• 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;
    }
    

      

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