题目描述
题目分析:
闫式DP分析:
不重不漏, 集合划分为更小的子集
表示
f(i, j, k, c) 表示在(i, j)位置取的物品为k件,且取得最后一件物品价值为c的方案数
初始化:
f(1,1,1w(1,1)) = 1
f(1,1,0,-1) = 0
划分
1.不取
f(i - 1, j, k, c)
f(i, j - 1, k, c)
2.取(w(i, j) == c) 1<= u <= 12 1<= v <= 13
f(i - 1, j, u - 1, c)
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 55, MOD = 1000000007;
int n, m, k, res;
int w[N][N];
int f[N][N][13][14];
int main()
{
cin>>n>>m>>k;
for (int i = 1; i <= n; i ++ )
for (int j = 1; j <= m; j ++ )
{
cin>>w[i][j];
w[i][j] ++;
}
f[1][1][1][w[1][1]] = 1;
f[1][1][0][0] = 1;
for (int i = 1; i <= n; i ++)
for(int j = 1; j <= m; j ++)
{
if(i == 1 && j == 1) continue;
for (int u = 0; u <= k; u ++ )
for (int v = 0; v <= 13; v ++ )
{
int &val = f[i][j][u][v];
//不取
val = (val + f[i - 1][j][u][v]) % MOD;
val = (val + f[i][j - 1][u][v]) % MOD;
//取
if(u > 0 && w[i][j] == v)
for (int c = 0; c < v; c ++ )
{
val = (val + f[i - 1][j][u - 1][c]) % MOD;
val = (val + f[i][j - 1][u - 1][c]) % MOD;
}
}
}
for (int i = 0; i <= 13; i ++ ) res = (res + f[n][m][k][i]) % MOD;
cout<<res<<endl;
return 0;
}