FZu Problem 2236 第十四个目标 (线段树 + dp)

题目链接：

　　FZu  Problem 2236 第十四个目标

题目描述：

　　给出一个n个数的序列，问这个序列内严格递增序列有多少个？不要求连续

解题思路：

　　又遇到了用线段树来优化dp的题目，线段树节点里面保存所表达区间里面的方案数。先离散化序列(升序排列)，建树，然后按照没有sort前的顺序向线段树里面加点，每次查询小于该数的方案数目+1, 就是当前节点插入进去能影响的方案数目。在线段树对应位置加上新增加的数目即可。

 #include <cstdio>
 #include <queue>
 #include <stack>
 #include <cmath>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 using namespace std;

 #define lson 2*root
 #define rson 2*root+1
 typedef __int64 LL;
 const LL mod = ;
 const LL INF= 1e14+;
 const int maxn = ;

 struct node
 {
     int l, r;
     LL num;
     int mid ()
     {
         return (l + r) / ;
     }
 } tree[maxn*];
 LL a[maxn], b[maxn];

 void build (int root, int l, int r)
 {
     tree[root].l = l;
     tree[root].r = r;
     tree[root].num = ;
     if (l == r)
         return ;

     build (lson, l, tree[root].mid());
     build (rson, tree[root].mid()+, r);
 }
 void Insert (int root, LL x, int y)
 {
     if (tree[root].l == tree[root].r && tree[root].l == y)
     {
         tree[root].num =(tree[root].num + x) % mod;
         return ;
     }

     if (tree[root].mid() >= y)
         Insert (lson, x, y);
     else
         Insert (rson, x, y);
     tree[root].num = (tree[lson].num + tree[rson].num) % mod;
 }
 LL query (int root, int l, int r)
 {
     if (tree[root].l == l && tree[root].r == r)
         return tree[root].num;

     if (tree[root].mid() >= r)
         return query (lson, l, r);
     else if (tree[root].mid() < l)
         return query (rson, l, r);
     else
     {
         LL num = ;
         num += query (lson, l, tree[root].mid());
         num += query (rson, tree[root].mid()+, r);
         return num % mod;
     }
 }

 int main ()
 {
     int n;

     while (scanf ("%d", &n) != EOF)
     {
         for (int i=; i<n; i++)
         {
             scanf ("%I64d", &a[i]);
             b[i] = a[i];
         }

         sort (a, a+n);
         int m = unique (a, a+n) - a;

         LL ans = ;
         build (, , m);
         for (int i=; i<n; i++)
         {
             LL nu, tmp = lower_bound (a, a+m, b[i]) - a;
             if (tmp == )
                 nu = ;
             else
                 nu = query (, , tmp-);

             Insert (, nu+, tmp);
         }

         printf ("%I64d\n", query (, , m));
     }
     return ;
 }
 ///4 1 2 2 3