Mex
Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 623 Accepted Submission(s): 209
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.
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.
0 1 3
5
1 0 2 0 1
0
24
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.
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 ;
}