• POJ 1753 Flip Game 高斯消元


    先用高斯消元化简,然后枚举自由变元

    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <queue>
    #include <stack>
    #include <map>
    #include <set>
    #include <climits>
    #include <iostream>
    #include <string>
    
    using namespace std;
     
    #define MP make_pair
    #define PB push_back
    typedef long long LL;
    typedef unsigned long long ULL;
    typedef vector<int> VI;
    typedef pair<int, int> PII;
    typedef pair<double, double> PDD;
    const int INF = INT_MAX / 3;
    const double eps = 1e-8;
    const LL LINF = 1e17;
    const double DINF = 1e60;
    const int maxn = 250;
    const int dx[] = {0, 0, 1, -1};
    const int dy[] = {1, -1, 0, 0};
    int a[maxn][maxn], n;
    bool fx[maxn];
    char buf[16][16];
    
    void Gauss() {
        for(int i = 0; i < n * n; i++) {
            int k = i;
            while(a[k][i] == 0 && k < n * n) k++;
            if(k >= n * n) break;
            for(int j = 0; j <= n * n; j++) swap(a[i][j], a[k][j]);
            for(int j = 0; j < n * n; j++) if(j != i && a[j][i] != 0) {
                for(int k = 0; k <= n * n; k++) {
                    a[j][k] ^= a[i][k];
                }
            }
        }
    }
    
    void pp() {
        for(int i = 0; i < n * n; i++) {
            for(int j = 0; j <= n * n; j++) {
                printf("%d ", a[i][j]);
            }
            puts("");
        }
    }
    
    int solve() {
        memset(a, 0, sizeof(a));
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < n; j++) {
                if(buf[i][j] == 'w') a[i * n + j][n * n] = 1;
                else a[i * n + j][n * n] = 0;
            }
        }
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < n; j++) {
                int u = i * n + j;
                a[u][u] = 1;
                for(int k = 0; k < 4; k++) {
                    int nx = i + dx[k], ny = j + dy[k], nu = nx * n + ny;
                    if(nx >= 0 && nx < n && ny >= 0 && ny < n) {
                        a[u][nu] = a[nu][u] = 1;
                    }
                }
            }
        }
        Gauss();
        memset(fx, 0, sizeof(fx));
        int ans = INF, fcnt = 0;
        bool bad = false;
        for(int i = 0; i < n * n; i++) {
            int nsum = 0;
            for(int j = 0; j  < n * n; j++) nsum += a[i][j];
            if(nsum == 0 && a[i][n * n] == 1) bad = true;
            if(nsum == 0) fx[i] = true, fcnt++;
        }
        if(bad) return INF;
        for(int s = 0; s < (1 << fcnt); s++) {
            int nowans = 0;
            for(int i = 0; i < n * n - fcnt; i++) {
                int col = 0;
                for(int j = 0; j < n * n; j++) if(fx[j] && a[i][j]) {
                    if(s & (1 << (n * n - j - 1))) col ^= 1;
                }
                if(a[i][n * n] != col) nowans++; 
            }
            int bcnt = 0;
            for(int i = 0; i < fcnt; i++) if(s & (1 << i)) bcnt++;
            ans = min(nowans + bcnt, ans);
        }
        return ans;
    }
    
    int main() {
        while(scanf("%s%s%s%s", buf[0], buf[1], buf[2], buf[3]) != EOF) {
            n = 4;
            int ans = solve();
            for(int i = 0; i < 4; i++) {
                for(int j = 0; j < 4; j++) {
                    if(buf[i][j] == 'w') buf[i][j] = 'b';
                    else buf[i][j] = 'w';
                }
            }
            ans = min(ans, solve());
            if(ans < INF) printf("%d
    ", ans);
            else puts("Impossible");
        }
        return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/rolight/p/4066466.html
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