HDU 4747 Mex

Problem 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

30 1 351 0 2 0 10

 

Sample Output

524
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

我一开始看到这道题，只能想出n^2的算法（尴尬）。

于是看了题解，如果将Σ看作n个同起点的mex求和，所有同起点的mex是单调不减的（重要）。

如何转移呢，对于样例2

1 0 2 0 1

将1删除就表示将mex(2...x)都变成不大于1的数，x即1下一次出现的位置-1。

可以运用单调性，查找，预处理类似于链表。

剩下的就是线段树区间修改，维护和，最值。

注意对于a[i]大于n，可以直接舍去，比较方便。

#include<algorithm>#include<iostream>#include<cstring>#include<cstdio>using namespace std;const int N=200005;int n,data[N],t[N],mex[N],pos[N],nxt[N];long long ans;struct node{int l,r,tag,mx;long long sum;}a[4*N];void build(int num,int l,int r){a[num].l=l,a[num].r=r,a[num].tag=-1;if(l==r){a[num].sum=(long long)mex[l],a[num].mx=mex[l];return ;}int mid=(l+r)/2;build(2*num,l,mid);build(2*num+1,mid+1,r);a[num].sum=a[2*num].sum+a[2*num+1].sum;a[num].mx=max(a[2*num].mx,a[2*num+1].mx);}void update(int num){if(a[num].l!=a[num].r){a[2*num].tag=a[2*num+1].tag=a[num].tag;a[2*num].sum=(long long)(a[2*num].r-a[2*num].l+1)*a[2*num].tag;a[2*num+1].sum=(long long)(a[2*num+1].r-a[2*num+1].l+1)*a[2*num+1].tag;a[2*num].mx=a[2*num+1].mx=a[num].mx;}a[num].tag=-1;}void chg(int num,int l,int r,int x)//将区间[l,r]修改为x {if(a[num].l>r||a[num].r<l)return ;if(a[num].l>=l&&a[num].r<=r){a[num].tag=x,a[num].mx=x;a[num].sum=(long long)(a[num].r-a[num].l+1)*x;return ;}if(a[num].tag>=0)update(num);chg(2*num,l,r,x);chg(2*num+1,l,r,x);a[num].sum=a[2*num].sum+a[2*num+1].sum;a[num].mx=max(a[2*num].mx,a[2*num+1].mx);}int query(int num,int x)//查找第一个等于x的位置 {if(a[num].l==a[num].r)return a[num].l;if(a[num].tag>=0)update(num);if(a[2*num].mx>=x)return query(2*num,x);return query(2*num+1,x);}int main(){while(scanf("%d",&n)&&n){ans=0;memset(t,0,sizeof(t));for(int i=1;i<=n;i++){scanf("%d",&data[i]);if(data[i]>n)data[i]=n;}for(int i=1;i<=n;i++)if(data[i]<=n){t[data[i]]++;for(int j=mex[i-1];j<=n;j++)if(t[j]==0){mex[i]=j;break;}}elsemex[i]=mex[i-1];build(1,1,n);for(int i=0;i<=n;i++)pos[i]=n+1;for(int i=n;i>=1;i--)if(data[i]<=n){nxt[i]=pos[data[i]];pos[data[i]]=i;}ans+=a[1].sum;for(int i=1;i<=n;i++){chg(1,i,i,0);if(data[i]<=a[1].mx){int v=query(1,data[i]);if(v<=nxt[i]-1)chg(1,v,nxt[i]-1,data[i]);}ans+=a[1].sum;}printf("%lld\n",ans);}return 0;}