HDU 4747 Mex

 

Time Limit: 5000MS   Memory Limit: 65535KB   64bit IO Format: %I64d & %I64u

Description

Mex is a function on a set of integers, which is universally used for impartial game theorem. For a non-negative integer set S, mex(S) is defined as the least non-negative integer which is not appeared in S. Now our problem is about mex function on a sequence. 

Consider a sequence of non-negative integers {ai}, we define mex(L,R) as the least non-negative integer which is not appeared in the continuous subsequence from aL to aR, inclusive. Now we want to calculate the sum of mex(L,R) for all 1 <= L <= R <= n.

Input

The input contains at most 20 test cases. 
For each test case, the first line contains one integer n, denoting the length of sequence. 
The next line contains n non-integers separated by space, denoting the sequence. 
(1 <= n <= 200000, 0 <= ai <= 10^9) 
The input ends with n = 0.

Output

For each test case, output one line containing a integer denoting the answer.

Sample Input

3
0 1 3
5
1 0 2 0 1
0

Sample Output

5
24

Hint

 
For the first test case: mex(1,1)=1, mex(1,2)=2, mex(1,3)=2, mex(2,2)=0, mex(2,3)=0,mex(3,3)=0. 1 + 2 + 2 + 0 +0 +0 = 5.

Source

2013 ACM/ICPC Asia Regional Hangzhou Online

 

首先O(n)复杂度求出 mex(1,i)，可以知道 mex(i,i),mex(i,i+1)到 mex(i,n)是递增的。
然后使用线段树维护，需要不断删除前面的数。
若删掉第一个数 a[1]，那么在下一个 a[1]出现前的大于 a[1]的 mex 值都要变成 a[1]
然后需要线段树实现区间修改和区间求和。

 

 1 #include<iostream>
 2 #include<algorithm>
 3 #include<cstring>
 4 #include<cmath>
 5 #include<cstdio>
 6 #define ls l,mid,rt<<1
 7 #define rs mid+1,r,rt<<1|1
 8 #define warning keep silent
 9 using namespace std;
10 const int mxn=200020;
11 struct node{
12     long long sum;
13     int max;
14     int mark;
15 }t[mxn<<2];
16 int a[mxn];
17 int n;
18 long long ans=0;
19 int cnt[mxn];
20 int mex[mxn],pos[mxn];
21 int nxt[mxn];
22 void pushdown(int l,int r,int rt){
23     if(t[rt].mark!=-1){
24         t[rt<<1].mark=t[rt<<1|1].mark=t[rt].mark;
25         t[rt<<1].max=t[rt<<1|1].max=t[rt].mark;
26         int mid=(l+r)>>1;
27         t[rt<<1].sum=(long long)t[rt].mark*(mid-l+1);
28         t[rt<<1|1].sum=(long long)t[rt].mark*(r-mid);
29         t[rt].mark=-1;
30     }
31     return;
32 }
33 void Build(int l,int r,int rt){
34     t[rt].mark=-1;
35     if(l==r){
36         t[rt].max=t[rt].sum=mex[l];
37         return;
38     }
39     int mid=(l+r)>>1;
40     Build(ls);Build(rs);
41     t[rt].sum=t[rt<<1].sum+t[rt<<1|1].sum;
42     t[rt].max=max(t[rt<<1].max,t[rt<<1|1].max);
43     return;
44 }
45 void change(int v,int L,int R,int l,int r,int rt){
46     if(L<=l && r<=R){
47         t[rt].mark=v;
48         t[rt].max=v;
49         t[rt].sum=(long long)v*(r-l+1);
50         return;
51     }
52     pushdown(l,r,rt);
53     int mid=(l+r)>>1;
54     if(L<=mid)change(v,L,R,ls);
55     if(R>mid)change(v,L,R,rs);
56     t[rt].sum=t[rt<<1].sum+t[rt<<1|1].sum;
57     t[rt].max=max(t[rt<<1].max,t[rt<<1|1].max);
58     return;
59 }
60 int posi(int v,int l,int r,int rt){
61     if(l==r)return l;
62     int mid=(l+r)>>1;
63     pushdown(l,r,rt);
64     if(t[rt<<1].max>v)return posi(v,ls);
65     else return posi(v,rs);
66 }
67 int main(){
68     while(scanf("%d",&n) && n){
69         memset(t,0,sizeof t);
70         memset(cnt,0,sizeof cnt);
71         int i,j;
72         for(i=1;i<=n;i++){scanf("%d",&a[i]);if(a[i]>n)a[i]=n;};
73         int last=0;
74         for(i=1;i<=n;i++){
75             cnt[a[i]]=1;
76             while(cnt[last])last++;
77             mex[i]=last;
78         }
79         for(i=0;i<=n;i++)pos[i]=n+1;//
80         for(i=n;i;i--){//预处理，找到相同值的下一个出现位置
81             nxt[i]=pos[a[i]];
82             pos[a[i]]=i;
83         }
84         Build(1,n,1);
85         ans=0;
86         for(i=1;i<=n;i++){
87             ans+=t[1].sum;
88             change(0,i,i,1,n,1);
89             if(a[i]<t[1].max){
90                 int l=posi(a[i],1,n,1);
91                 int r=nxt[i]-1;
92                 if(l<=r)change(a[i],l,r,1,n,1);
93             }
94         }
95         printf("%lld\n",ans);
96     }
97     return 0;
98 }

 

posted @ 2016-08-21 22:34  SilverNebula  阅读(216)  评论(0)  编辑  收藏  举报
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