HDU 3359 Kind of a Blur(高斯消元)

题意：

       H * W (W,H <= 10) 的矩阵A的某个元素A[i][j]，从它出发到其他点的曼哈顿距离小于等于D的所有值的和S[i][j]除上可达点的数目，构成了矩阵B。给定矩阵B，求矩阵A。

 

 

题目先给宽再给高。。。坑我了一个小时

 

code

/*

       暴力确定每个位置有到那些位置的曼哈顿距离小于D

       然后对你n*m个未知数，n*m个方程进行高斯消元

*/

#include <iostream>

#include <cstdio>

#include <cmath>

#include <cstring>

using namespace std;



const int MAXN = 111;

int n, m, dx, cnt;

double A[MAXN][MAXN], ans[MAXN];

inline void build (int x, int y, int d) {

    int k = 0;

    for (int i = 0, tol = 0; i < n; ++i)

        for (int j = 0; j < m; ++j) {

            if (abs (i - x) + abs (j - y) <= dx)

                A[d][tol++] = 1, ++k;

            else

                A[d][tol++] = 0;

        }

    A[d][cnt] *= k;

}

void Gauss() {

    int i, j, k;

    double tmp, big;

    for (i = 0; i < cnt; i++) {

        for (big = 0, j = i; j < cnt; j++) {

            if (abs (A[j][i]) > big) {

                big = abs (A[j][i]);

                k = j;

            }

        }

        if (k != i) {

            for (j = 0; j <= cnt; j++)

                swap (A[i][j], A[k][j]);

        }

        for (j = i + 1; j < cnt; j++) {

            if (A[j][i]) {

                tmp = -A[j][i] / A[i][i];

                for (k = i; k <= cnt; k++)

                    A[j][k] += tmp * A[i][k];

            }

        }

    }

    for (i = cnt - 1; i >= 0; i--) {

        tmp = 0;

        for (j = i + 1; j < cnt; j++)

            tmp += A[i][j] * ans[j];

        ans[i] = (A[i][j] - tmp) / A[i][i];

    }

}

int main() {

    int cs = 0;

    while (scanf ("%d %d %d", &m, &n, &dx), n) {

        if (cs == 0) cs = 1;

        else

            putchar (10);

        cnt = n * m;

        for (int i = 0, tol = 0; i < n; ++i)

            for (int j = 0; j < m; ++j)

                scanf ("%lf", &A[tol++][cnt]);

        for (int i = 0, tol = 0; i < n; ++i)

            for (int j = 0; j < m; ++j, ++tol)

                build (i, j, tol);

        Gauss ();

        for (int i = 0, cnt = 0; i < n; i++) {

            for (int j = 0; j < m; j++)

                printf ("%8.2f", ans[cnt++]);

            putchar (10);

        }

    }

    return 0;

}

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