Educational Codeforces Round 103 (Rated for Div. 2) D. Journey dp

传送门

题意： 给$n+1$个点，每两个点之间有一个字符$L(表示i+1连向i)$或者$R(表示i连向i+1)$，每秒可以移动一次，每移动一次每两个点之间的方向会互换，问每个点能到达的点个数最多是多少。

思路： 显然不管我们从这个点到了哪个点，我们都可以原路返回，这告诉我们向左向右是独立的。所以我们用$dp$维护每个点向左向右走的最远距离即可。这里只写出了向右的最远距离的转移方程。$$\{\begin{array}{l}dp[i]=max(dp[i],dp[i-2]+2),\text{if (s[i]==L,s[i-1]==R)}\\ dp[i]=max(dp[i],1),\text{if s[i]==L}\end{array}$$

//#pragma GCC optimize(2)
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<map>
#include<cmath>
#include<cctype>
#include<vector>
#include<set>
#include<queue>
#include<algorithm>
#include<sstream>
#include<ctime>
#include<cstdlib>
#define X first
#define Y second
#define L (u<<1)
#define R (u<<1|1)
#define pb push_back
#define mk make_pair
#define Mid (tr[u].l+tr[u].r>>1)
#define Len(u) (tr[u].r-tr[u].l+1)
#define random(a,b) ((a)+rand()%((b)-(a)+1))
#define db puts("---")
using namespace std;

//void rd_cre() { freopen("d://dp//data.txt","w",stdout); srand(time(NULL)); }
//void rd_ac() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//AC.txt","w",stdout); }
//void rd_wa() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//WA.txt","w",stdout); }

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;

const int N=1000010,mod=1e9+7,INF=0x3f3f3f3f;
const double eps=1e-6;

int n;
int l[N],r[N];
char s[N];

int main()
{
    
    
//	ios::sync_with_stdio(false);
//	cin.tie(0);

    int _; scanf("%d",&_);
    while(_--)
    {
    
    
        scanf("%d%s",&n,s+1);
        l[0]=0;
        for(int i=1;i<=n;i++)
        {
    
    
            if(i>=2&&s[i]=='L'&&s[i-1]=='R') l[i]=max(l[i],l[i-2]+2);
            if(s[i]=='L') l[i]=max(l[i],1);
            else l[i]=0;
        }
        r[n]=0;
        for(int i=n-1;i>=0;i--)
        {
    
    
            if(i<=n-2&&s[i+1]=='R'&&s[i+2]=='L') r[i]=max(r[i],r[i+2]+2);
            if(s[i+1]=='R') r[i]=max(r[i],1);
            else r[i]=0;
        }
        for(int i=0;i<=n;i++) printf("%d ",l[i]+r[i]+1),l[i]=r[i]=0;
    }





	return 0;
}
/*

*/