HDU 4747 Mex （2013杭州网络赛1010题，线段树）

Mex

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 623    Accepted Submission(s): 209

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 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

 

 

Recommend

liuyiding

 

 

 

题目定义了mex(i,j)表示，没有在i到j之间出现的最小的非负整数。

 

 

求所有组合的i,j（i<=j）的和

就是求mex(1,1) + mex(1,2)+....+mex(1,n)

+mex(2,2) + mex(2,3) + ...mex(2,n)

+mex(3,3) + mex(3,4)+...+mex(3,n)

+ mex(n,n)

 

可以知道mex(i,i),mex(i,i+1)到mex(i,n)是递增的。

 

首先很容易求得mex(1,1),mex(1,2)......mex(1,n)

因为上述n个数是递增的。

 

然后使用线段树维护，需要不断删除前面的数。

比如删掉第一个数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]

因为是单调递增的，所以找到第一个 mex > a[1]的位置，到下一个a[1]出现位置，这个区间的值变成a[1].

 

然后需要线段树实现区间修改和区间求和。

 

/* ***********************************************
Author        :kuangbin
Created Time  :2013-9-17 21:15:02
File Name     :G:\2013ACM练习\2013网络赛\2013杭州网络赛\1010.cpp
************************************************ */

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;

const int MAXN = 200010;
struct Node
{
    int l,r;
    long long sum;//区间和
    int mx;//最大值
    int lazy;//懒惰标记，表示赋值为相同的
}segTree[MAXN*3];
void push_up(int i)
{
    if(segTree[i].l == segTree[i].r)return;
    segTree[i].sum = segTree[i<<1].sum + segTree[(i<<1)|1].sum;
    segTree[i].mx = max(segTree[i<<1].mx,segTree[(i<<1)|1].mx);
}
void Update_Same(int i,int v)
{
    segTree[i].sum = (long long)v*(segTree[i].r - segTree[i].l + 1);
    segTree[i].mx = v;
    segTree[i].lazy = 1;
}
void push_down(int i)
{
    if(segTree[i].l == segTree[i].r)return;
    if(segTree[i].lazy)
    {
        Update_Same(i<<1,segTree[i].mx);
        Update_Same((i<<1)|1,segTree[i].mx);
        segTree[i].lazy = 0;
    }
}
int mex[MAXN];
void Build(int i,int l,int r)
{
    segTree[i].l = l;
    segTree[i].r = r;
    segTree[i].lazy = 0;
    if(l == r)
    {
        segTree[i].mx = mex[l];
        segTree[i].sum = mex[l];
        return;
    }
    int mid = (l + r)>>1;
    Build(i<<1,l,mid);
    Build((i<<1)|1,mid+1,r);
    push_up(i);
}
//将区间[l,r]的数都修改为v
void Update(int i,int l,int r,int v)
{
    if(segTree[i].l == l && segTree[i].r == r)
    {
        Update_Same(i,v);
        return;
    }
    push_down(i);
    int mid = (segTree[i].l + segTree[i].r)>>1;
    if(r <= mid)
    {
        Update(i<<1,l,r,v);
    }
    else if(l > mid)
    {
        Update((i<<1)|1,l,r,v);
    }
    else
    {
        Update(i<<1,l,mid,v);
        Update((i<<1)|1,mid+1,r,v);
    }
    push_up(i);
}
//得到值>= v的最左边位置
int Get(int i,int v)
{
    if(segTree[i].l == segTree[i].r)
        return segTree[i].l;
    push_down(i);
    if(segTree[i<<1].mx > v)
        return Get(i<<1,v);
    else return Get((i<<1)|1,v);
}
int a[MAXN];
map<int,int>mp;
int next[MAXN];
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int n;
    while(scanf("%d",&n) && n)
    {
        for(int i = 1;i <= n;i++)
            scanf("%d",&a[i]);
        mp.clear();
        int tmp = 0;
        for(int i = 1;i <= n;i++)
        {
            mp[a[i]] = 1;
            while(mp.find(tmp) != mp.end())tmp++;
            mex[i] = tmp;
        }
        mp.clear();
        for(int i = n;i >= 1;i--)
        {
            if(mp.find(a[i]) == mp.end())next[i] = n+1;
            else next[i] = mp[a[i]];
            mp[a[i]] = i;
        }
        Build(1,1,n);
        long long sum = 0;
        for(int i = 1;i <= n;i++)
        {
            sum += segTree[1].sum;
            if(segTree[1].mx > a[i])
            {
                int l = Get(1,a[i]);
                int r = next[i];
                if(l < r)
                    Update(1,l,r-1,a[i]);
            }
            Update(1,i,i,0);
        }
        printf("%I64d\n",sum);

    }
    return 0;
}