hdu6052To my boyfriend(子矩阵计数处理)
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To my boyfriend
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 975 Accepted Submission(s): 450
Problem Description
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
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
Input
The 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).
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).
Output
Each case outputs a number that holds 9 decimal places.
Sample Input
12 31 2 12 1 2
Sample Output
1.666666667Hint6(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)计算子矩阵颜色期望值,每个矩阵数字代表一个颜色最多有10000种,也就是不同颜色子矩阵数量/子矩阵首先考虑子矩阵数量,子矩阵可以枚举两个i~n,j~m,子矩阵为i*j的数量。然后考虑颜色,对于每个颜色单独考虑,也就是每个点要单独考虑,将点作为关键值考虑,每次计算的矩阵至少要包括这个点然后就是考虑子矩阵的上下左右域了,对于每个颜色点,先用row[c][j]数组,代表颜色为c,第j列的行号。接下来考虑点的value时下域就不要考虑了,直接是n-x+1,上域有不同,如果每个点向上出现了已经标记过的点,那么上域到此行结束,左右域也是取决于是否有标记过的点出现在和上域相应的列上,否则到此结束。在这里左右域有个优化,如果点是当前列最后一行,那么不用考虑上面的点了,因为上面的点的左右域只会小于。这样计算出每个点的左右下域,就能直接算出子矩阵的不同矩阵颜色值=左*右*下=(y-l[i])*(r[i]-y)*(n-x+1),再求出所有的点的值就是子矩阵的不同颜色数量。计算过程:先对每个点的前x个点的左右域初始化为0和m+1(最大化),然后对前y点的左域和后y点的右域更新,每个点的上域肯定是更新后的最后一个点,也就是第y列的点的行号确认。最后再从x-1到h行对点的左右域进行拉伸,找到公共的左右域,这样就找出来所有的域参考:http://www.cnblogs.com/nicetomeetu/p/7261147.htmlhttp://blog.csdn.net/calabash_boy/article/details/76272704?yyue=a21bo.50862.201879#include <cstdio>#include <cstring>#include <algorithm>#include <vector>#include <queue>#include <iostream>using namespace std;int l[110],r[110],row[10010][110];int map[110][110];int n,m;long long ans;void cal(int x,int y,int c){ for(int i=1;i<=x;i++) { l[i]=0; r[i]=m+1; } for(int i=1;i<y;i++) { l[row[c][i]]=i; } for(int i=m;i>y;i--) { r[row[c][i]]=i; } int h=row[c][y]; for(int i=x-1;i>h;i--) { l[i]=max(l[i],l[i+1]); r[i]=min(r[i],r[i+1]); } for(int i=x;i>h;i--) { ans+=(long long)(y-l[i])*(r[i]-y)*(n-x+1); }}int main(){ int t; scanf("%d",&t); //int c; int c; while(t--) { memset(row,0,sizeof(row)); scanf("%d%d",&n,&m); ans=0; for(int i=1;i<=n;i++) { for(int j=1;j<=m;j++) { scanf("%d",&c); cal(i,j,c); row[c][j]=i; } } long long sum=0; for(int i=1;i<=n;i++) { for(int j=1;j<=m;j++) { sum+=i*j; } } //cout<<ans<<endl; //cout<<sum<<endl; printf("%.9f\n",(double)ans/sum); } return 0;}
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