017.记忆化递归

优化暴力递归

递归层数太深会超时+栈溢出

对策

• 记忆化递归
• 剪枝

int f(int x){
    if(x==1||x==2)return 1;
    return f(x-1)+f(x-2);
}

记忆化

int F[101]={-1};
int f(int x){
    if(x==1||x==2)return 1;
    if(F[x]!=-1)return F[x];
    return F[x]=f(x-1)+f(x-2);
}

例题

luogu P1464
经典记忆化递归题目

ll A,B,C;
ll m[30][30][30];
ll w(ll a,ll b,ll c){
    if(a<=0||b<=0||c<=0)return 1;
    if(a>20||b>20||c>20)return w(20,20,20);
    if(m[a][b][c]!=-1)return m[a][b][c];
    if(a<b&&b<c)return m[a][b][c]=w(a,b,c-1)+w(a,b-1,c-1)-w(a,b-1,c);
    return m[a][b][c]=w(a-1,b,c)+w(a-1,b-1,c)+w(a-1,b,c-1)-w(a-1,b-1,c-1);
}
void solve(){
    memset(m,-1,sizeof(m));
    while(cin>>A>>B>>C){
        if(A==-1&&C==-1&&B==-1)return;
        cout<<"w("<<A<<", "<<B<<", "<<C<<") = "<<w(A,B,C)<<'\n';
    }
}

luogu P1164

正解是dp,不过暴力递归遍历+记忆化+剪枝也是ac了

int ans=0;
int n,m;
int a[101],f[101][10005];
void dfs(int sum,int indx){
    if(f[indx][m-sum]){
        ans+=f[indx][m-sum];
        return;
    }
    if(sum==m){
        ans++;
        return;
    }
    if(sum>m)return;
    int t=ans;
    for(int i=indx;i<n;++i){
        dfs(sum+a[i],i+1);
    }
    f[indx][m-sum]=ans-t;
}
void solve(){
    read(n,m);
    for(int i=0;i<n;++i)read(a[i]);
    memset(f,0,sizeof(f));
    dfs(0,0);
    wr(ans);
}