friend namespace difficult cer break con str div org

Dear Liao

I never forget the moment I met with you. You carefully asked me: "I have a very difficult problem. Can you teach me?". I replied with a smile, "of course". You replied:"Given a matrix, I randomly choose a sub-matrix, what is the expectation of the number of **different numbers** it contains?"

Sincerely yours,
Guo

InputThe first line of input contains an integer T(T≤8) indicating the number of test cases.
Each case contains two integers, n and m (1≤n, m≤100), the number of rows and the number of columns in the grid, respectively.
The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains g_i,j (0≤ g_i,j < n*m).OutputEach case outputs a number that holds 9 decimal places.Sample Input

1
2 3
1 2 1
2 1 2

Sample Output

1.666666667

        
 

Hint

6(size = 1) + 14(size = 2) + 4(size = 3) + 4(size = 4) + 2(size = 6) = 30  / 18 = 6(size = 1) + 7(size = 2) + 2(size = 3) + 2(size = 4) + 1(size = 6)


題目大意：給你n行m列的二維數組，數組中的每一個數值表示顏色。問二維數組構成的矩形中有不同的顏色和除以二維數組構成的不同矩形數



解題思路：首先要算出一個顏色的不同上下左右邊界，假定下邊界固定了。然後求出上邊界，如果上邊界的相同位置和本身的顏色相同就結束這個搜索，當i==x是只要求出左邊界就行了。最後求出左邊界和右邊界算出矩陣的個數即可。枚舉二維數組中中每一顏色加和就能得到二維數組構成的矩形中有不同的顏色和。

二維數組構成的不同矩形數=n*(n+1)*m*(m+1)/4

AC代碼：

 1 #include <iostream>
 2 #include <bits/stdc++.h>
 3 #define LL long long
 4 using namespace std;
 5 int n,m;
 6 LL color[105][105];
 7 LL count(int x,int y)
 8 {
 9     LL res=0;
10     int L=1,R=m,c=color[x][y];
11     for(int i=x;i>=1;i--)
12     {
 
13         if(i<x&&c==color[i][y]) break;
14         int l=y,r=y;
15         for(int j=y-1;j>=max(1,L);j--)
16         {
17             if(c==color[i][j]) break;
18             l=j;
19         }
20         L=max(L,l);
21         if(i==x)
22         {
23             res+=(LL)(n-x+1LL)*(y-L+1LL)*(R-y+1LL);
24             continue;
25         }
26         for(int j=y+1;j<=min(R,m);j++)
27         {
28             if(c==color[i][j]) break;
29             r=j;
30         }
31         R=min(R,r);
32         res+=(LL)(n-x+1LL)*(y-L+1LL)*(R-y+1LL);
33     }
34     return res;
35 }
36 int main()
37 {
38     int t;
39     while(~scanf("%d",&t))
40     {
41         while(t--)
42         {
43             scanf("%d%d",&n,&m);
44             LL ss=n*(n+1)/2*m*(m+1)/2;
45             for(int i=1;i<=n;i++)
46             {
47                 for(int j=1;j<=m;j++)
48                 {
49                     scanf("%lld",&color[i][j]);
50                 }
51             }
52             LL ans=0;
53             for(int i=1;i<=n;i++)
54             {
55                 for(int j=1;j<=m;j++)
56                 {
57                     //printf("%lld\n",count(i,j));
58                     ans+=count(i,j);
59                 }
60             }
61             //printf("%lld %lld\n",ans,ss);
62             printf("%.9f\n",ans*1.0/ss);
63         }
64     }
65     return 0;
66 }

View Code

第二場 hdu 6052 To my boyfriend