } ans=min (ans,cnt); }    returnans;}voidInitintN) {Rep0 (i,0, N) {        intID =i; A[id][id]=1; if(Id >0) a[id-1][id] =1; if(ID <var-1) a[id+1][id] =1; }}inttmp[ -];intMain () {intn = -, I,id =0; Equ=var=N; Rep0 (i,0, -) {scanf ("%d", &a[i][var]);    } init (n); //debug ();printf"%d\n", Gauss ()); return 0;}

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POJ 3185 the water bowls