Hdu 5245 Joyful【滚动数组+概率Dp】

Joyful

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1413    Accepted Submission(s): 622

Problem Description

Sakura has a very magical tool to paint walls. One day, kAc asked Sakura to paint a wall that looks like an M×N matrix. The wall has M×N squares in all. In the whole problem we denotes (x,y) to be the square at the x-th row, y-th column. Once Sakura has determined two squares (x1,y1) and (x2,y2), she can use the magical tool to paint all the squares in the sub-matrix which has the given two squares as corners.

However, Sakura is a very naughty girl, so she just randomly uses the tool for K times. More specifically, each time for Sakura to use that tool, she just randomly picks two squares from all the M×N squares, with equal probability. Now, kAc wants to know the expected number of squares that will be painted eventually.

 

Input

The first line contains an integer T(T≤100), denoting the number of test cases.

For each test case, there is only one line, with three integers M,N and K.
It is guaranteed that 1≤M,N≤500, 1≤K≤20.

 

Output

For each test case, output ''Case #t:'' to represent the t-th case, and then output the expected number of squares that will be painted. Round to integers.

 

Sample Input

23 3 14 4 2

 

Sample Output

Case #1: 4Case #2: 8
Hint
The precise answer in the first test case is about 3.56790123.
 

题目大意：

给你一个N*M的矩阵，进行K次随机操作，每个点被选中的概率都是等同的，每次操作随机选两个点，作为一个矩阵的对角，然后将这个矩阵涂上颜色，问K次操作之后，期望被涂上色的点的个数。点可以被重复涂抹。

思路：

我们没法考虑最终的形状，要不然时间复杂度非常大，所以我们不妨考虑一个点在K次操作之后是否会贡献其本身的价值（即这个点在k次操作之后是否被涂抹了）。

那么设定Dp【i】【x】【y】【2】：

①Dp【i】【x】【y】【0】表示操作了i次之后，（x，y）这个点没有被涂的概率。

②Dp【i】【x】【y】【1】表示操作了i次之后，（x，y）这个点被涂了的概率。

那么状态转移方程并不难写出，考虑到double类型的数据，O（n*m*k）的空间复杂度有些大（实际最开始第一发就交的这个，然后MLE了之后才改的）。我们滚动一下数组再写出状态转移方程有：

这里P【i】【j】表示涂抹一次，（i，j）这个点被涂抹到的概率。

其计算方式也很简单，简单容斥几个公式就行了。这里具体参考一下Ac代码吧。

那么ans=ΣDp【n】【x】【y】【1】；

Ac代码（800+ms过的）：

#include<stdio.h>#include<string.h>#include<math.h>using namespace std;#define ll long long intdouble p[501][501];double dp[2][501][501][3];ll Cal(ll a,ll b){    return a*a*b*b;}int main(){    int kase=0;    int t;    scanf("%d",&t);    while(t--)    {        ll n,m,k;        memset(dp,0,sizeof(dp));        scanf("%lld%lld%lld",&n,&m,&k);        for(ll i=1;i<=n;i++)        {            for(ll j=1;j<=m;j++)            {                p[i][j]+=Cal(i-1,m);                p[i][j]+=Cal(n-i,m);                p[i][j]+=Cal(n,j-1);                p[i][j]+=Cal(n,m-j);                p[i][j]-=Cal(i-1,j-1);                p[i][j]-=Cal(i-1,m-j);                p[i][j]-=Cal(n-i,j-1);                p[i][j]-=Cal(n-i,m-j);                double fenmu=n*m*n*m;                p[i][j]/=fenmu;                p[i][j]=1-p[i][j];            }        }        for(int i=1;i<=n;i++)        {            for(int j=1;j<=m;j++)            {                dp[0][i][j][0]=1.0;            }        }        for(int z=1;z<=k;z++)        {            for(int i=1;i<=n;i++)            {                for(int j=1;j<=m;j++)                {                    dp[1][i][j][0]+=dp[0][i][j][0]*(1.0-p[i][j]);                    dp[1][i][j][1]+=dp[0][i][j][0]*(p[i][j]);                    dp[1][i][j][1]+=dp[0][i][j][1]*(p[i][j]);                    dp[1][i][j][1]+=dp[0][i][j][1]*(1.0-p[i][j]);                    dp[0][i][j][0]=dp[1][i][j][0];                    dp[0][i][j][1]=dp[1][i][j][1];                    dp[1][i][j][0]=0;                    dp[1][i][j][1]=0;                }            }        }        double output=0;        for(int i=1;i<=n;i++)        {            for(int j=1;j<=m;j++)            {                output+=dp[0][i][j][1];            }        }        printf("Case #%d: ",++kase);        printf("%.0f\n",output);    }}