POJ 3185 The Water bowls Gaussian elimination + Enumeration free variable _ state compression

The Water bowls

Time Limit: 1000MS		Memory Limit: 65536K
Total submissions: 5085		accepted: 1991

Description the cows have a line of water bowls from which they. The bowls can is either right-side-up (properly oriented to serve refreshing cool water) or upside-down (a position which Holds no water). They want all water bowls to is right-side-up and thus use their wide to flip snouts.

Their snouts, though, are so wide this they flip not only one bowl but also the bowls on either side of this bowl (a total of three or--in the case of either end bowl--two bowls).

Given the initial state of the bowls (1=undrinkable, 0=drinkable--it even looks like a bowl), and what is the minimum number of bowl flips necessary to turn all bowls right-side-up?

Input Line 1: "A single" with space-separated integers

Output Line 1:the minimum number of bowl flips necessary to flip all bowls right-side-up (i.e., to 0). For the inputs given, it'll always be possible to find some combination of flips of that would manipulate the bowls to 20 0 ' S.

Sample Input

0 0 1 1 1 0 0 1 1 0 1 1 0 0 0 0 0 0 0-0

Sample Output

3

Hint Explanation of the sample:

Flip Bowls 4, 9, and them all drinkable:
0 0 1 1 1 0 0 1 1 0 1 1 0 0 0 0 0 0 0 [Initial State]
0 0 0 0 0 0 0 1 1 0 1 1 0 0 0 0 0 0 0 0 [after flipping Bowl 4]
0 0 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 [after flipping Bowl 9]

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 [after flipping Bowl 11]

20 integers of 0 or 1, known to change the number of position I, will change the number of i-1 and i+1 positions at the same time, changing the effect--0->1 or 1->0. Ask you the least action to make 20 integers all 0.

Train of thought: 20 equations, 20 variable--Gaussian elimination + makeup pressure enumeration free variable can be.

AC Code:

#include <cstdio> #include <cstring> #include <algorithm> #define MAXN #define INF 0x3f3f3f3f using n
Amespace std;
int A[MAXN][MAXN];
int FREE_REC[MAXN];
int X[MAXN];
int Equ, VAR;
    void Init_a () {equ = var = 20;
    memset (A, 0, sizeof (a));
        for (int i = 1; i < i++) {scanf ("%d", &a[i][var]), A[i][var] ^= 0;
        A[i][i] = 1;
        A[I][I-1] = 1;
    if (I <) a[i][i+1] = 1;
    an int Gauss () {int max_r, K;
    int col = 0;
    int num = 0;
        for (k = 0; k < equ && Col < var; k++, col++) {max_r = k;
        for (int i = k+1 i < equ i++) if (A[i][col] > a[max_r][i]) max_r = i;
        if (Max_r!= k) for (int i = col; i < var+1; i++) Swap (a[k][i), a[max_r][i]);
            if (a[k][col] = = 0) {k--;
            free_rec[num++] = col;
        Continue for (int i = k+1 I < EQu i++) if (A[i][col]!= 0) for (int j = col; J < Var+1; J +) A[i][j] ^= a[k
    ][J];
    for (int i = k+1 i < equ i++) if (A[i][col)!= 0) return-1;
    if (Var > k) return var-k;
return 0;
    } void Solve (int S) {int state = (1<<s);
    int ans = INF;
        for (int i = 0; i < state; i++) {int cnt = 0;
                for (int j = 0; J < S; J +) {if ((1<<j) & i) {cnt++;
            X[FREE_REC[J]] = 1;
        else X[free_rec[j]] = 0;
            for (int j = var-s-1 J >= 0; j--) {int temp = A[j][var];
            for (int l = j+1 L < var; l++) if (a[j][l]) temp ^= x[l];
            X[J] = temp; CNT + + X[j]?
        1:0;
    ans = min (ans, CNT);
printf ("%d\n", ans); int main () {WHIle (scanf ("%d", &a[0][20])!= EOF) {a[0][0] = 1, a[0][1] = 1, a[0][20] ^= 0;
        Init_a ();
    Solve (Gauss ());
return 0;
 }