ZOJ3822 ACM-ICPC 2014 亞洲地區賽牡丹江賽區現場賽D題Domination 機率DP，zoj3822acm-icpc

ZOJ3822 ACM-ICPC 2014 亞洲地區賽牡丹江賽區現場賽D題Domination 機率DP，zoj3822acm-icpc

DominationTime Limit: 8 Seconds      Memory Limit: 131072 KB      Special Judge

Edward is the headmaster of Marjar University. He is enthusiastic about chess and often plays chess with his friends. What's more, he bought a large decorative chessboard with N rows and M columns.

Every day after work, Edward will place a chess piece on a random empty cell. A few days later, he found the chessboard was dominated by the chess pieces. That means there is at least one chess piece in every row. Also, there is at least one chess piece in every column.

"That's interesting!" Edward said. He wants to know the expectation number of days to make an empty chessboard of N × M dominated. Please write a program to help him.

Input

There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:

There are only two integers N and M (1 <= N, M <= 50).

Output

For each test case, output the expectation number of days.

Any solution with a relative or absolute error of at most 10-8 will be accepted.

Sample Input

21 32 2

Sample Output

3.0000000000002.666666666667

    這道題算是一個機率DP的裸題吧，很裸的演算法，題目意思是告訴你有一個n*m的地區，佔領一個格子可以控制這個行和列，求控制所有行列所需要佔領格子的期望，直接推匯出狀態轉移方程即可，對於算機率，我設dp[i][j][k]表示的意思為當我用k步控制了i行j列的機率，那麼我們可以知道，這時候下一步則有4不狀態轉移，分別為繼續取已控制的i行j列的格子，取未控制的i行，已控制的j列的格子，取已控制的i行和未控制的j列的格子，取未控制的i行j列的格子四種不同的情況，對於每種情況，狀態轉移方程如下：

1、繼續取已控制的i行j列的格子

dp[i][j][k+1]+=dp[i][j][k]*(i*j-k)/(n*m-k);
2、取未控制的i行，已控制的j列的格子
dp[i+1][j][k+1]+=dp[i][j][k]*(n-i)*j/(n*m-k);
3、取已控制的i行和未控制的j列的格子
dp[i][j+1][k+1]+=dp[i][j][k]*(m-j)*i/(n*m-k);
4、取未控制的i行j列的格子
dp[i+1][j+1][k+1]+=dp[i][j][k]*(n-i)*(m-j)/(n*m-k);
從而最終求出控制n行m列需要步數的機率，由期望的定義式EX=n1p1+n2p2+...即可求出最終期望，具體程式如下：

#include<cstdio>#include<iostream>#include<cstring>#include<cmath>#include<cstdlib>#include<algorithm>#include<map>#include<vector>#include<queue>using namespace std;double dp[55][55][55*55],a[55][55];const double eps=1e-8;int maxx(int a,int b){    if(a>b)return a;    return b;}int main(){  //  freopen("in.txt","r",stdin);    int t;    cin>>t;    while(t--)    {        int n,m;        scanf("%d%d",&n,&m);        memset(dp,0,sizeof(dp));        memset(a,0,sizeof(a));        dp[1][1][1]=1;        a[1][1]=1;        for(int i=1;i<=n;i++)        {            for(int j=1;j<=m;j++)            {                for(int k=1;k<=a[i][j];k++)                {                    if(i==n&&j==m)                        break;                    dp[i][j][k+1]+=dp[i][j][k]*(i*j-k)/(n*m-k);                    if(i*j>k)                        a[i][j]=maxx(a[i][j],k+1);                }                for(int k=1;k<=a[i][j];k++)                {                    dp[i+1][j][k+1]+=dp[i][j][k]*(n-i)*j/(n*m-k);                    a[i+1][j]=maxx(a[i+1][j],k+1);                    dp[i][j+1][k+1]+=dp[i][j][k]*(m-j)*i/(n*m-k);                    a[i][j+1]=maxx(a[i][j+1],k+1);                    dp[i+1][j+1][k+1]+=dp[i][j][k]*(n-i)*(m-j)/(n*m-k);                    a[i+1][j+1]=maxx(a[i+1][j+1],k+1);                }            }        }        double sum=0;        for(int i=1;i<=a[n][m];i++)            sum+=dp[n][m][i]*i;        printf("%.12f\n",sum);    }    return 0;}