题意:给出N*M大小的矩阵,每次放置一颗棋子,问每行每列都至少有1颗棋子的期望放置次数。
范围:N,M<=50
解法:8秒题,比较容易想到2500*2500的复杂度解法,DP[I][J][K] 表示已经I行J列在K步后已经有棋子的概率,则可以转移:
DP[I][J][K+1] = DP[I][J][K] * (I*J-K)/(N*M-K)
DP[I+1][J][K+1] = DP[I][J][K] * ((N-I)*J)/(N*M-K)
DP[I][J+1][K+1] = DP[I][J][K] * (I*(M-J))/(N*M-K)
DP[I+1][J+1][K+1] = DP[I][J][K] * ((N-I)*(M-J))/(N*M-K)
需注意的是,如果这样转移,最后求得的概率总和不是1(因为定义是K步后棋子覆盖了I行J列的概率,所以K很大的时候几乎都是1)
求期望的时候,需要用恰好在K步完全覆盖的概率*K ,然后累加得到ans。
恰好在K步完全覆盖的概率是 DP[I][J][K]-DP[I][J][K-1]
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<iostream>
#include<stdlib.h>
#include<set>
#include<map>
#include<queue>
#include<vector>
#include<bitset>
#pragma comment(linker, "/STACK:1024000000,1024000000")
template <class T>
bool scanff(T &ret){ //Faster Input
char c; int sgn; T bit=0.1;
if(c=getchar(),c==EOF) return 0;
while(c!='-'&&c!='.'&&(c<'0'||c>'9')) c=getchar();
sgn=(c=='-')?-1:1;
ret=(c=='-')?0:(c-'0');
while(c=getchar(),c>='0'&&c<='9') ret=ret*10+(c-'0');
if(c==' '||c=='\n'){ ret*=sgn; return 1; }
while(c=getchar(),c>='0'&&c<='9') ret+=(c-'0')*bit,bit/=10;
ret*=sgn;
return 1;
}
#define inf 1073741823
#define llinf 4611686018427387903LL
#define PI acos(-1.0)
#define lth (th<<1)
#define rth (th<<1|1)
#define rep(i,a,b) for(int i=int(a);i<=int(b);i++)
#define drep(i,a,b) for(int i=int(a);i>=int(b);i--)
#define gson(i,root) for(int i=ptx[root];~i;i=ed[i].next)
#define tdata int testnum;scanff(testnum);for(int cas=1;cas<=testnum;cas++)
#define mem(x,val) memset(x,val,sizeof(x))
#define mkp(a,b) make_pair(a,b)
#define findx(x) lower_bound(b+1,b+1+bn,x)-b
#define pb(x) push_back(x)
using namespace std;
#define NN 100100
typedef long long ll;
typedef double ld;
ld dp[66][66][2666];
int main(){
tdata{
int n,m;
scanff(n);
scanff(m);
rep(i,0,n)
rep(j,0,m)
rep(k,0,n*m)dp[i][j][k]=0;
dp[0][0][0]=ld(1);
rep(i,0,n)
rep(j,0,m)
rep(k,0,i*j){
dp[i+1][j+1][k+1]+=(dp[i][j][k]*ld(n-i)*ld(m-j)/ld(n*m-k));
dp[i+1][j][k+1]+=(dp[i][j][k]*(ld(n-i)*ld(j))/ld(n*m-k));
dp[i][j+1][k+1]+=(dp[i][j][k]*(ld(i)*ld(m-j))/ld(n*m-k));
dp[i][j][k+1]+=(dp[i][j][k]*(ld(i)*ld(j)-k)/ld(n*m-k));
//printf("dp[%d][%d][%d]=%.20lf\n",i,j,k,dp[i][j][k]);
}
ld ans=0.0;
rep(i,1,n*m){
ans+=(dp[n][m][i]-dp[n][m][i-1])*ld(i);
}
printf("%.9f\n",ans);
}
return 0;
}
/*
172.559362734591
*/