Codeforces1312E Array Shrinking 区间DP

题意

给你一个数组\(a\)，只要满足\(a_i=a_{i+1}\)就可以将这两个元素合并成一个值为\(a_i+1\)的元素，问数组最小长度。

解题思路

记得之前某场的F和这题差不多，当时好像是相邻且相等就可以移除这两个数问最小长度。

看到\(n\)的范围就想到区间DP了，感觉是一道挺裸的区间DP板子题。

AC代码

#include <bits/stdc++.h>
using namespace std;
  
typedef long long ll;
typedef pair<int,int> pi;
 
#define x first
#define y second
  
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define rall(x) (x).rbegin(),(x).rend()
#define endl '\n'
  
const double PI=acos(-1.0);
 
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int rnd(int l,int r){return l+rng()%(r-l+1);}
  
namespace IO{
    bool REOF = 1; //为0表示文件结尾
    inline char nc() {
        static char buf[100000], *p1 = buf, *p2 = buf;
        return p1 == p2 && REOF && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2) ? (REOF = 0, EOF) : *p1++;
    }
  
    template<class T>
    inline bool read(T &x) {
        char c = nc();bool f = 0; x = 0;
        while (c<'0' || c>'9')c == '-' && (f = 1), c = nc();
        while (c >= '0'&&c <= '9')x = (x << 3) + (x << 1) + (c ^ 48), c = nc();
        if(f)x=-x;
        return REOF;
    }
  
    template<typename T, typename... T2>
    inline bool read(T &x, T2 &... rest) {
        read(x);
        return read(rest...);
    }
  
  
    inline bool need(char &c) { return ((c >= 'a') && (c <= 'z')) || ((c >= '0') && (c <= '9')) || ((c >= 'A') && (c <= 'Z')); }
    // inline bool need(char &c) { return ((c >= 'a') && (c <= 'z')) || ((c >= '0') && (c <= '9')) || ((c >= 'A') && (c <= 'Z')) || c==' '; }
  
    inline bool read_str(char *a) {
        while ((*a = nc()) && need(*a) && REOF)++a; *a = '\0';
        return REOF;
    }
  
    inline bool read_dbl(double &x){
        bool f = 0; char ch = nc(); x = 0;
        while(ch<'0'||ch>'9')  {f|=(ch=='-');ch=nc();}
        while(ch>='0'&&ch<='9'){x=x*10.0+(ch^48);ch=nc();}
        if(ch == '.') {
            double tmp = 1; ch = nc();
            while(ch>='0'&&ch<='9'){tmp=tmp/10.0;x=x+tmp*(ch^48);ch=nc();}
        }
        if(f)x=-x;
        return REOF;
    }
  
    template<class TH> void _dbg(const char *sdbg, TH h){ cerr<<sdbg<<'='<<h<<endl; }
  
    template<class TH, class... TA> void _dbg(const char *sdbg, TH h, TA... a) {
        while(*sdbg!=',')cerr<<*sdbg++;
        cerr<<'='<<h<<','<<' '; _dbg(sdbg+1, a...);
    }
     
    template<class T> ostream &operator<<(ostream& os, vector<T> V) {
        os << "["; for (auto vv : V) os << vv << ","; return os << "]";
    }
  
    template<class T> ostream &operator<<(ostream& os, set<T> V) {
        os << "["; for (auto vv : V) os << vv << ","; return os << "]";
    }

    template<class T> ostream &operator<<(ostream& os, map<T,T> V) {
        os << "["; for (auto vv : V) os << vv << ","; return os << "]";
    }
 
    template<class L, class R> ostream &operator<<(ostream &os, pair<L,R> P) {
        return os << "(" << P.st << "," << P.nd << ")";
    }
     
    #define debug(...) _dbg(#__VA_ARGS__, __VA_ARGS__)
}
  
using namespace IO;
const int maxn=2e5+5;
const int maxv=2e5+5;
const int mod=998244353; // 998244353 1e9+7
const int INF=1e9+7; // 1e9+7 0x3f3f3f3f 0x3f3f3f3f3f3f3f3f
const double eps=1e-12;
  
int dx[4]={0,1,0,-1};
//int dx[8]={1,0,-1,1,-1,1,0,-1};
int dy[4]={1,0,-1,0};
//int dy[8]={1,1,1,0,0,-1,-1,-1};
 
// #define ls (x<<1)
// #define rs (x<<1|1)
// #define mid ((l+r)>>1)
// #define lson ls,l,mid
// #define rson rs,mid+1,r

// int tot,head[maxn];
// struct Edge{
// 	int v,nxt;
//     Edge(){}
//     Edge(int _v,int _nxt):v(_v),nxt(_nxt){}
// }e[maxn<<1];
// void init(){
// 	tot=1;
// 	memset(head,0,sizeof(head));
// }
// void addedge(int u,int v){
// 	e[tot]=Edge(v,head[u]); head[u]=tot++;
// 	e[tot]=Edge(u,head[v]); head[v]=tot++;
// }
// void addarc(int u,int v){
//     e[tot]=Edge(v,head[u]); head[u]=tot++;
// }
  
/**
 * **********     Backlight     **********
 * 仔细读题
 * 注意边界条件
 * 记得注释输入流重定向
 * 没有思路就试试逆向思维
 * 加油，奥利给
 */

int n,a[505],dp[505][505],v[505][505];

void solve(){
    read(n);
    for(int i=1;i<=n;i++)read(a[i]),dp[i][i]=1,v[i][i]=a[i];

    for(int l=2;l<=n;l++){
        for(int i=1;i<=n-l+1;i++){
            int j=i+l-1;
            dp[i][j]=INF;
            for(int k=i;k<=j-1;k++){
                if(dp[i][j]>=dp[i][k]+dp[k+1][j]){
                    dp[i][j]=dp[i][k]+dp[k+1][j];
                    if(dp[i][k]==1 && dp[k+1][j]==1 && v[i][k]==v[k+1][j]){
                        dp[i][j]=1;
                        v[i][j]=v[i][k]+1;
                    }
                }
            }
        }
    }

    printf("%d\n",dp[1][n]);
}
 
int main()
{
    // freopen("in.txt","r",stdin);
    // ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    // int _T; read(_T); for(int _=1;_<=_T;_++)solve();
    // while(read(n))solve();
    solve();
    return 0;
}