2015 ACM长春赛区签到题

星期天打的网络赛,虽然没我什么事(┬_┬)感觉差距好大。。。这个学期再不能贪玩了,好好学ACM,争取不拉队友的后腿。这道题是一道线段树的裸题,并没有什么好讲的,写个题解留个纪念。。

The Water Problem

Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 574    Accepted Submission(s): 460


Problem Description
In Land waterless, water is a very limited resource. People always fight for the biggest source of water. Given a sequence of water sources with a1,a2,a3,...,anrepresenting the size of the water source. Given a set of queries each containing 2 integers l and r, please find out the biggest water source between al and ar.
 
Input
First you are given an integer T(T10) indicating the number of test cases. For each test case, there is a number n(0n1000) on a line representing the number of water sources. n integers follow, respectively a1,a2,a3,...,an, and each integer is in {1,...,106}. On the next line, there is a number q(0q1000) representing the number of queries. After that, there will be q lines with two integers l and r(1lrn) indicating the range of which you should find out the biggest water source.
 
Output
For each query, output an integer representing the size of the biggest water source.
 
Sample Input
3 1 100 1 1 1 5 1 2 3 4 5 5 1 2 1 3 2 4 3 4 3 5 3 1 999999 1 4 1 1 1 2 2 3 3 3
 
Sample Output
100 2 3 4 4 5 1 999999 999999 1
 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 using namespace std;
 6 #define INF 0x7fffffff
 7 int maxv=-INF;
 8 struct node
 9 {
10     int l,r;
11     int maxv;
12     int mid()
13     {
14         return (l+r)/2;
15     }
16 }tree[800020];
17 void buildtree(int root,int l,int r)
18 {
19     tree[root].l=l;
20     tree[root].r=r;
21     tree[root].maxv=-INF;
22     if(l!=r)
23     {
24         buildtree(2*root+1,l,(l+r)/2);
25         buildtree(2*root+2,(l+r)/2+1,r);
26     }
27 }
28 void insert(int root,int i,int v)
29 {
30     if(tree[root].l==tree[root].r)
31     {
32         tree[root].maxv=v;
33         return;
34     }
35     tree[root].maxv=max(tree[root].maxv,v);
36     if(i<=tree[root].mid())
37         insert(root*2+1,i,v);
38     else
39         insert(root*2+2,i,v);
40 }
41 void query(int root,int s,int e)
42 {
43     if(maxv>=tree[root].maxv)
44         return;
45     if(tree[root].l==s&&tree[root].r==e)
46     {
47         maxv=max(tree[root].maxv,maxv);
48         return;
49     }
50     if(e<=tree[root].mid())
51         query(2*root+1,s,e);
52     else if(s>tree[root].mid())
53         query(2*root+2,s,e);
54     else
55     {
56         query(2*root+1,s,tree[root].mid());
57         query(2*root+2,tree[root].mid()+1,e);
58     }
59 }
60 int main()
61 {
62     int T,n,m,i,j;
63     cin>>T;
64     while(T--)
65     {
66         cin>>n;
67         buildtree(0,1,n);
68         for(i=1;i<=n;i++)
69         {
70             int h;
71             cin>>h;
72             insert(0,i,h);
73         }
74         cin>>m;
75         for(i=0;i<m;i++)
76         {
77             int s,e;
78             cin>>s>>e;
79             maxv=-INF;
80             query(0,s,e);
81             printf("%d
",maxv);
82         }
83     }
84     return 0;
85 }
View Code
原文地址:https://www.cnblogs.com/zero-zz/p/4811392.html