HDOJ 4747 Mex

非常好的线段树题。。。。此题必定会火。。。。。

Mex

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 955    Accepted Submission(s): 320

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

3
0 1 3
5
1 0 2 0 10

 

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

 

Recommend

liuyiding

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <map>

#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1

using namespace std;

const int maxn=200005;

typedef long long int LL;

int a[maxn],mex[maxn],next[maxn],mx[maxn<<2],sig[maxn<<2],cnt;
LL sum[maxn<<2];
bool vis[maxn];

void pushUP(int rt)
{
    mx[rt]=max(mx[rt<<1],mx[rt<<1|1]);
    sum[rt]=sum[rt<<1]+sum[rt<<1|1];
}

void pushDOWN(int rt,int len)
{
    if(~sig[rt])
    {
        sig[rt<<1]=sig[rt<<1|1]=sig[rt];
        mx[rt<<1]=mx[rt<<1|1]=sig[rt];
        sum[rt<<1]=sig[rt]*(len-len>>1);
        sum[rt<<1|1]=sig[rt]*(len>>1);
        sig[rt]=-1;
    }
}

void build(int l,int r,int rt)
{
    if(l==r)
    {
        mx[rt]=mex[cnt];
        sum[rt]=mex[cnt++];
        return ;
    }
    int m=(r+l)>>1;
    build(lson);
    build(rson);
    pushUP(rt);
}

LL query(int L,int R,int l,int r,int rt)
{
    if(L<=l&&r<=R)
    {
        return sum[rt];
    }
    pushDOWN(rt,r-l+1);
    int m=(l+r)>>1;
    LL ret=0;
    if(L<=m) ret+=query(L,R,lson);
    if(R>m) ret+=query(L,R,rson);
    return ret;
}

int getID(int L,int R,int v,int l,int r,int rt)
{
    if(l==r)
        return l;
    pushDOWN(rt,r-l+1);
    int m=(r+l)>>1;
    if(L<=m&&mx[rt<<1]>v) return getID(L,R,v,lson);
    else if(R>m&&mx[rt<<1|1]>v) return getID(L,R,v,rson);
    return R+1;
}

void update(int L,int R,int c,int l,int r,int rt)
{
    if(L<=l&&r<=R)
    {
        sig[rt]=c;
        mx[rt]=c;
        sum[rt]=c*(r-l+1);
        return ;
    }
    pushDOWN(rt,r-l+1);
    int m=(r+l)>>1;
    if(L<=m) update(L,R,c,lson);
    if(R>m) update(L,R,c,rson);
    pushUP(rt);
}

int main()
{
    int n;
    while(scanf("%d",&n)!=EOF&&n)
    {
        for(int i=1;i<=n;i++)
            scanf("%d",a+i);
        memset(vis,false,sizeof(vis));
        mex[1]=0;
        if(a[1]<n) vis[a[1]]=true;
        while(vis[mex[1]]) mex[1]++;
        for(int i=2;i<=n;i++)
        {
            if(a<n) vis[a]=true;
            mex=mex[i-1];
            while(vis[mex]) mex++;
        }
        map<int,int> mII;
        for(int i=n;i>=1;i--)
        {
            if(mII.find(a)==mII.end()) next=n+1;
            else next=mII[a];
            mII[a]=i;
        }
        cnt=1;LL ans=0;
        memset(mx,0,sizeof(mx));
        memset(sum,0,sizeof(sum));
        memset(sig,-1,sizeof(sig));
        build(1,n,1);
        for(int i=1;i<=n;i++)
        {
            ans+=query(i,n,1,n,1);
            int st=getID(i+1,next-1,a,1,n,1);
            int ed=next-1;
            if(st<=ed) update(st,ed,a,1,n,1);
        }
        printf("%I64d\n",ans);
    }
    return 0;
}

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