[USACO 2016Dec] Team Building

[题目链接]

         https://www.lydsy.com/JudgeOnline/problem.php?id=4742

[算法]

        动态规划

        用Fi,j,k表示约翰的前i头牛和保罗的前j头牛匹配 ， 共选了k头 ， 有多少种方案

        转移详见代码

       时间复杂度 ： O(N ^ 2K)

[代码]

       

#include<bits/stdc++.h>
using namespace std;
#define MAXN 1010
#define MAXT 12
const int P = 1e9 + 9;

int n , m , t;
int a[MAXN] , b[MAXN];
int dp[MAXN][MAXN][MAXT];

template <typename T> inline void chkmax(T &x,T y) { x = max(x,y); }
template <typename T> inline void chkmin(T &x,T y) { x = min(x,y); }
template <typename T> inline void read(T &x)
{
    T f = 1; x = 0;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
    for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
    x *= f;
}

int main()
{
        
        read(n); read(m); read(t);
        for (int i = 1; i <= n; i++) read(a[i]);
        for (int i = 1; i <= m; i++) read(b[i]);
        sort(a + 1 , a + n + 1);
        sort(b + 1 , b + m + 1);
        dp[0][0][0] = 1;
        for (int i = 0; i <= n; i++)
        {
                for (int j = 0; j <= m; j++)
                {
                        for (int k = 0; k <= t; k++)
                        {
                                if (i + j == 0) continue;
                                if (i == 0) dp[i][j][k] = dp[i][j - 1][k];
                                else if (j == 0) dp[i][j][k] = dp[i - 1][j][k];
                                else dp[i][j][k] = dp[i - 1][j][k] + dp[i][j - 1][k] - dp[i - 1][j - 1][k];
                                dp[i][j][k] = (dp[i][j][k] % P + P) % P;
                                if (k > 0 && a[i] > b[j]) dp[i][j][k] += dp[i - 1][j - 1][k - 1];
                                dp[i][j][k] = (dp[i][j][k] % P + P) % P;
                        }
                }
        }
        printf("%d
" , dp[n][m][t]);
        
        return 0;
    
}