HDU 1698 Just a Hook 解题报告

题意：给定一个长度为n的区间，动态改变某个区间的值，问你最后整个区间的合是多少

解题思路：第一个区间覆盖的线段树问题，很自豪自己YY出了延迟标记（覆盖需要利用的延迟标记）哈哈，就是利用的延迟标记覆盖区间，然后树根的值就是整个区间的值，每次覆盖后pushup就行

解题代码：

 1 #include <stdio.h>
 2 #include <string.h>
 3 #include <stdlib.h>
 4 #define MAXN 100005
 5 struct node
 6 {
 7     int left ,right,mid;
 8     int num ;
 9     int cover;
10 } tree[MAXN*4];
11 int L(int c)
12 {
13     return 2*c;
14 }
15 int R(int c)
16 {
17     return 2*c + 1;
18 }
19 void up(int c)
20 {
21     tree[c].num = tree[L(c)].num + tree[R(c)].num;
22 }
23 void build(int c, int p ,int v )
24 {
25     tree[c].left = p ;
26     tree[c].right = v ;
27     tree[c].mid = (p+v)/2;
28     tree[c].num = (v-p+1) ;
29     tree[c].cover = 1 ;
30     if(p == v)
31     {
32         return ;
33     }
34     build(L(c),p,tree[c].mid);
35     build(R(c),tree[c].mid+1,v);
36 }
37 
38 void update(int c, int p , int v , int value)
39 {
40     if(p <= tree[c].left && v >= tree[c].right )
41     {
42         tree[c].num = (tree[c].right - tree[c].left+1) * value ;
43         tree[c].cover = value;
44         return ;
45     }
46     if(tree[c].cover != 0 )
47     {
48         tree[L(c)].cover = tree[c].cover;
49         tree[L(c)].num =(tree[L(c)].right - tree[L(c)].left+1) * tree[L(c)].cover;
50         tree[R(c)].cover = tree[c].cover;
51         tree[R(c)].num =(tree[R(c)].right - tree[R(c)].left+1) * tree[R(c)].cover;
52         tree[c].cover = 0 ;
53     }
54     if(v <= tree[c].mid)  update(L(c),p,v,value);
55     else if(p > tree[c].mid) update(R(c),p,v,value);
56     else
57     {
58         update(L(c),p,tree[c].mid,value);
59         update(R(c),tree[c].mid+1,v,value);
60     }
61     up(c);
62 }
63 int main()
64 {
65     int T ;
66     scanf("%d",&T);
67     for(int CASE = 1; CASE <= T ; CASE ++  )
68     {
69         int n ;
70         scanf("%d",&n);
71         build(1,1,n);
72         int m ;
73         scanf("%d",&m);
74         for(int i = 1; i <= m; i ++)
75         {
76             int a,b,value;
77             scanf("%d %d %d",&a,&b,&value);
78             update(1,a,b,value);
79             // printf("%d %d\n",tree[2].num,tree[3].num);
80         }
81 
82         printf("Case %d: The total value of the hook is %d.\n",CASE,tree[1].num);
83     }
84     return  0 ;
85 }

View Code

ps：一看第一46ms，本来以为自己 1300 ms+是线段树写搓了，用胡浩的代码交上去既然也是1000+ , 应该是有什么更快的数据结构

posted on 2013-07-31 20:41 dark_dream 阅读(226) 评论(0) 编辑收藏举报