luogu P3842 [TJOI2007]线段 线性dp

总感觉之前做过一个类似的

#include<iostream>
#include<cmath>
using namespace std;
const int N=2e4+10;
int r[N],l[N];
int dp[N][2];
int n;
int main()
{
	cin>>n;
	for(int i=1;i<=n;i++)
		cin>>l[i]>>r[i];
	//停在右边 
	dp[1][1]=r[1]-1;
	//停在左边 
	dp[1][0]=r[1]+r[1]-l[1]-1;
	for(int i=2;i<=n;i++)
	{
		dp[i][0]=min(dp[i-1][0]+abs(l[i-1]-r[i])+r[i]-l[i]+1,dp[i-1][1]+abs(r[i-1]-r[i])+r[i]-l[i]+1);
		dp[i][1]=min(dp[i-1][0]+abs(l[i-1]-l[i])+r[i]-l[i]+1,dp[i-1][1]+abs(r[i-1]-l[i])+r[i]-l[i]+1);
	}
	cout<<min(dp[n][0]+n-l[n],dp[n][1]+n-r[n])<<endl;
	return 0;
}