[Usaco2007 Jan]Balanced Lineup

嘟嘟嘟

一道RMQ板子题，分别维护最大值和最小值，不解释。

 #include<cstdio>
 #include<iostream>
 #include<algorithm>
 #include<cmath>
 #include<cstring>
 #include<cstdlib>
 #include<cctype>
 #include<stack>
 #include<queue>
 #include<vector>
 using namespace std;
 #define enter puts("")
 #define space putchar(' ')
 #define Mem(a) memset(a, 0, sizeof(a))
 typedef long long ll;
 typedef double db;
 const int INF = 0x3f3f3f3f;
 const db eps  =1e-;
 const int maxn = 5e4 + ;
 inline ll read()
 {
     ll ans = ;
     char ch = getchar(), last = ' ';
     while(!isdigit(ch)) {last = ch; ch = getchar();}
     while(isdigit(ch)) {ans = ans *  + ch - ''; ch = getchar();}
     if(last == '-') ans = -ans;
     return ans;
 }
 inline void write(ll x)
 {
     if(x < ) putchar('-'), x = -x;
     if(x >= ) write(x / );
     putchar(x %  + '');
 }
 
 int n, m, a[maxn];
 
 int dp[maxn][][], b[maxn];
 void RMQ()
 {
     for(int i = ; i <= n; ++i) dp[i][][] = dp[i][][] = a[i];
     for(int j = ; ( << j) <= n; ++j)
         for(int i = ; i + ( << j) -  <= n; ++i)
         {
             dp[i][j][] = max(dp[i][j - ][], dp[i + ( << (j - ))][j - ][]);
             dp[i][j][] = min(dp[i][j - ][], dp[i + ( << (j - ))][j - ][]);
         }
     int x = ;
     for(int i = ; i <= n; ++i)
     {
         b[i] = x;
         if(( << (x + )) <= (i + )) x++;
     }
 }
 int query(int L, int R, bool flag)
 {
     int k = b[R - L + ];
     if(flag) return min(dp[L][k][flag], dp[R - ( << k) + ][k][flag]);    //应该是R - (1 << k) + 1,不是R - k + 1……
     else return max(dp[L][k][flag], dp[R - ( << k) + ][k][flag]);
 }
 
 int main()
 {
     n = read(); m = read();
     for(int i = ; i <= n; ++i) a[i] = read();
     RMQ();
     for(int i = ; i <= m; ++i)
     {
         int L = read(), R = read();
         write(query(L, R, ) - query(L, R, )); enter;
     }
     return ;
 }