CodeForces 279C Ladder (RMQ + dp)

题意:给定一个序列,每次一个询问,问某个区间是不是先增再降的。

析:首先先取处理以 i 个数向左能延伸到哪个数,向右能到哪个数,然后每次用RQM来查找最大值,分别向两边延伸,是否是覆盖区间。

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>
#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 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 LL LNF = 1e16;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 10;
const int 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], f[maxn], g[maxn];
int dp[maxn][20];

inline int Max(int l, int r){ return a[l] >= a[r] ? l : r; }

void init(){
  for(int i = 0; i < n; ++i)  dp[i][0] = i;
  for(int j = 1; (1<<j) <= n; ++j)
    for(int i = 0; i + (1<<j) <= n; ++i)
      dp[i][j] = Max(dp[i][j-1], dp[i+(1<<j-1)][j-1]);
}

int query(int l, int r){
  int k = log(r-l+1.0) / log(2.0);
  return Max(dp[l][k], dp[r-(1<<k)+1][k]);
}

int main(){
  scanf("%d %d", &n, &m);
  f[0] = 0;  g[n-1] = n-1;
  for(int i = 0; i < n; ++i)  scanf("%d", a+i);
  for(int i = 1; i < n; ++i)  f[i] = a[i] >= a[i-1] ? f[i-1] : i;
  for(int i = n-2; i >= 0; --i)  g[i] = a[i] >= a[i+1] ? g[i+1] : i;
  init();
  while(m--){
    int l, r;
    scanf("%d %d", &l, &r);
    --l,  --r;
    int k = query(l, r);
    if(f[k] <= l && g[k] >= r)  puts("Yes");
    else   puts("No");
  }
  return 0;
}

  

原文地址:https://www.cnblogs.com/dwtfukgv/p/6848249.html