#HDU 4747 Mex 【通过区间处理next变化统计总值】

题目：

Mex

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 2517    Accepted Submission(s): 816

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

30 1 351 0 2 0 10

 

Sample Output

524
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

思路： Mex 是一个函数， 指定一个集合 A ，Mex(A) 为在集合A中未出现的最小的非负整数。 //例：Mex（0,1,2,5）=3

题目给定长度为N的数列，求其所有连续子串的Mex和

N最大为2*10^5 ，子串共有 k * 10 ^ 10 若枚举必然超时， 故使用DP求解。

建立一棵线段树，点 i 所存的信息为若某字串起点位置为 1 ，终点位置 为 i ，该字串的Mex值。能在O（n）时间内求出，且易得，该线段树单调递增。

线段树用于求和，则其父节点所存信息为以 1 为起点 的所有子串 Mex和。 在O（1）内求出。

接着递推DP（2），即以2为起点的所有子串Mex和。易得，在串中删去 位置为 1  的数字（设为 data[i] ）后，由于Mex单调递增，故，在下一个 data[i] 出现前，所有大于 data[i]  的值都将更新为 data[i]。

即需要另外两颗线段树，一个用于存储 对于每一个 位置，下一个与其值相同的位置在哪里 即next数组。 另一个用于求第一个大于等于 data[i] 的位置在哪里。（该操作可以使用lower_bound函数二分完成 ，效率同样为 log 级别）

