bzoj 1609[Usaco2008 Feb]Eating Together麻烦的聚餐【dp】

设up[i][j]为第i位升序为j的最小修改数，down为降序

#include<iostream>
#include<stdio.h>
using namespace std;
int n,a[30005],up[30005][4],down[30005][4];
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        up[i][3]=min(up[i-1][1],min(up[i-1][2],up[i-1][3]))+1;
        up[i][2]=min(up[i-1][1],up[i-1][2])+1;
        up[i][1]=up[i-1][1]+1;
        up[i][a[i]]--;
        down[i][1]=min(down[i-1][1],min(down[i-1][2],down[i-1][3]))+1;
        down[i][2]=min(down[i-1][2],down[i-1][3])+1;
        down[i][3]=down[i-1][3]+1;
        down[i][a[i]]--;
    }
    int ans=up[n][1];
    for(int i=1;i<=3;i++)
        ans=min(ans,min(up[n][i],down[n][i]));
    printf("%d\n",ans);
	return 0;
}