sdut 3565 Feed the monkey dp

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题目链接

思路:

dp[i][j][k][a][b] 表示i个B j个P K 个A 选择第a种水果连续选择了b个的种数,即可得到转移方程.

最后我们记录所有以其中任何一种水果结尾的连续的i个的种类.

以此时放B为例:

我们枚举可连续放B的个数b,可得到如下的转移方程:

1.继续放B 则 dp[i+1][j][k][a][b+1]=(dp[i+1][j][k][a][b+1]+dp[i][j][k][a][b])

2.放P

dp[i][j+1][k][1][1]=(dp[i][j+1][k][1][1]+dp[i][j][k][a][b])%MOD;

3.放A

dp[i][j][k+1][2][1]=(dp[i][j][k+1][2][1]+dp[i][j][k][a][b])%MOD;

PS：

一开始这个题我真的没想明白要怎么去转移,当时思维有点混乱,只想了5维dp,想了一下复杂度觉得还可以,转移的时候就开始乱想了,搜了一发题解发现都是用记忆化搜索写的,本人最弱的就是记忆化搜索,除了会套数位dp的板子.

#include<bits/stdc++.h>#define Ri(a) scanf("%d", &a)#define Rl(a) scanf("%lld", &a)#define Rf(a) scanf("%lf", &a)#define Rs(a) scanf("%s", a)#define Pi(a) printf("%d\n", (a))#define Pf(a) printf("%lf\n", (a))#define Pl(a) printf("%lld\n", (a))#define Ps(a) printf("%s\n", (a))#define W(a) while(a--)#define CLR(a, b) memset(a, (b), sizeof(a))#define MOD 1000000007#define inf 0x3f3f3f3f#define exp 0.00000001#define  pii  pair<int, int>#define  mp   make_pair#define  pb   push_backusing namespace std;typedef long long ll;const int maxn=1e5+10;int dp[55][55][55][5][55];int n1,n2,n3,d1,d2,d3;int main(){int t;Ri(t);W(t){Ri(n1),Ri(n2),Ri(n3),Ri(d1),Ri(d2),Ri(d3);CLR(dp,0);dp[1][0][0][0][1]=1;dp[0][1][0][1][1]=1;dp[0][0][1][2][1]=1;for(int i=0;i<=n1;i++){for(int j=0;j<=n2;j++){for(int k=0;k<=n3;k++){if(i==0&&j==0&&k==0)continue;for(int a=0;a<=2;a++){if(a==0){for(int b=1;b<=min(i,d1);b++){if(b+1<=d1)dp[i+1][j][k][a][b+1]=(dp[i+1][j][k][a][b+1]+dp[i][j][k][a][b])%MOD;dp[i][j+1][k][1][1]=(dp[i][j+1][k][1][1]+dp[i][j][k][a][b])%MOD;dp[i][j][k+1][2][1]=(dp[i][j][k+1][2][1]+dp[i][j][k][a][b])%MOD;}}if(a==1){for(int b=1;b<=min(j,d2);b++){if(b+1<=d2)dp[i][j+1][k][a][b+1]=(dp[i][j+1][k][a][b+1]+dp[i][j][k][a][b])%MOD;dp[i+1][j][k][0][1]=(dp[i+1][j][k][0][1]+dp[i][j][k][a][b])%MOD;dp[i][j][k+1][2][1]=(dp[i][j][k+1][2][1]+dp[i][j][k][a][b])%MOD;}}if(a==2){for(int b=1;b<=min(k,d3);b++){if(b+1<=d3)dp[i][j][k+1][a][b+1]=(dp[i][j][k+1][a][b+1]+dp[i][j][k][a][b])%MOD;dp[i+1][j][k][0][1]=(dp[i+1][j][k][0][1]+dp[i][j][k][a][b])%MOD;dp[i][j+1][k][1][1]=(dp[i][j+1][k][1][1]+dp[i][j][k][a][b])%MOD;}}}}}}ll ans=0;for(int i=1;i<=d1;i++)ans=(ans+dp[n1][n2][n3][0][i])%MOD;for(int i=1;i<=d2;i++)ans=(ans+dp[n1][n2][n3][1][i])%MOD;for(int i=1;i<=d3;i++)ans=(ans+dp[n1][n2][n3][2][i])%MOD;Pl(ans);}return 0;}

记忆化搜索:

#include<bits/stdc++.h>  #define ll long long  #define mod 1000000007  using namespace std;  ll dp[55][55][55][5][55];  int n1,n2,n3,d1,d2,d3;  ll dfs(int a,int b,int c,int t,int e)  {      if(a<0||b<0||c<0)          return 0;      if(t==0&&e>d1||t==1&&e>d2||t==2&&e>d3)          return 0;      if(a==0&&b==0&&c==0)          return 1;      if(dp[a][b][c][t][e]!=-1)          return dp[a][b][c][t][e];      ll ans=0;      if(t==0)      {          ans=(ans+dfs(a-1,b,c,0,e+1))%mod;          ans=(ans+dfs(a,b-1,c,1,1))%mod;          ans=(ans+dfs(a,b,c-1,2,1))%mod;          return dp[a][b][c][t][e]=ans;      }      if(t==1)      {          ans=(ans+dfs(a-1,b,c,0,1))%mod;          ans=(ans+dfs(a,b-1,c,1,e+1))%mod;          ans=(ans+dfs(a,b,c-1,2,1))%mod;          return dp[a][b][c][t][e]=ans;      }      if(t==2)      {          ans=(ans+dfs(a-1,b,c,0,1))%mod;          ans=(ans+dfs(a,b-1,c,1,1))%mod;          ans=(ans+dfs(a,b,c-1,2,e+1))%mod;          return dp[a][b][c][t][e]=ans;      }    }  int main()  {      int T;      cin>>T;      while(T--)      {          scanf("%d%d%d%d%d%d",&n1,&n2,&n3,&d1,&d2,&d3);          memset(dp,-1,sizeof(dp));          ll ans=0;              ans=(ans+dfs(n1-1,n2,n3,0,1))%mod;              ans=(ans+dfs(n1,n2-1,n3,1,1))%mod;              ans=(ans+dfs(n1,n2,n3-1,2,1))%mod;          cout<<ans<<endl;      }      return 0;  }

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