spoj 1716 Can you answer these queries III（线段树）

和I相比有了单点更新，所以不能只记录一个前缀和，而是要在线段树上多维护一个sum，表示这个结点的区间和，然后其他的就和I一样了。

  1 #include <iostream>
  2 #include <cstring>
  3 #include <cstdio>
  4 using namespace std;
  5 
  6 const int N = 50001;
  7 int a[N];
  8 
  9 struct Node 
 10 {
 11     int l, r, sum;
 12     int maxl, maxr, maxn;
 13 } node[N << 2];
 14 
 15 void pushup( int i )
 16 {
 17     int lc = i << 1, rc = lc | 1;
 18     node[i].sum = node[lc].sum + node[rc].sum;
 19     node[i].maxl = max( node[lc].maxl, node[lc].sum + node[rc].maxl );
 20     node[i].maxr = max( node[rc].maxr, node[rc].sum + node[lc].maxr );
 21     node[i].maxn = max( node[lc].maxn, node[rc].maxn );
 22     node[i].maxn = max( node[i].maxn, node[lc].maxr + node[rc].maxl );
 23 }
 24 
 25 void build( int i, int l, int r )
 26 {
 27     node[i].l = l, node[i].r = r;
 28     if ( l == r )
 29     {
 30         node[i].sum = node[i].maxl = node[i].maxr = node[i].maxn = a[l];
 31         return ;
 32     }
 33     int lc = i << 1, rc = lc | 1, mid = ( l + r ) >> 1;
 34     build( lc, l, mid );
 35     build( rc, mid + 1, r );
 36     pushup(i);
 37 }
 38 
 39 void update( int i, int pos, int val )
 40 {
 41     if ( node[i].l == pos && node[i].r == pos )
 42     {
 43         node[i].sum = node[i].maxl = node[i].maxr = node[i].maxn = val;
 44         return ;
 45     }
 46     int mid = ( node[i].l + node[i].r ) >> 1;
 47     if ( pos <= mid )
 48     {
 49         update( i << 1, pos, val );
 50     }
 51     else
 52     {
 53         update( i << 1 | 1, pos, val );
 54     }
 55     pushup(i);
 56 }
 57 
 58 Node query( int i, int l, int r )
 59 {
 60     if ( node[i].l == l && node[i].r == r )
 61     {
 62         return node[i];
 63     }
 64     int lc = i << 1, rc = lc | 1, mid = ( node[i].l + node[i].r ) >> 1;
 65     if ( r <= mid )
 66     {
 67         return query( lc, l, r );
 68     }
 69     else if ( l > mid )
 70     {
 71         return query( rc, l, r );
 72     }
 73     else
 74     {
 75         Node ln = query( lc, l, mid ), rn = query( rc, mid + 1, r ), res;
 76         res.sum = ln.sum + rn.sum;
 77         res.maxl = max( ln.maxl, ln.sum + rn.maxl );
 78         res.maxr = max( rn.maxr, rn.sum + ln.maxr );
 79         res.maxn = max( ln.maxn, rn.maxn );
 80         res.maxn = max( res.maxn, ln.maxr + rn.maxl );
 81         return res;
 82     }
 83 }
 84 
 85 int main ()
 86 {
 87     int n, m;   
 88     while ( scanf("%d", &n) != EOF )
 89     {
 90         for ( int i = 1; i <= n; i++ )
 91         {
 92             scanf("%d", a + i);
 93         }
 94         build( 1, 1, n );
 95         scanf("%d", &m);
 96         while ( m-- )
 97         {
 98             int op, x, y;
 99             scanf("%d%d%d", &op, &x, &y);
100             if ( op == 0 )
101             {
102                 update( 1, x, y );
103             }
104             else
105             { 
106                 Node tmp = query( 1, x, y );
107                 printf("%d\n", tmp.maxn);
108             }
109         }
110     }
111     return 0;
112 }

 

posted @ 2015-07-31 08:47  hxy_has_been_used  阅读(178)  评论(0)  编辑  收藏  举报