代码恢复训练 2024.7.2.
链接 (codeforces)
链接 (luogu)
一道很基础的区间 dp。
只讲状态定义，\(dp_{i,j}\) 表示 \(i \sim j\) 区间需要的最少消除次数。
时间复杂度 \(O(n^2)\)。
点击查看代码
/*
Tips:
你数组开小了吗？
你MLE了吗？
你觉得是贪心，是不是该想想dp？
一个小时没调出来，是不是该考虑换题？
打 cf 不要用 umap！！！记住，rating 是身外之物。该冲正解时冲正解！Problem:算法：思路：*/
#include<bits/stdc++.h>
using namespace std;
//#define map unordered_map
#define re register
#define ll long long
#define forl(i,a,b) for(re ll i=a;i<=b;i++)
#define forr(i,a,b) for(re ll i=a;i>=b;i--)
#define forll(i,a,b,c) for(re ll i=a;i<=b;i+=c)
#define forrr(i,a,b,c) for(re ll i=a;i>=b;i-=c)
#define lc(x) x<<1
#define rc(x) x<<1|1
#define mid ((l+r)>>1)
#define cin(x) scanf("%lld",&x)
#define cout(x) printf("%lld",x)
#define lowbit(x) (x&-x)
#define pb push_back
#define pf push_front
#define IOS ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
#define endl '\n'
#define QwQ return 0;
#define db long double
#define ull unsigned long long
#define lcm(x,y) x/__gcd(x,y)*y
#define Sum(x,y) 1ll*(x+y)*(y-x+1)/2
#define aty cout<<"Yes\n";
#define atn cout<<"No\n";
#define cfy cout<<"YES\n";
#define cfn cout<<"NO\n";
#define xxy cout<<"yes\n";
#define xxn cout<<"no\n";
#define printcf(x) x?cout<<"YES\n":cout<<"NO\n";
#define printat(x) x?cout<<"Yes\n":cout<<"No\n";
#define printxx(x) x?cout<<"yes\n":cout<<"no\n";
ll t;
ll n;
ll a[510];
ll dp[510][510];
void solve()
{cin>>n;forl(i,1,n)cin>>a[i];forl(i,1,n)forl(j,1,n)dp[i][j]=1e18;forl(i,1,n)dp[i][i]=1;forl(i,1,n-1){if(a[i]==a[i+1])dp[i][i+1]=1;elsedp[i][i+1]=2;}forl(len,2,n-1){forl(i,1,n-len){if(a[i]==a[i+len])dp[i][i+len]=min(dp[i][i+len],dp[i+1][i+len-1]);forl(j,i,i+len-1)dp[i][i+len]=min(dp[i][i+len],dp[i][j]+dp[j+1][i+len]);}}cout<<dp[1][n]<<endl;
}
int main()
{IOS;t=1;// cin>>t;while(t--)solve();/******************//*while(L<q[i].l) *//* del(a[L++]);*//*while(L>q[i].l) *//* add(a[--L]);*//*while(R<q[i].r) *//* add(a[++R]);*//*while(R>q[i].r) *//* del(a[R--]);*//******************/QwQ;
}
