GTY‘s gay friends
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 264 Accepted Submission(s): 57
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 a i
, 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≤100000
), 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 i th
number a i
( 1≤a i ≤n
) indicates GTY‘s i th
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..r−l+1]
in [l,r]
, 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
转自官方题解：http://bestcoder.hdu.edu.cn/
1003 GTY‘s gay friends 一个区间是排列只需要区间和为len(len+1)2 (len 为区间长度),且互不相同,对于第一个问题我们用前缀和解决,对于第二个问题,预处理每个数的上次出现位置,记它为pre,互不相同即区间中pre的最大值小于左端点,使用线段树或Sparse Table即可在O(n)/O(nlogn) 的预处理后 O(logn)/O(1) 回答每个询问.不过我们还有更简单的hash做法,对于[1..n] 中的每一个数随机一个64位无符号整型作为它的hash值,一个集合的hash值为元素的异或和,预处理[1..n] 的排列的hash和原序列的前缀hash异或和,就可以做到线性预处理,O(1) 回答询问.
1 #include<iostream>
2 #include<cstring>
3 #include<cstdlib>
4 #include<cstdio>
5 #include<algorithm>
6 #include<cmath>
7 #include<queue>
8 #include<map>
9 #include<set>
10 #include<stack>
11 #include<string>
12
13 #define N 1000005
14 #define M 105
15 #define mod 10000007
16 //#define p 10000007
17 #define mod2 1000000000
18 #define ll long long
19 #define LL long long
20 #define eps 1e-6
21 #define inf 100000000
22 #define maxi(a,b) (a)>(b)? (a) : (b)
23 #define mini(a,b) (a)<(b)? (a) : (b)
24
25 using namespace std;
26
27 ll n,m;
28 ll a[N];
29 ll f[N];
30 ll pr[N];
31 ll sum[N];
32
33 ll mapre[4*N];
34
35 ll build(ll i,ll l,ll r)
36 {
37 if(l==r){
38 mapre[i]=pr[l];
39 //printf(" i=%I64d l=%I64d mapre=%I64d pr=%I64d\n",i,l,mapre[i],pr[l]);
40 return mapre[i];
41 }
42 ll mid=(l+r)/2;
43 ll lm=build(2*i,l,mid);
44 ll rm=build(2*i+1,mid+1,r);
45 mapre[i]=max(lm,rm);
46 //printf(" i=%I64d mapre=%I64d\n",i,mapre[i]);
47 return mapre[i];
48 }
49
50 ll query(ll i,ll L,ll R,ll l,ll r)
51 {
52 if(l>=L && r<=R){
53 return mapre[i];
54 }
55 ll mid=(l+r)/2;
56 ll lm,rm;
57 lm=0;rm=0;
58 if(L<=mid){
59 lm=query(i*2,L,R,l,mid);
60 }
61 if(R>mid){
62 rm=query(i*2+1,L,R,mid+1,r);
63 }
64 return max(lm,rm);
65 }
66
67 void ini()
68 {
69 ll i;
70 sum[0]=0;
71 memset(f,0,sizeof(f));
72 for(i=1;i<=n;i++){
73 scanf("%I64d",&a[i]);
74 sum[i]=sum[i-1]+a[i];
75 }
76 //for(i=0;i<=n;i++){
77 // printf(" i=%I64d sum=%I64d\n",i,sum[i]);
78 // }
79 for(i=1;i<=n;i++){
80 pr[i]=f[ a[i] ];
81 f[ a[i] ]=i;
82 }
83 // for(i=0;i<=n;i++){
84 // printf(" i=%I64d pr=%I64d\n",i,pr[i]);
85 // }
86 build(1,1,n);
87 }
88
89 void solve()
90 {
91 ll x,y;
92 ll ss;
93 ll len;
94 ll qu;
95 while(m--){
96 scanf("%I64d%I64d",&x,&y);
97 len=y-x+1;
98 ss=sum[y]-sum[x-1];
99 //printf(" ss=%I64d sum=%I64d\n",ss,len*(len+1)/2);
100 if(ss==(len*(len+1)/2)){
101 qu=query(1,x,y,1,n);
102 if(qu<x){
103 printf("YES\n");
104 }
105 else{
106 printf("NO\n");
107 }
108 }
109 else{
110 printf("NO\n");
111 }
112 }
113 }
114
115 void out()
116 {
117
118 }
119
120 int main()
121 {
122 //freopen("data.in","r",stdin);
123 //freopen("data.out","w",stdout);
124 //scanf("%d",&T);
125 //for(int ccnt=1;ccnt<=T;ccnt++)
126 // while(T--)
127 //scanf("%d%d",&n,&m);
128 while(scanf("%I64d%I64d",&n,&m)!=EOF)
129 {
130 ini();
131 solve();
132 out();
133 }
134 return 0;
135 }
BestCoder Round #29 1003 （hdu 5172） GTY's gay friends [线段树 判不同 预处理 好题]