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 = ;
 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*];
 void push_up(int i)
 {
     if(segTree[i].l == segTree[i].r)return;
     segTree[i].sum = segTree[i<<].sum + segTree[(i<<)|].sum;
     segTree[i].mx = max(segTree[i<<].mx,segTree[(i<<)|].mx);
 }
 void Update_Same(int i,int v)
 {
     segTree[i].sum = (long long)v*(segTree[i].r - segTree[i].l + );
     segTree[i].mx = v;
     segTree[i].lazy = ;
 }
 void push_down(int i)
 {
     if(segTree[i].l == segTree[i].r)return;
     if(segTree[i].lazy)
     {
         Update_Same(i<<,segTree[i].mx);
         Update_Same((i<<)|,segTree[i].mx);
         segTree[i].lazy = ;
     }
 }
 int mex[MAXN];
 void Build(int i,int l,int r)
 {
     segTree[i].l = l;
     segTree[i].r = r;
     segTree[i].lazy = ;
     if(l == r)
     {
         segTree[i].mx = mex[l];
         segTree[i].sum = mex[l];
         return;
     }
     int mid = (l + r)>>;
     Build(i<<,l,mid);
     Build((i<<)|,mid+,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)>>;
     if(r <= mid)
     {
         Update(i<<,l,r,v);
     }
     else if(l > mid)
     {
         Update((i<<)|,l,r,v);
     }
     else
     {
         Update(i<<,l,mid,v);
         Update((i<<)|,mid+,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<<].mx > v)
         return Get(i<<,v);
     else return Get((i<<)|,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 = ;i <= n;i++)
             scanf("%d",&a[i]);
         mp.clear();
         int tmp = ;
         for(int i = ;i <= n;i++)
         {
             mp[a[i]] = ;
             while(mp.find(tmp) != mp.end())tmp++;
             mex[i] = tmp;
         }
         mp.clear();
         for(int i = n;i >= ;i--)
         {
             if(mp.find(a[i]) == mp.end())next[i] = n+;
             else next[i] = mp[a[i]];
             mp[a[i]] = i;
         }
         Build(,,n);
         long long sum = ;
         for(int i = ;i <= n;i++)
         {
             sum += segTree[].sum;
             if(segTree[].mx > a[i])
             {
                 int l = Get(,a[i]);
                 int r = next[i];
                 if(l < r)
                     Update(,l,r-,a[i]);
             }
             Update(,i,i,);
         }
         printf("%I64d\n",sum);

     }
     return ;
 }