hdu 5316 Magician (线段树)

题目链接：

　　hdu 5316 Magician

题目描述：

　　有n个精灵，每个精灵都有一个魔法值，现在有两个操作：

　　(0, a, b)查询[a, b]序列中的一个完美序列的最大和，完美序列就是数组中相邻数字的下标奇偶性不同。

　　(1, a, b)更新下标为a的精灵魔法值为b。

对的，没错就是这个意思，但是比赛的时候就是题意卡的死死的有没有啊，就是各种不懂有没有啊！！！

  1 #include <cstdio>
  2 #include <cstring>
  3 #include <iostream>
  4 #include <algorithm>
  5 using namespace std;
  6 typedef long long LL;
  7 #define Max(a, b, c, d) max(max(a, b), max(c, d))
  8 const LL INF = 1000000000;
  9 const LL maxn = 400000;
 10 struct point
 11 {
 12     LL odd_e, odd_o, even_o, even_e;
 13     LL Maxn ()
 14     {
 15         return Max(odd_e, odd_o, even_e, even_o);
 16     }
 17 };
 18 struct node
 19 {
 20     LL l, r;
 21     point p;
 22     LL Mid ()
 23     {
 24         return (l+r)/2;
 25     }
 26 } tree[maxn];
 27 point Add (point x, point y)
 28 {
 29     point s;
 30     s.even_e = Max(x.even_e, y.even_e, x.even_e+y.odd_e,  x.even_o+y.even_e);
 31     s.even_o = Max(x.even_o, y.even_o, x.even_e+y.odd_o,  x.even_o+y.even_o);
 32     s.odd_e  = Max(x.odd_e,  y.odd_e,  x.odd_e +y.odd_e,  x.odd_o+y.even_e);
 33     s.odd_o  = Max(x.odd_o,  y.odd_o,  x.odd_o +y.even_o, x.odd_e+y.odd_o);
 34     return s;
 35 }
 36 void bulid (LL root, LL l, LL r)
 37 {
 38     tree[root].l = l;
 39     tree[root].r = r;
 40
 41     if (l == r)
 42     {
 43         tree[root].p.even_e = -INF;
 44         tree[root].p.even_o = -INF;
 45         tree[root].p.odd_e = -INF;
 46         tree[root].p.odd_o = -INF;
 47         LL num;
 48         scanf ("%lld", &num);
 49         if (l%2)
 50             tree[root].p.odd_o = num;
 51         else
 52             tree[root].p.even_e = num;
 53         return ;
 54     }
 55     bulid (2*root+1, l, tree[root].Mid());
 56     bulid (2*root+2, tree[root].Mid()+1, r);
 57     tree[root].p = Add (tree[2*root+1].p, tree[2*root+2].p);
 58 }
 59 void update (LL root, LL x, LL y)
 60 {
 61     if (tree[root].l == tree[root].r)
 62     {
 63         tree[root].p.even_e = -INF;
 64         tree[root].p.even_o = -INF;
 65         tree[root].p.odd_e = -INF;
 66         tree[root].p.odd_o = -INF;
 67         if ( x%2 == 0 )
 68             tree[root].p.even_e = y;
 69         else
 70             tree[root].p.odd_o = y;
 71         return ;
 72     }
 73     if (x <= tree[root].Mid())
 74         update(2*root+1, x, y);
 75     else
 76         update(2*root+2, x, y);
 77     tree[root].p = Add (tree[2*root+1].p, tree[2*root+2].p);
 78 }
 79 point query (LL root, LL l, LL r)
 80 {
 81     if (tree[root].l == l && tree[root].r == r)
 82         return tree[root].p;
 83     if (r <= tree[root].Mid())
 84         query (2*root+1, l, r);
 85     else if (tree[root].Mid() < l)
 86         query (2*root+2, l, r);
 87     else
 88     {
 89         point ls, rs;
 90         ls = query (2*root+1, l, tree[root].Mid());
 91         rs = query (2*root+2, tree[root].Mid()+1, r);
 92         return Add (ls, rs);
 93     }
 94 }
 95 int main ()
 96 {
 97     LL t;
 98     scanf ("%lld", &t);
 99
100     while (t --)
101     {
102         LL n, q;
103         scanf ("%lld %lld", &n, &q);
104         bulid(0, 1, n);
105         while (q --)
106         {
107             LL type, a, b;
108             scanf ("%lld %lld %lld", &type, &a, &b);
109             if (type == 0)
110             {
111                 point s = query (0, a, b);
112                 printf ("%lld\n", s.Maxn());
113             }
114             else
115                 update (0, a, b);
116         }
117     }
118     return 0;
119 }