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

传送门

题意： 给n+1n+1n+1个点，每两个点之间有一个字符L(表示i+1连向i)L(表示i+1连向i)L(表示i+1连向i)或者R(表示i连向i+1)R(表示i连向i+1)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dpdpdp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。{dp[i]=max(dp[i],dp[i−2]+2),if (s[i]==L,s[i-1]==R)dp[i]=max(dp[i],1),if s[i]==L\begin{cases} 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{cases} {dp[i]=max(dp[i],dp[i−2]+2),if (s[i]==L,s[i-1]==R)dp[i]=max(dp[i],1),if s[i]==L

//#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;
}
/**/

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

Educational Codeforces Round 103 (Rated for Div. 2)A~E解题报告
Educational Codeforces Round 103 (Rated for Div. 2) A. K-divisible Sum 原题信息解题思路 AC代码 #include <b ...
Educational Codeforces Round 103 (Rated for Div. 2)
Educational Codeforces Round 103 (Rated for Div. 2) http://codeforces.com/contest/1476 A K-divisible ...
Educational Codeforces Round 103 (Rated for Div. 2)前四题
Educational Codeforces Round 103 (Rated for Div. 2) 第二次被别人hack,悲 A - K-divisible Sum 题意给定两个整数 n,kn, ...
Educational Codeforces Round 114 (Rated for Div. 2) （A ~ F）全题解
整理的算法模板合集: ACM模板点我看算法全家桶系列!!! 实际上是一个全新的精炼模板整合计划 Educational Codeforces Round 114 (Rated for Div. 2) ...
Educational Codeforces Round 106 (Rated for Div. 2)（A ~ E）题解（每日训练 Day.16 ）
整理的算法模板合集: ACM模板点我看算法全家桶系列!!! 实际上是一个全新的精炼模板整合计划目录 Educational Codeforces Round 106 (Rated for Div. ...
Educational Codeforces Round 37 (Rated for Div. 2) 1
Educational Codeforces Round 37 (Rated for Div. 2) A.Water The Garden 题意:Max想给花园浇水.花园可被视为长度为n的花园床,花园 ...
Educational Codeforces Round 90 (Rated for Div. 2)（A, B, C, D, E）
Educational Codeforces Round 90 (Rated for Div. 2) Donut Shops 思路分三种情况: a==c/ba == c / ba==c/b这个时候两 ...
Educational Codeforces Round 89 (Rated for Div. 2)（A, B, C, D）
Educational Codeforces Round 89 (Rated for Div. 2) A. Shovels and Swords 思路题意非常简单,就是得到最多的物品嘛,我们假定a, ...
Educational Codeforces Round 114 (Rated for Div. 2) D. The Strongest Build 暴力 + bfs
传送门文章目录题意: 思路: 题意: 你有nnn个装备槽,每个槽里面有cic_ici个力量加成,对于每个槽只能选一个力量加成,现在给你mmm个力量组合[b1,b2,...,bn][b_1,b_2 ...

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

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

最新文章

热门文章