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HDU 5172 GTY's gay friends 线段树

时间:2015-02-16 18:06:23      阅读:250      评论:0      收藏:0      [点我收藏+]

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GTY‘s gay friends

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

【Problem Description】
GTY has n gay friends. To manage them conveniently, every morning he ordered all his gay friends to stand in a line. Every gay friend has a characteristic value ai , to express how manly or how girlish he is. You, as GTY‘s assistant, have to answer GTY‘s queries. In each of GTY‘s queries, GTY will give you a range [l,r] . Because of GTY‘s strange hobbies, he wants there is a permutation [1..r−l+1] in [l,r]. You need to let him know if there is such a permutation or not.
 
【Input】
Multi test cases (about 3) . The first line contains two integers n and m ( 1≤n,m≤1000000 ), indicating the number of GTY‘s gay friends and the number of GTY‘s queries. the second line contains n numbers seperated by spaces. The ith number ai ( 1≤ai≤n ) indicates GTY‘s ith gay friend‘s characteristic value. The next m lines describe GTY‘s queries. In each line there are two numbers l and r seperated by spaces ( 1≤l≤r≤n ), indicating the query range.
 
【Output】
For each query, if there is a permutation [1..rl+1]技术分享
in [l,r]技术分享
, print ‘YES‘, else print ‘NO‘.
 
【Sample Input】
8 5 
2 1 3 4 5 2 3 1 
1 3 
1 1 
2 2 
4 8 
1 5 
3 2 
1 1 1 
1 1 
1 2

【Sample Output】

YES 
NO 
YES 
YES 
YES 
YES
NO

 

【题意】

给出一个数列,询问连续的从l开始到r为止的数是否刚好能够组成从1开始到r-l+1的数列。

 
【分析】
每一次询问都是一个区间询问。
对于每一个区间询问,需要判断区间内的数是否刚好可以组成1到k的连续数列,主要的判断标准有两个:
1.区间数字的总和与(1+k)*k/2相等;
2.保证区间内所有数都只出现一次。
 
第一个可以在读入数据时用前缀和解决。
第二个就要用到线段树了,读入时预处理记录下与当前数相同的数最近一次出现的位置。询问l~r的区间时,检索每个数的最近出现位置位置,若得到的所有结果都在区间左端的左边,那就说明区间中所有的数都是不重复出现的,则满足条件。这里就是用线段树判断区间最大值小于区间左端的过程了。
 
技术分享
 1 /* ***********************************************
 2 MYID    : Chen Fan
 3 LANG    : G++
 4 PROG    : HDU5172
 5 ************************************************ */
 6 
 7 #include <iostream>
 8 #include <cstdio>
 9 #include <cstring>
10 #include <algorithm>
11 
12 using namespace std;
13 
14 const int N=1e6+10;
15 
16 int last[N],a,sum[N];
17  
18 typedef struct treetyp
19 {
20     int a,b,l,r,data;
21 } treetype;
22 treetype tree[2*N];
23 int treetail;
24 
25 void maketree(int l,int r)
26 {
27     treetail++;
28     int now=treetail;
29     tree[now].a=l;
30     tree[now].b=r;
31     if (l<r)
32     {
33         tree[now].l=treetail+1;
34         maketree(l,(l+r)/2);
35         tree[now].r=treetail+1;
36         maketree((l+r)/2+1,r);
37     }
38 }
39 
40 void add(int n,int i,int data)
41 {
42     if (tree[n].data<data) tree[n].data=data;
43     if (i==tree[n].a&&i==tree[n].b) return ;
44     else if (i<=(tree[n].a+tree[n].b)/2) add(tree[n].l,i,data);
45          else add(tree[n].r,i,data);
46 }
47 
48 int res;
49 
50 void search(int n,int a,int b)
51 {
52     if (tree[n].a>=a&&tree[n].b<=b)
53     {
54         if (res<tree[n].data) res=tree[n].data;
55         return ;
56     }
57     if (tree[n].a==tree[n].b) return ;
58     if (a<=(tree[n].a+tree[n].b)/2) search(tree[n].l,a,b);
59     if (b>=(tree[n].a+tree[n].b)/2+1) search(tree[n].r,a,b);
60 }
61 
62 int main()
63 {
64     int n,m;
65     while(scanf("%d%d",&n,&m)==2)
66     {
67         memset(sum,0,sizeof(sum));
68         memset(last,0,sizeof(last));
69 
70         treetail=0;
71         maketree(1,n);
72         for (int i=1;i<=n;i++) 
73         {
74             scanf("%d",&a);
75             sum[i]=sum[i-1]+a;
76             add(1,i,last[a]);
77             last[a]=i;
78         }
79 
80         for (int i=1;i<=m;i++)
81         {
82             int l,r;
83             scanf("%d%d",&l,&r);
84             if ((r-l+1)*(r-l+2)/2==sum[r]-sum[l-1])
85             {
86                 res=0;
87                 search(1,l,r);
88                 if (res<l) printf("YES\n");
89                 else printf("NO\n");
90             } else printf("NO\n");
91         }
92     }
93 
94     return 0;
95 }
View Code

 

HDU 5172 GTY's gay friends 线段树

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原文地址:http://www.cnblogs.com/jcf94/p/4294284.html

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