hdu3359 Kind of a Blur

Posted 2020-08-31 ws_ccd

因为变化出来的是平均数，那么就可以对每一个变化出来的列方程，直接高斯消元就行了。

 1 #include<bits/stdc++.h>
 2 #define N 100005
 3 #define LL long long
 4 #define inf 0x3f3f3f3f
 5 #define ls c[x][0]
 6 #define rs c[x][1]
 7 using namespace std;
 8 inline int ra()
 9 {
10     int x=0,f=1; char ch=getchar();
11     while (ch<‘0‘ || ch>‘9‘) {if (ch==‘-‘) f=-1; ch=getchar();}
12     while (ch>=‘0‘ && ch<=‘9‘) {x=x*10+ch-‘0‘; ch=getchar();}
13     return x*f;
14 }
15 const int maxn=105;  
16 const double eps=1e-8;  
17 double a[maxn][maxn];  
18 double ave[15][15];  
19 double b[maxn];  
20 void guass(int n, int m)
21 {
22     int equ=n*m,var=n*m;
23     for (int row=1,col=1; row<=equ && col<=var; row++,col++)
24     {
25         int maxr=row;
26         for (int i=row; i<=equ; i++)
27             if (fabs(a[i][col])-fabs(a[maxr][col])>eps)
28                 maxr=i;
29         if (fabs(a[maxr][col])<eps) 
30         {
31             row--; continue;
32         }
33         if (maxr!=row)
34             for (int j=col; j<=var+1; j++)
35                 swap(a[maxr][j],a[row][j]);
36         for (int i=row+1; i<=equ; i++)
37             if (fabs(a[i][col])>eps)
38             {
39                 double s=a[i][col]/a[row][col];
40                 for (int j=col; j<=var+1; j++)
41                     a[i][j]-=a[row][j]*s;
42             }
43     }
44     for (int i=equ; i>=1; i--)
45     {
46         double tmp=a[i][var+1];
47         for (int j=i+1; j<=var; j++)
48             tmp-=a[i][j]*b[j];
49         b[i]=tmp/a[i][i];
50     }
51 }
52 int main()
53 {
54     int n,m,d; int first=1;
55     while (~scanf("%d%d%d",&m,&n,&d))
56     {
57         memset(a,0,sizeof(a));
58         if (n+m+d==0) break;
59         for (int i=1; i<=n; i++)
60             for (int j=1; j<=m; j++)
61                 scanf("%lf",&ave[i][j]);
62         for (int k=1; k<=n*m; k++)
63         {
64             int x=(k-1)/m+1,y=k%m; if (!y) y=m;
65             double num=0;
66             for (int i=1; i<=n; i++)
67                 for (int j=1; j<=m; j++)
68                     if (abs(i-x)+abs(j-y)<=d)
69                         a[k][(i-1)*m+j]=1,num++;
70             a[k][n*m+1]=num*ave[x][y];
71         }
72         guass(n,m);
73         if (first) first=0; else cout<<endl;
74         for (int i=1; i<=n*m; i++)
75         {
76             printf("%8.2lf",b[i]);
77             if (i%m==0) printf("\n");
78         }
79     }
80     return 0;
81 }