迭代更新至 DP（N）

#define _CRT_SECURE_NO_WARNINGS#include <iostream>#include <stdio.h>#include <string.h>#include <stdlib.h>#include <math.h>#include <vector>#include <algorithm>#define MAXN 300001long long t[3 * MAXN], a[3 * MAXN];int lazy[3 * MAXN];int lazys[3 * MAXN];long long data[3 * MAXN];int nexts[3 * MAXN];long long ts[3 * MAXN];int n;using namespace std;void BuildTree(int l, int r, int x){if (l == r){t[x] = a[l];return;}int m = (l + r) >> 1;BuildTree(l, m, x << 1);BuildTree(m + 1, r, x << 1 | 1);t[x] = t[x << 1] + t[x << 1 | 1];return;}void PushDown(int l, int r, int x){int m = (l + r) >> 1;if (lazy[x] != -1){t[x << 1] = lazy[x] * (m - l + 1);t[x << 1 | 1] = lazy[x] * (r - m);lazy[x << 1 | 1] = lazy[x];lazy[x << 1] = lazy[x];lazy[x] = -1;}return;}void Modify(int pos, int val, int l, int r, int x){if (pos == l && r == l){t[x] += val;return;}int m = (l + r) >> 1;if (pos <= m){Modify(pos, val, l, m, x << 1);}else Modify(pos, val, m + 1, r, x << 1 | 1);t[x] = t[x << 1] + t[x << 1 | 1];return;}long long Query(int L, int R, int l, int r, int x){if (L == l && R == r)return t[x];PushDown(l, r, x);int m = (l + r) >> 1;if (R <= m)return Query(L, R, l, m, x << 1);else if (L>m)return Query(L, R, m + 1, r, x << 1 | 1);else return Query(L, m, l, m, x << 1) + Query(m + 1, R, m + 1, r, x << 1 | 1);}void SegModify(int L, int R, int val, int l, int r, int x){if (l == L&&r == R){t[x] = (R - L + 1)*val;lazy[x] = val;return;}PushDown(l, r, x);int m = (l + r) >> 1;if (R <= m)SegModify(L, R, val, l, m, x << 1);else if (L>m)SegModify(L, R, val, m + 1, r, x << 1 | 1);else {SegModify(L, m, val, l, m, x << 1);SegModify(m + 1, R, val, m + 1, r, x << 1 | 1);}t[x] = t[x << 1] + t[x << 1 | 1];return;}void BuildTrees(int l, int r, int x){if (l == r){ts[x] = a[l];return;}int m = (l + r) >> 1;BuildTrees(l, m, x << 1);BuildTrees(m + 1, r, x << 1 | 1);ts[x] = max(ts[x << 1], ts[x << 1 | 1]);return;}void PushDowns(int l, int r, int x){int m = (l + r) >> 1;if (lazys[x] != -1){ts[x << 1] = lazys[x];ts[x << 1 | 1] = lazys[x];lazys[x << 1 | 1] = lazys[x];lazys[x << 1] = lazys[x];lazys[x] = -1;}return;}void Modifys(int pos, int val, int l, int r, int x){if (pos == l && r == l){ts[x] += val;return;}int m = (l + r) >> 1;if (pos <= m){Modifys(pos, val, l, m, x << 1);}else Modifys(pos, val, m + 1, r, x << 1 | 1);ts[x] = max(ts[x << 1], ts[x << 1 | 1]);return;}long long Querys(int L, int R, int l, int r, int x){if (L == l && R == r)return ts[x];PushDowns(l, r, x);int m = (l + r) >> 1;if (R <= m)return Querys(L, R, l, m, x << 1);else if (L>m)return Querys(L, R, m + 1, r, x << 1 | 1);else return max(Querys(L, m, l, m, x << 1), Querys(m + 1, R, m + 1, r, x << 1 | 1));}void SegModifys(int L, int R, int val, int l, int r, int x){if (l == L&&r == R){ts[x] = val;lazys[x] = val;return;}PushDowns(l, r, x);int m = (l + r) >> 1;if (R <= m)SegModifys(L, R, val, l, m, x << 1);else if (L>m)SegModifys(L, R, val, m + 1, r, x << 1 | 1);else {SegModifys(L, m, val, l, m, x << 1);SegModifys(m + 1, R, val, m + 1, r, x << 1 | 1);}ts[x] = max(ts[x << 1], ts[x << 1 | 1]);return;}int finds(int val, int l, int r, int x){if (l == r){return l;}PushDowns(l, r, x);if (ts[x << 1] >= val){return finds(val, l, (l + r) / 2, x << 1);}else if (ts[x << 1 | 1] >= val){return finds(val, (l + r) / 2 + 1, r, x << 1 | 1);}else{return n + 1;}}int main(){int nn, mm;while (cin >> nn){long long ans = 0;if (nn == 0){return 0;}n = 1;while (n < nn){n *= 2;}memset(t, 0, sizeof(t));memset(lazy, -1, sizeof(lazy));memset(lazys, -1, sizeof(lazys));for (size_t i = 1; i <= nn; i++){scanf("%d", &data[i]);if (data[i]>200000){data[i] = 200001;}}int bes = 0;for (size_t i = 0; i <= n * 2 + 1; i++){a[i] = -1;}for (size_t i = 1; i <= nn; i++){t[data[i]] = 1;while (t[bes]){bes++;}a[i] = bes;}memset(t, 0, sizeof(t));BuildTrees(1, n, 1);bes = 0;memset(a, 0, sizeof(a));for (size_t i = 1; i <= nn; i++){t[data[i]] = 1;while (t[bes]){bes++;}a[i] = bes;}memset(t, 0, sizeof(t));for (size_t i = 1; i <= nn; i++){nexts[i] = nn + 1;}for (size_t i = 1; i <= nn; i++){nexts[t[data[i]]] = i;t[data[i]] = i;}memset(t, 0, sizeof(t));BuildTree(1, n, 1);ans += Query(1, n, 1, n, 1);for (size_t i = 1; i < nn; i++){int tofind = data[i];SegModify(i, i, -1, 1, n, 1);SegModifys(i, i, -1, 1, n, 1);int mid;mid = finds(tofind, 1, n, 1);if (mid<nexts[i]){SegModify(mid, nexts[i] - 1, data[i], 1, n, 1);SegModifys(mid, nexts[i] - 1, data[i], 1, n, 1);}ans += Query(i + 1, nn, 1, n, 1);}cout << ans << "\n";}return 0;}