HDU 4747 Mex（线段树）（2013 ACM/ICPC Asia Regional Hangzhou Online）
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.
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.
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.
题目大意：定义mex(l,r)为S[l, r]中没出现的最小的非负数。求$ \sum mex(i,j) $，0≤i≤j≤n
思路：我总觉得自己思路有点奇葩大家将就着看。。
考虑每一个子段S[l,n]，如果S[l,n]的第一个0出现在 i （ i 在 l 和 n 之间），那么S[l, i-1]的mex值都为0。
按这个思路搞，考虑S[i]为从以S[i](1≤i≤n)开头还没确定mex值的子段数，那么初始化为S[i] = n-i+1（数组从1开始计数）
/* 补充说明一下s[i]
就是什么都没做的时候，比如s[1] = n
然后只有一个0在位置3
那么就可以确定mex[1,1]和mex[1,2]为0了，mex[1,i],i≥3都至少为1，但还不知道他们会不会大于1，所以s[1] = n - 2
然后只有一个1在位置7
那么就可以确定mex[1,i]，3≤i＜7，都为1（都只有0没有1），而i≥7的mex[1,i]都至少为2（他们都含有0和1），所以s[1] = n - 7
s[i]就是从i开始的mex[i,x]在第 p 阶段还没有确定值只知道有s[i]个mex[i,x]至少大于p
然后只有一个0在位置3
那么就可以确定mex[1,1]和mex[1,2]为0了，mex[1,i],i≥3都至少为1，但还不知道他们会不会大于1，所以s[1] = n - 2
然后只有一个1在位置7
那么就可以确定mex[1,i]，3≤i＜7，都为1（都只有0没有1），而i≥7的mex[1,i]都至少为2（他们都含有0和1），所以s[1] = n - 7
s[i]就是从i开始的mex[i,x]在第 p 阶段还没有确定值只知道有s[i]个mex[i,x]至少大于p
*/
那么从0开始考虑，如果只有一个0，出现在了x位置，那么s[1,x-1]的子段的mex值都为0，所以S[i] = n-x+1（i < x），大于 x 的 s[i]都为0（大于x的子段不存在0，他们的最小非负数都为0）
如果有两个0，那么设第一个0位置为x，第二个0位置为y，那么s[i] = n-x+1（i < x），s[i] = n - y + 1（x≤i<y），大于 y 的 s[i]都为0
有多个0也一样，处理完0之后，得到的sum{s[i]}就是最少为1的mex子段数
然后从1开始往上处理，对某一个数在位置x，s[i] = min(n-x+1, s[i])。
每处理完一个数，就得到一个$ \sum s[i] $ ，依次可以得到最少为2的mex字段数，最少为3的mex字段数……把这些都加起来就是答案。
说得有点抽象……举个例子1 0 2 0 1
初始化s[1] = 5, s[2] = 4, s[3] = 3, s[4] = 2, s[5] = 1
那么处理数字0，s[1] = 4, s[2] = 4, s[3] = 2, s[4] = 2, s[5] = 0, ans = 12
处理数字1，s[1] = 4, s[2] = 1, s[3] = 1, s[4] = 1, s[5] = 0, ans = 12 + 7 = 19
处理数字2，s[1] = 3, s[2] = 1, s[3] = 1, s[4] = 0, s[5] = 0, ans = 19 + 5 = 24
那么可以看出这个是区间赋值，采用线段树处理可以在O(nlogn)的时间内解出答案。
代码（1000MS）：
1 #include <iostream> 2 #include <cstring> 3 #include <cstdio> 4 #include <algorithm> 5 using namespace std; 6 typedef long long LL; 7 8 #define ll x * 2 9 #define rr x * 2 + 1 10 11 const int MAXN = 200010; 12 13 LL tree[MAXN * 4]; 14 int maxt[MAXN * 4], mint[MAXN * 4]; 15 int a[MAXN], n; 16 17 int head[MAXN], lcnt; 18 int pos[MAXN], next[MAXN]; 19 20 void init() { 21 memset(head, 0, (n + 1) * sizeof(int)); 22 lcnt = 1; 23 } 24 25 void add_link(int x, int i) { 26 pos[lcnt] = i; next[lcnt] = head[x]; head[x] = lcnt++; 27 } 28 29 void build(int x, int left, int right) { 30 if(left == right) tree[x] = maxt[x] = mint[x] = n - left + 1; 31 else { 32 int mid = (left + right) >> 1; 33 if(left <= mid) build(ll, left, mid); 34 if(mid < right) build(rr, mid + 1, right); 35 tree[x] = tree[ll] + tree[rr]; 36 maxt[x] = max(maxt[ll], maxt[rr]); 37 mint[x] = min(mint[ll], mint[rr]); 38 } 39 } 40 41 void update(int x, int left, int right, int a, int b, int val) { 42 if(a <= left && right <= b && mint[x] >= val) { 43 tree[x] = LL(val) * (right - left + 1); 44 maxt[x] = mint[x] = val; 45 } 46 else { 47 if(right == left) return ; 48 int mid = (left + right) >> 1; 49 if(maxt[x] == mint[x]) { 50 maxt[ll] = mint[ll] = maxt[x]; 51 tree[ll] = LL(mid - left + 1) * maxt[x]; 52 maxt[rr] = mint[rr] = maxt[x]; 53 tree[rr] = LL(right - (mid + 1) + 1) * maxt[x]; 54 } 55 if(a <= mid && maxt[ll] > val) update(ll, left, mid, a, b, val); 56 if(mid < b && maxt[rr] > val) update(rr, mid + 1, right, a, b, val); 57 tree[x] = tree[ll] + tree[rr]; 58 maxt[x] = max(maxt[ll], maxt[rr]); 59 mint[x] = min(mint[ll], mint[rr]); 60 } 61 } 62 63 LL solve() { 64 LL ret = 0; 65 build(1, 1, n); 66 for(int i = 0; i <= n && tree[1]; ++i) { 67 int last = 0; 68 for(int p = head[i]; p; p = next[p]) { 69 update(1, 1, n, last + 1, pos[p], n - pos[p] + 1); 70 last = pos[p]; 71 } 72 update(1, 1, n, last + 1, n, 0); 73 ret += tree[1]; 74 } 75 return ret; 76 } 77 78 int main() { 79 while(scanf("%d", &n) != EOF && n) { 80 for(int i = 1; i <= n; ++i) scanf("%d", &a[i]); 81 init(); 82 for(int i = n; i > 0; --i) if(a[i] <= n) add_link(a[i], i); 83 cout<<solve()<<endl; 84 } 85 }
