BZOJ1609: [Usaco2008 Feb]Eating Together麻烦的聚餐

题目传送门

题解：dp题，要说的基本都放在注释里了，不难想

#include<bits/stdc++.h>#define ll long longconst int INF = 0x7fffffff;const double eps = 1e-5;using namespace std;int read(){int x = 0 , f = 1; char ch = getchar();while(ch<'0'||ch>'9') {if(ch=='-')f*=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+(ch-'0');ch=getchar();}return x * f;}int f[30001][4],a[30001],ans=INF,n;//f数组表示前i头牛以编号j结尾的最少修改次数 void dp(){memset(f,127,sizeof(f));f[0][1] = f[0][2] = f[0][3] = 0;for(int i = 1 ; i <= n ; i++)for(int j = 1 ; j <= 3 ; j++)for(int k = 1 ; k <= j ; k++)if(a[i] == j)f[i][j] = min(f[i][j],f[i-1][k]);//无需修改 else f[i][j] = min(f[i][j],f[i-1][k]+1);//需要修改 for(int i = 1 ; i <= 3 ; i++)ans = min(ans,f[n][i]);}int main(){n = read();for(int i = 1 ; i <= n ; i++)a[i] = read();dp();for(int i = 1 ; i <= (n>>1) ; i++)swap(a[i],a[n-i+1]);dp();printf("%d",ans);}