HDU ACM 4027 Can you answer these queries?(线段树)

http://acm.hdu.edu.cn/showproblem.php?pid=4027

题意:有N条船,每条船都有各自的耐久度.有一种武器能攻击一个范围内的船.还要能计算一定范围内所有船剩余耐久的和.

　　耐久度去整数.耐久度的和不大于263

思路:每次更改都要更新范围内每条船的耐久度.

　　在攻击一定次数后,剩余耐久度肯定为1

　　所以结点中所有子节点耐久度均为1,则不需要继续更新

　　即发现Tree[o].sum == r - l + 1;

　　则不需要更新其子节点

 1 #include <iostream>
 2 #include <math.h>
 3 using namespace std;
 4 #define ol (o*2)
 5 #define or (o*2+1)
 6 const int MAX = 100000 + 10;
 7 struct Node{
 8     int l;
 9     int r;
10     __int64 sum;
11 }Tree[MAX*4];
12 void BuildTree(int o,int l,int r){
13     Tree[o].l = l;
14     Tree[o].r = r;
15     if(l == r){
16         scanf("%I64d",&Tree[o].sum);
17         return;
18     }
19     int mid = (l + r)/2;
20     BuildTree(ol,l,mid);
21     BuildTree(or,mid+1,r);
22     Tree[o].sum = Tree[ol].sum + Tree[or].sum;
23 }
24 void Update(int o,int l,int r){
25     if(Tree[o].sum == r - l + 1){//**********
26         return;
27     }
28     if(Tree[o].l == Tree[o].r){
29         Tree[o].sum = (__int64)sqrt(1.0*Tree[o].sum);
30         return;
31     }
32     int mid = (Tree[o].l + Tree[o].r)/2;
33     if(l > mid){
34         Update(or,l,r);
35     }
36     else{
37         if(r <= mid){
38             Update(ol,l,r);
39         }
40         else{
41             Update(ol,l,mid);
42             Update(or,mid+1,r);
43         }
44     }
45     Tree[o].sum = Tree[ol].sum + Tree[or].sum;
46 }
47 __int64 Query(int o,int l,int r){
48     if(Tree[o].l == l && Tree[o].r == r){
49         return Tree[o].sum;
50     }
51     int mid = (Tree[o].l + Tree[o].r)/2;
52     if(l > mid){
53         return Query(or,l,r);
54     }
55     else{
56         if(r <= mid){
57             return Query(ol,l,r);
58         }
59         else{
60             return Query(ol,l,mid) + Query(or,mid+1,r);
61         }
62     }
63 }
64 int main(){
65     int n;
66     int Case = 0;
67     while(cin>>n){
68         memset(Tree,0,sizeof(Tree));
69         Case++;
70         printf("Case #%d:\n",Case);
71         BuildTree(1,1,n);
72         int m;
73         cin>>m;
74         while(m--){
75             int k,x,y;
76             cin>>k>>x>>y;
77             if(x > y){
78                 swap(x,y);
79             }
80             if(k == 0){
81                 Update(1,x,y);
82             }
83             else{
84                 printf("%I64d\n",Query(1,x,y));
85             }
86         }
87         printf("\n");
88     }
89     return 0;
90 }

 

posted @ 2013-03-25 12:16  zx雄  阅读(154)  评论(0)  编辑  收藏  举报