CodeForces 279C Ladder (RMQ + dp)
Posted dwtfukgv
tags:
篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了CodeForces 279C Ladder (RMQ + dp)相关的知识,希望对你有一定的参考价值。
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; }
以上是关于CodeForces 279C Ladder (RMQ + dp)的主要内容,如果未能解决你的问题,请参考以下文章