POJ - 3185 The Water Bowls

模型是个线性模2方程组。高斯消元的话可能解不唯一，要找独立变元。

但是不难看出，枚举左端点按下的状态，之后可以影响端点就只有一个变量了，根据端点递推了，然后检查。（写的时候想的是枚举左右端点

/*********************************************************
*            ------------------                          *
*   author AbyssalFish                                   *
**********************************************************/
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<queue>
#include<vector>
#include<stack>
#include<vector>
#include<map>
#include<set>
#include<algorithm>
#include<cmath>
#include<numeric>
using namespace std;

const int N = 20;
int b[N];


//#define LOCAL
int main()
{
#ifdef LOCAL
    freopen("in.txt","r",stdin);
#endif
    for(int i = 0; i < N; i++) scanf("%d",b+i);
    int best = N;
    for(int S = 4; S--;){
        int x = S&1, y = S>>1;
        b[0] ^= x; b[1] ^= x;
        b[N-2] ^= y; b[N-1] ^= y;
        int v[N+3] = {};
        int rv = 0, ans = x+y;
        for(int i = 0; i < N; i++){
            if(b[i]^(rv^=v[i])){
                if(i < N-2){
                    v[i+3] ^= 1; rv ^= 1;
                    ans++;
                }
                else{
                    ans = best+1; break;
                }
            }
        }
        best = min(best,ans);
        b[0] ^= x; b[1] ^= x;
        b[N-2] ^= y; b[N-1] ^= y;
    }
    printf("%d\n", best);
    return 0;
}