HDU 4747 Mex(线段树)(2013 ACM/ICPC Asia Regional Hangzhou Online)
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.
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
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
typedef long long LL; #define ll x * 2
#define rr x * 2 + 1 const int MAXN = ; LL tree[MAXN * ];
int maxt[MAXN * ], mint[MAXN * ];
int a[MAXN], n; int head[MAXN], lcnt;
int pos[MAXN], next[MAXN]; void init() {
memset(head, , (n + ) * sizeof(int));
lcnt = ;
} void add_link(int x, int i) {
pos[lcnt] = i; next[lcnt] = head[x]; head[x] = lcnt++;
} void build(int x, int left, int right) {
if(left == right) tree[x] = maxt[x] = mint[x] = n - left + ;
else {
int mid = (left + right) >> ;
if(left <= mid) build(ll, left, mid);
if(mid < right) build(rr, mid + , right);
tree[x] = tree[ll] + tree[rr];
maxt[x] = max(maxt[ll], maxt[rr]);
mint[x] = min(mint[ll], mint[rr]);
}
} void update(int x, int left, int right, int a, int b, int val) {
if(a <= left && right <= b && mint[x] >= val) {
tree[x] = LL(val) * (right - left + );
maxt[x] = mint[x] = val;
}
else {
if(right == left) return ;
int mid = (left + right) >> ;
if(maxt[x] == mint[x]) {
maxt[ll] = mint[ll] = maxt[x];
tree[ll] = LL(mid - left + ) * maxt[x];
maxt[rr] = mint[rr] = maxt[x];
tree[rr] = LL(right - (mid + ) + ) * maxt[x];
}
if(a <= mid && maxt[ll] > val) update(ll, left, mid, a, b, val);
if(mid < b && maxt[rr] > val) update(rr, mid + , right, a, b, val);
tree[x] = tree[ll] + tree[rr];
maxt[x] = max(maxt[ll], maxt[rr]);
mint[x] = min(mint[ll], mint[rr]);
}
} LL solve() {
LL ret = ;
build(, , n);
for(int i = ; i <= n && tree[]; ++i) {
int last = ;
for(int p = head[i]; p; p = next[p]) {
update(, , n, last + , pos[p], n - pos[p] + );
last = pos[p];
}
update(, , n, last + , n, );
ret += tree[];
}
return ret;
} int main() {
while(scanf("%d", &n) != EOF && n) {
for(int i = ; i <= n; ++i) scanf("%d", &a[i]);
init();
for(int i = n; i > ; --i) if(a[i] <= n) add_link(a[i], i);
cout<<solve()<<endl;
}
}
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