• POJ1681 Painter's Problem 高消


    代码如下:

    #include <cstdlib>
    #include <cstring>
    #include <cstdio>
    #include <algorithm>
    #define TO(x, y) (x-1)*N+y
    using namespace std;
    
    char G[30][30];
    
    int dir[4][2] = {1, 0, -1, 0, 0, 1, 0, -1}, N;
    
    int C;
    
    inline bool judge(int x, int y)
    {
        if (x < 1 || x > N || y < 1 || y > N) {
            return false;
        }   
        return true;
    }
    
    void swap(int &a, int &b) 
    {
        int t = a;
        a = b;
        b = t;
    }
    
    struct Matrix
    {
        int a[300][300];
        void init() {
            int xx, yy, k, pos;
            memset(a, 0, sizeof (a));
            for (int i = 1; i <= N; ++i) {
                for (int j = 1; j <= N; ++j) {
                    k = TO(i, j);
                    a[k][k] = 1;
                    for (int d = 0; d < 4; ++d) {
                        xx = i + dir[d][0], yy = j + dir[d][1];
                        if (judge(xx, yy)) {
                            pos = TO(xx, yy);
                            a[pos][k] = 1;
                        }
                    }
                }
            }
        }
        void rswap(int x, int y, int s) {
            for (int j = s; j <= C+1; ++j) {
                swap(a[x][j], a[y][j]);
            }
        }
        void relax(int x, int y, int s) {
            for (int j = s; j <= C+1; ++j) {
                a[y][j] ^= a[x][j];
            }
        }
    }M;
    
    void solve(int R)
    { 
        int f1 = 0,  x1[300], t1, ans1 = 0x3fffffff;
        for (int i = R + 1; i <= C; ++i) {
            if (M.a[i][C+1]) f1 = 1;
        }
        if (f1) {
            puts("inf");
            return;
        }
        for (int s = 0; s < 1<<(C-R); ++s) {
            t1 = 0;
            memset(x1, 0, sizeof (x1)); 
            for (int i = 0; i < (C-R); ++i) {
                x1[R+i+1] = s & (1 << i) ? 1 : 0;
                if (x1[R+i+1]) ++t1;
            }
            for (int i = R; i >= 1; --i) {
                for (int j = i + 1; j <= C; ++j) {
                     x1[i] ^= (x1[j] * M.a[i][j]); 
                }
                x1[i] ^= M.a[i][C+1];
                if (x1[i]) ++t1;
            }
            ans1 = min(ans1, t1);
        }
        printf("%d\n", ans1);
    }
    
    void Gauss()
    {
        int i = 1, k;
        for (int j = 1; j <= C; ++j) { // 枚举上三角矩阵的对角线
            for (k = i; k <= C; ++k) {
                if (M.a[k][j])  break;
            }
            if (k > C) continue;
            if (k != i) {  // 如果该行就是第首行,那么我们直接进行消元,否则交换行,使得第一行为单位1
                M.rswap(k, i, j);
            }
            for (k = i + 1; k <= C; ++k) {  // 从下一行开始进行消元
                if (M.a[k][j]) {
                    M.relax(i, k, j);
                }
            }
            ++i;
        } 
        solve(i - 1);
    }
    
    int main()
    {
        int T;
        scanf("%d", &T);
        while (T--) {
            scanf("%d", &N);
            C = N*N;
            M.init();
            for (int i = 1; i <= N; ++i) {
                scanf("%s", G[i] + 1);
            }
            for (int i = 1; i <= N; ++i) {
                for (int j = 1; j <= N; ++j) {
                    M.a[TO(i, j)][C+1] = G[i][j] == 'w';
                }
            }
            Gauss();
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/Lyush/p/2610827.html
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