[bzoj4742][Usaco2016 Dec]Team Building

http://www.lydsy.com/JudgeOnline/problem.php?id=4742
dp[i][j][k]表示fJ前i只,FP前j只,各選了k只的方案數
初值:dp[i][j][0]=1
方程:
dp[i][j][k]=(dp[i-1][j][k]+dp[i][j-1][k]-dp[i-1][j-1][k]+(a[i]>b[j])*f[i-1][j-1][k-1])mod 1000000009
Orz w_yqts

#include <bits/stdc++.h>
using namespace std;
#define MOD 1000000009
#define N 1005
int a[N],b[N],f[N][N][11];
int n,m,K;
int main()
{
    cin>>n>>m>>K;
    for (int i=1;i<=n;++i) scanf("%d",&a[i]);sort(a+1,a+1+n);
    for (int i=1;i<=m;++i) scanf("%d",&b[i]);sort(b+1,b+1+m);
    f[0][0][0]=1;
    for (int i=1;i<=n;++i) f[i][0][0]=1;
    for (int i=1;i<=m;++i) f[0][i][0]=1;
    for (int i=1;i<=n;++i)
    for (int j=1;j<=m;++j)
    {
    f[i][j][0]=1;
    for (int k=1;k<=min(K,min(i,j));++k)
    {
        f[i][j][k]=0;
        f[i][j][k]=(f[i][j][k]+f[i-1][j][k])%MOD;
        f[i][j][k]=(f[i][j][k]+f[i][j-1][k])%MOD;
        f[i][j][k]=(f[i][j][k]-f[i-1][j-1][k]+MOD)%MOD;
        if (a[i]>b[j] && k) f[i][j][k]=(f[i][j][k]+f[i-1][j-1][k-1])%MOD;
    }
    }
    cout<<f[n][m][K]<<endl;
}