HDU1698Just a Hook(线段树 + 区间修改 + 求和）

题目链接

分析：1-N区间内初始都是1，然后q个询问，每个询问修改区间【a，b】的值为2或3或者1，统计最后整个区间的和

本来想刷刷手速，结果还是写了一个小时，第一个超时，因为输出的时候去每个区间查找了，直接输出tree[1].value就可以了 =_=

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <cstdio>
 5 using namespace std;
 6 const int Max = 100000;
 7 int hook[Max + 10];
 8 struct node
 9 {
10     int l,r;
11     int value;
12     int tag;
13 };
14 node tree[3 * Max];
15 void buildTree(int left, int right, int k)
16 {
17     tree[k].l = left;
18     tree[k].r = right;
19     tree[k].tag = -1;
20     if(left == right)
21     {
22         tree[k].value = 1;
23         return;
24     }
25     int mid = (left + right ) / 2;
26     buildTree(left, mid, k * 2);
27     buildTree(mid + 1, right, k * 2 + 1);
28     tree[k].value = tree[k * 2].value + tree[k * 2 + 1].value;
29 }
30 void upDate(int left, int right, int k, int newp)
31 {
32     if(tree[k].l ==left && tree[k].r == right)
33     {
34         tree[k].value = (right - left + 1) * newp;
35         tree[k].tag = newp;
36         return;
37     }
38     if(tree[k].tag != -1)
39     {
40         tree[k * 2].tag = tree[k * 2 + 1].tag = tree[k].tag;
41         tree[k * 2].value = (tree[k * 2].r - tree[k * 2].l + 1) * tree[k].tag;
42         tree[k * 2 + 1].value = (tree[k * 2 + 1].r - tree[k * 2 + 1].l + 1) * tree[k].tag;
43         tree[k].tag = -1;
44     }
45     int mid = (tree[k].l + tree[k].r) / 2;
46     if(right <= mid)
47     {
48         upDate(left, right, k * 2, newp);
49     }
50     else if(mid < left)
51     {
52         upDate(left, right, k * 2 + 1, newp);
53     }
54     else
55     {
56         upDate(left, mid, k * 2, newp);
57         upDate(mid + 1, right, k * 2 + 1, newp);
58     }
59     tree[k].value = tree[k * 2].value + tree[k * 2 + 1].value;
60 }
61 int Search(int k)
62 {
63     if(tree[k].tag != -1)
64     {
65         tree[k * 2].tag = tree[k * 2 + 1].tag = tree[k].tag;
66         tree[k * 2].value = (tree[k * 2].r - tree[k * 2].l + 1) * tree[k].tag;
67         tree[k * 2 + 1].value = (tree[k * 2 + 1].r - tree[k * 2 + 1].l + 1) * tree[k].tag;
68         tree[k].tag = -1;
69     }
70     if(tree[k].l == tree[k].r)
71         return tree[k].value;
72     return Search(k * 2) + Search(k * 2 + 1);
73 }
74 int main()
75 {
76     int test,n;
77     scanf("%d", &test);
78     for(int t = 1; t <= test; t++)
79     {
80         int q,a,b,newp;
81         scanf("%d", &n);
82         buildTree(1, n, 1);
83         scanf("%d", &q);
84         for(int i = 1; i <= q; i++)
85         {
86             scanf("%d%d%d", &a, &b, &newp);
87             upDate(a, b, 1, newp);
88         }
89         printf("Case %d: The total value of the hook is %d.\n", t, tree[1].value);
90     }
91     return 0;
92 }
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posted @ 2016-03-03 15:05  zhaop  阅读(217)  评论(0)  编辑  收藏  举报