【HDU】3359 Kind of a Blur

题意：给出一个h*w的矩阵，每个值等于与它哈密顿距离不超过d的平均值，求原矩阵。

构造方程组，高斯消元，注意行/列，消元要取最大的。

 1 #include<cstdio>

 2 #include<cstring>

 3 #include<cmath>

 4 #include<algorithm>

 5 #define MAXN 110

 6 using namespace std;

 7 double a[MAXN][MAXN], g[MAXN][MAXN], x[MAXN];

 8 int pos[MAXN][MAXN];

 9 int n, m, d, cnt;

10 int MHT(int x1, int y1, int x2, int y2) {

11     return abs(x1 - x2) + abs(y1 - y2);

12 }

13 void Gauss() {

14     int i, j, k;

15     double tmp, big;

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

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

18             if (abs(g[j][i]) > big) {

19                 big = abs(g[j][i]);

20                 k = j;

21             }

22         }

23         if (k != i) {

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

25                 swap(g[i][j], g[k][j]);

26         }

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

28             if (g[j][i]) {

29                 tmp = -g[j][i] / g[i][i];

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

31                     g[j][k] += tmp * g[i][k];

32             }

33         }

34     }

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

36         tmp = 0;

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

38             tmp += g[i][j] * x[j];

39         x[i] = (g[i][j] - tmp) / g[i][i];

40     }

41 }

42 int main() {

43     int i, j, k, l;

44     bool flag = true;

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

46         if (flag)

47             flag = false;

48         else

49             putchar('\n');

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

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

52                 scanf("%lf", &a[i][j]);

53                 pos[i][j] = cnt++;

54             }

55         }

56         memset(g, 0, sizeof(g));

57         for (i = 0; i < n; i++) {

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

59                 for (k = 0; k < n; k++) {

60                     for (l = 0; l < m; l++) {

61                         if (MHT(i, j, k, l) <= d) {

62                             g[pos[i][j]][pos[k][l]] = 1;

63                             g[pos[i][j]][cnt] += a[i][j];

64                         }

65                     }

66                 }

67             }

68         }

69         Gauss();

70         for (i = 0; i < n; i++) {

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

72                 printf("%8.2lf", x[i * m + j]);

73             putchar('\n');

74         }

75     }

76     return 0;

77 }