BZOJ1048: [HAOI2007]分割矩阵

 1 #include<cstdio>
 2 
 3 #include<cstdlib>
 4 
 5 #include<cmath>
 6 
 7 #include<cstring>
 8 
 9 #include<algorithm>
10 
11 #include<iostream>
12 
13 #include<vector>
14 
15 #include<map>
16 
17 #include<set>
18 
19 #include<queue>
20 
21 #include<string>
22 
23 #define inf 1000000000
24 
25 #define maxn 11
26 
27 #define maxm 200000+5
28 
29 #define eps 1e-10
30 
31 #define ll long long
32 
33 #define pa pair<int,int>
34 
35 #define for0(i,n) for(int i=0;i<=(n);i++)
36 
37 #define for1(i,n) for(int i=1;i<=(n);i++)
38 
39 #define for2(i,x,y) for(int i=(x);i<=(y);i++)
40 
41 #define for3(i,x,y) for(int i=(x);i>=(y);i--)
42 
43 #define for4(i,x) for(int i=head[x],y=e[i].go;i;i=e[i].next,y=e[i].go)
44 
45 #define for5(n,m) for(int i=1;i<=n;i++)for(int j=1;j<=m;j++)
46 
47 #define mod 1000000007
48 #define sqr(x) (x)*(x)
49 
50 using namespace std;
51 
52 inline int read()
53 
54 {
55 
56     int x=0,f=1;char ch=getchar();
57 
58     while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
59 
60     while(ch>=‘0‘&&ch<=‘9‘){x=10*x+ch-‘0‘;ch=getchar();}
61 
62     return x*f;
63 
64 }
65 int n,m,k,s[maxn][maxn];
66 double ave,f[maxn][maxn][maxn][maxn][maxn];
67 inline double dp(int x1,int y1,int x2,int y2,int z)
68 {
69     double &t=f[x1][y1][x2][y2][z];
70     if(t<inf)return t;
71     if(!z)return t=sqr(s[x2][y2]-s[x1-1][y2]-s[x2][y1-1]+s[x1-1][y1-1]-ave);
72     t=inf;
73     for0(i,z-1)
74      {
75          for2(j,x1,x2-1)t=min(t,dp(x1,y1,j,y2,i)+dp(j+1,y1,x2,y2,z-1-i));
76          for2(j,y1,y2-1)t=min(t,dp(x1,y1,x2,j,i)+dp(x1,j+1,x2,y2,z-1-i));
77      }
78     //cout<<x1<<‘ ‘<<y1<<‘ ‘<<x2<<‘ ‘<<y2<<‘ ‘<<z<<‘ ‘<<t<<endl;
79     return t;
80 }
81 
82 int main()
83 
84 {
85 
86     freopen("input.txt","r",stdin);
87 
88     freopen("output.txt","w",stdout);
89 
90     n=read();m=read();k=read();
91     for1(i,n)for1(j,m)s[i][j]=s[i-1][j]+s[i][j-1]-s[i-1][j-1]+read();
92     ave=(double)s[n][m]/(double)k;
93     for1(i1,n)for1(i2,m)for1(i3,n)for1(i4,m)for0(i5,k)f[i1][i2][i3][i4][i5]=inf;
94     dp(1,1,n,m,k-1);
95     printf("%.2f\n",sqrt((double)f[1][1][n][m][k-1]/(double)k));
96 
97     return 0;
98 
99 }