HDU 2836 Traversal 简单DP + 树状数组

题意：给你一个序列，问相邻两数高度差绝对值小于等于H的子序列有多少个。

dp[i]表示以i为结尾的子序列有多少，易知状态转移方程为：dp[i] = sum( dp[j] ) + 1;( abs( height[i] - height[j] ) <= H )

由abs( height[i] - height[j] ) <= H 可得 height[i] - H <= height[j] <= height[i] + H

将序列中的数离散化，每个height对应一个id, 用树状数组求区间[ height[i] - H的id, height[i] + H的id ]内dp[j]的和，并且每次把新得到的dp[i]更新到树状数组中height[i]的id对应的位置。

 #include <cstdio>
 #include <cstring>
 #include <cstdlib>
 #include <algorithm>

 using namespace std;

 const int MAXN = ;
 const int MOD = ;

 int dp[MAXN];
 int height[MAXN];
 int num[MAXN];
 int C[MAXN];
 int n, d;

 int lowbit( int x )
 {
     return x & ( -x );
 }

 int Query( int x )
 {
     int res = ;
     while ( x >  )
     {
         res += C[x];
         res %= MOD;
         x -= lowbit(x);
     }
     return res;
 }

 void Add( int x, int v )
 {
     while ( x <= n )
     {
         C[x] += v;
         C[x] %= MOD;
         x += lowbit(x);
     }
     return;
 }

 int main()
 {
     while ( ~scanf( "%d%d", &n, &d ) )
     {
         for ( int i = ; i <= n; ++i )
         {
             scanf( "%d", &height[i] );
             num[i] = height[i];
         }

         sort( num + , num + n +  );
         int cnt = unique( num + , num + n +  ) - num - ;

         memset( C, , sizeof(C) );
         int ans = ;
         dp[] = ;
         for ( int i = ; i <= n; ++i )
         {
             int id = lower_bound( num + , num + cnt + , height[i] ) - num;
             int left = lower_bound( num + , num + cnt + , height[i] - d ) - num;
             int right = upper_bound( num + , num + cnt + , height[i] + d ) - num - ;
             dp[i] = ( Query( right ) - Query( left -  ) +  ) % MOD;
             ans += dp[i];
             Add( id, dp[i] );
         }

         if ( ans >= n ) ans -= n;
         else ans = ;
         printf("%d\n", ans % MOD );
     }
     return ;
 }