There are n+1n+1 cities, numbered from 00 to nn. nn roads connect these cities, the ii-th road connects cities i−1i−1 and ii (i∈[1,n]i∈[1,n]).
Each road has a direction. The directions are given by a string of nn characters such that each character is either L or R. If the ii-th character is L, it means that the ii-th road initially goes from the city ii to the city i−1i−1; otherwise it goes from the city i−1i−1 to the city ii.
A traveler would like to visit as many cities of this country as possible. Initially, they will choose some city to start their journey from. Each day, the traveler must go from the city where they currently are to a neighboring city using one of the roads, and they can go along a road only if it is directed in the same direction they are going; i. e., if a road is directed from city ii to the city i+1i+1, it is possible to travel from ii to i+1i+1, but not from i+1i+1 to ii. After the traveler moves to a neighboring city, all roads change their directions to the opposite ones. If the traveler cannot go from their current city to a neighboring city, their journey ends; it is also possible to end the journey whenever the traveler wants to.
The goal of the traveler is to visit as many different cities as possible (they can visit a city multiple times, but only the first visit is counted). For each city ii, calculate the maximum number of different cities the traveler can visit during exactly one journey if they start in the city ii.
The first line contains one integer tt (1≤t≤1041≤t≤104) — the number of test cases.
Each test case consists of two lines. The first line contains one integer nn (1≤n≤3⋅1051≤n≤3⋅105). The second line contains the string ss consisting of exactly nn characters, each character is either L or R.
It is guaranteed that the sum of nn over all test cases does not exceed 3⋅1053⋅105.
For each test case, print n+1n+1 integers. The ii-th integer should be equal to the maximum number of different cities the traveler can visit during one journey if this journey starts in the ii-th city.
2 6 LRRRLL 3 LRL
1 3 2 3 1 3 2 1 4 1 4
分析
题目给出的背景有点迷，我们设为上下两部分来算
b 0 1 2 3 4 5 6
c 0 1 2 3 4 5 6
由于LRRRLL
b1链接c0，c2连接b1，这样每次想走某一条边，就只能通过已经连的进行
接着就是类似并查集的操作，倒着把连接的边数算到前面
然后再从前往后依次输出并更新
可惜就差那么几秒
就写完了
唉
代码
https://codeforces.com/contest/1476/submission/105938876
#include <cstdio> #include <cstring> #include <algorithm> #include <vector> #include <math.h> #include <string.h> #include <algorithm> #include <iostream> #include <string> #include <sstream> #include <iostream> #include <time.h> #include <queue> #include <list> #include <map> #include <set> #include <vector> #include <stack> #include <string.h> #include <bitset> #define sf scanf #define pf printf #define lf double #define p123 printf("123\n"); #define pn printf("\n"); #define pk printf(" "); #define p(n) printf("%d",n); #define pln(n) printf("%d\n",n); #define s(n) scanf("%d",&n); #define ss(n) scanf("%s",n); #define ps(n) printf("%s",n); #define sld(n) scanf("%lld",&n); #define pld(n) printf("%lld",n); #define slf(n) scanf("%lf",&n); #define plf(n) printf("%lf",n); #define sc(n) scanf("%c",&n); #define pc(n) printf("%c",n); #define gc getchar(); #define ll long long #define re(n,a) memset(n,a,sizeof(n)); #define len(a) strlen(a) #define eps 1e-13 #define zero(x) (((x) > 0? (x):(-x)) < eps) using namespace std; char a[300005]; int b[300005]; int c[300005]; int bv[300005]; int cv[300005]; int main(){ int t; s(t) int n; while(t --){ s(n) gc; for(int i = 0; i <= n;i++){ b[i] = i; c[i] = i; bv[i] = 1; cv[i] = 1; } for(int i = 1; i <= n; i ++){ a[i] = gc; if(a[i] == 'L'){ b[i] = i-1; }else if(a[i] == 'R'){ c[i] = i-1; } } for(int i = n; i >= 1; i --){ if(b[i] == i){ cv[i] ++; bv[i-1] = cv[i]; }else{ bv[i] ++; cv[i-1] = bv[i]; } } /*for(int i = 0; i <= n; i ++){ p(b[i]) pk } pn for(int i = 0; i <= n; i ++){ p(c[i]) pk } pn*/ for(int i = 0; i <= n-1; i ++){ p(bv[i]) pk if(b[i+1] == i){ bv[i+1] = cv[i]; }else{ cv[i+1] = bv[i]; } } p(bv[n]) pk pn } return 0; }
