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;
}