link:https://codeforces.com/contest/1114/problem/D
题意:
给定一个数组,有不同的颜色,你可以从任意一个位置开始,改变颜色,相邻的是同一种颜色的位子的颜色也要跟着改变,问最少需要改变几次颜色。
思路:
我一开始想的是去掉相邻重复后,假设有k个,那么答案就是k-1个,然后可以使结果更优的就是不相邻相同的,还能使结果更优的是一层再套一层相同的,就比如1,2,3,2,1。现场不知道怎么数这个套了几层,经过高人代码点拨,就是记忆化递归。
#include <algorithm> #include <iterator> #include <iostream> #include <cstring> #include <cstdlib> #include <iomanip> #include <bitset> #include <cctype> #include <cstdio> #include <string> #include <vector> #include <stack> #include <cmath> #include <queue> #include <list> #include <map> #include <set> #include <cassert> /* ⊂_ヽ \\ Λ_Λ 来了老弟 \('ㅅ') > ⌒ヽ / へ\ / / \\ レ ノ ヽ_つ / / / /| ( (ヽ | |、\ | 丿 \ ⌒) | | ) / 'ノ ) Lノ */ using namespace std; #define lson (l , mid , rt << 1) #define rson (mid + 1 , r , rt << 1 | 1) #define debug(x) cerr << #x << " = " << x << "\n"; #define pb push_back #define pq priority_queue typedef long long ll; typedef unsigned long long ull; //typedef __int128 bll; typedef pair<ll ,ll > pll; typedef pair<int ,int > pii; typedef pair<int,pii> p3; //priority_queue<int> q;//这是一个大根堆q //priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q #define fi first #define se second //#define endl '\n' #define OKC ios::sync_with_stdio(false);cin.tie(0) #define FT(A,B,C) for(int A=B;A <= C;++A) //用来压行 #define REP(i , j , k) for(int i = j ; i < k ; ++i) #define max3(a,b,c) max(max(a,b), c); #define min3(a,b,c) min(min(a,b), c); //priority_queue<int ,vector<int>, greater<int> >que; const ll oo = 1ll<<17; const ll mos = 0x7FFFFFFF; //2147483647 const ll nmos = 0x80000000; //-2147483648 const int inf = 0x3f3f3f3f; const ll inff = 0x3f3f3f3f3f3f3f3f; //18 const int mod = 1e9+7; const double esp = 1e-8; const double PI=acos(-1.0); const double PHI=0.61803399; //黄金分割点 const double tPHI=0.38196601; template<typename T> inline T read(T&x){ x=0;int f=0;char ch=getchar(); while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar(); while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); return x=f?-x:x; } /*-----------------------showtime----------------------*/ const int maxn = 5e3+9; int a[maxn],b[maxn],dp[maxn][maxn]; int cal(int le,int ri){ if(dp[le][ri]>=0)return dp[le][ri]; if(le >= ri) return 0; if(b[le]==b[ri])return dp[le][ri] = cal(le+1,ri-1) + 1; else return dp[le][ri] = max(cal(le+1,ri), cal(le,ri-1)); } int main(){ int n; scanf("%d", &n); memset(dp, -1, sizeof(dp)); for(int i=1; i<=n; i++) scanf("%d", &a[i]); int tot = 0; for(int i=1,j; i<=n; i++){ for(j=i; a[i]==a[j] && j <= n; j++); b[++tot] = a[i]; i = j-1; } cout<<tot-1-cal(1,tot)<<endl; return 0; }View Code