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  • POJ 1681 高斯消元

    题目连接http://poj.org/problem?id=1681

    基本上同上一题。求解一的个数。

    #include <cstdio>
    #include <cstdlib>
    #include <cstring>
    #include <cmath>
    #include <cctype>
    #include <vector>
    #include <stack>
    #include <queue>
    #include <map>
    #include <algorithm>
    #include <iostream>
    #include <string>
    #include <set>
    #define X first
    #define Y second
    #define sqr(x) (x)*(x)
    #pragma comment(linker,"/STACK:102400000,102400000")
    using namespace std;
    const double PI = acos(-1.0);
    map<int, int>::iterator it;
    typedef long long LL ;
    template<typename T> void checkmin(T &x, T y) {x = min(x, y);}
    template<typename T> void checkmax(T &x, T y) {x = max(x, y);}
    
    const double eps = 1e-7;
    const int N = 255;
    int u[N], n;
    bool l[N];
    int a[N][N], as[N];
    int gauss() {
        int i, j, k, r = 0;
        double tmp;
        for(i = 0; i < n; i++)
            l[i] = 0;
        for(i = 0; i < n; i++) {
            for(j = r; j < n; j++)
                if(a[j][i]) {
                    for(k = i; k <= n; k++)
                        swap(a[j][k], a[r][k]);
                    break;
                }
            if(a[r][i] == 0)continue;
            for(j = 0; j < n; j++)
                if(j != r && a[j][i]) {
                    for(k = i; k <= n; k++)
                        a[j][k] ^= a[r][k];
                }
            l[i] = 1; r++;
        }
        for(i = 0; i < n; i++) {
            if(l[i]) {
                for(j = 0; j < n; j++)
                    if(a[j][i])
                        as[i] = a[j][n];
            }
        }
        for(i = r; i < n; i++)if(a[i][n])return -1;
        int cnt = 0;
        for(int i = 0; i < n; ++i)cnt += as[i];
        return cnt;
    }
    void pf() {
        for(int i = 0; i < n; ++i) {
            for(int j = 0; j <= n; ++j) {
                printf("%d ", a[i][j]);
            } puts("");
        }
    }
    char s[22];
    
    int dx[] = {0, 1, 0, -1};
    int dy[] = {1, 0, -1, 0};
    int b[22][22];
    int m;
    bool in(int x, int y) {
        return x >= 0 && x < m && y >= 0 && y < m;
    }
    int main() {
        int T;
        scanf("%d", &T);
        while(T--) {
            scanf("%d", &m);
            n = m * m;
            memset(a, 0, sizeof(a));
            for(int i = 0; i < m; ++i) {
                scanf("%s", s);
                for(int j = 0; j < m; ++j) {
                    a[i*m+j][n] = ((s[j] == 'w') ? 1 : 0);
                }
            }
            for(int i = 0; i < m; ++i) {
                for(int j = 0; j < m; ++j) {
                    int p = i * m + j;
                    int q = p;
                    a[p][p] = 1;
                    for(int k = 0; k < 4; ++k) {
                        int nx = dx[k] + i;
                        int ny = dy[k] + j;
                        if(in(nx, ny)) {
                            p = nx * m + ny;
                            a[p][q] = 1;
                        }
                    }
                }
            }
            int res = gauss();
            if(~res) {
                printf("%d
    ", res);
            }
            else {
                printf("inf
    ");
            }
        }
        return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/cxw199204/p/3329441.html
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