UVa 1608 Non-boring sequences (分治)

题意：给你一个长度为n序列，如果这个任意连续子序列的中都有至少出现一次的元素，那么就称这个序列是不无聊的，判断这个序列是不是无聊的。

析：首先如果整个序列中有一个只出过一次的元素，假设是第 p 个，那么我就可以看他左边和右边的序列是不是不无聊，也就是判断 1~p-1  和 p+1 ~ n，这可以用分治来进行处理。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define all 1,n,1
#define FOR(x,n)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 200000 + 10;
const LL mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
  return r > 0 && r <= n && c > 0 && c <= m;
}

int a[maxn];
int l[maxn], r[maxn];
map<int, int> mp;

bool dfs(int L, int R){
  if(L >= R)  return true;
  int t = R - L >> 1;
  for(int i = 0; i <= t; ++i){
    if(l[i+L] < L && r[i+L] > R)  return dfs(L, i+L-1) && dfs(i+1+L, R);
    if(l[R-i] < L && r[R-i] > R)  return dfs(L, R-i-1) && dfs(R-i+1, R);
  }
  return false;
}

int main(){
  int T;
  cin >> T;
  while(T--){
      scanf("%d", &n);
    mp.cl;
    for(int i = 1; i <= n; ++i){
      scanf("%d", a+i);
      if(mp.count(a[i]))  l[i] = mp[a[i]];
      else  l[i] = 0;
      mp[a[i]] = i;
    }
    mp.cl;
    for(int i = n; i; --i){
      if(mp.count(a[i]))  r[i] = mp[a[i]];
      else r[i] = n+1;
      mp[a[i]] = i;
    }
    puts(dfs(1, n) ? "non-boring" : "boring");
  }
  return 0;
}