本文主要是介绍【Intel Code Challenge Elimination Round (Div1 + Div2, combined) C】【正难则反并查集】n数按照次序删除每次删除后最大联通块之和，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
You are given an array consisting of n non-negative integers a1, a2, ..., an.
You are going to destroy integers in the array one by one. Thus, you are given the permutation of integers from 1 to n defining the order elements of the array are destroyed.
After each element is destroyed you have to find out the segment of the array, such that it contains no destroyed elements and the sum of its elements is maximum possible. The sum of elements in the empty segment is considered to be 0.
The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the length of the array.
The second line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 109).
The third line contains a permutation of integers from 1 to n — the order used to destroy elements.
Print n lines. The i-th line should contain a single integer — the maximum possible sum of elements on the segment containing no destroyed elements, after first i operations are performed.
4 1 3 2 5 3 4 1 2
5 4 3 0
5 1 2 3 4 5 4 2 3 5 1
6 5 5 1 0
8 5 5 4 4 6 6 5 5 5 2 8 7 1 3 4 6
18 16 11 8 8 6 6 0
Consider the first sample:
- Third element is destroyed. Array is now 1 3 * 5. Segment with maximum sum 5 consists of one integer 5.
- Fourth element is destroyed. Array is now 1 3 * * . Segment with maximum sum 4 consists of two integers 1 3.
- First element is destroyed. Array is now * 3 * * . Segment with maximum sum 3 consists of one integer 3.
- Last element is destroyed. At this moment there are no valid nonempty segments left in this array, so the answer is equal to 0.
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }
#define MS(x,y) memset(x,y,sizeof(x))
#define MC(x,y) memcpy(x,y,sizeof(x))
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b>a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b<a)a = b; }
const int N = 1e5 + 10, M = 0, Z = 1e9 + 7, inf = 0x3f3f3f3f;
template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
int n;
int a[N];
int p[N];
int f[N];
LL sum[N];
LL ans[N];
int find(int x)
{return f[x] == x ? x : f[x] = find(f[x]);
}
int main()
{while (~scanf("%d", &n)){for (int i = 1; i <= n; ++i)scanf("%d", &a[i]), f[i] = 0;for (int i = 1; i <= n; ++i)scanf("%d", &p[i]);LL ANS = 0;ans[n] = 0;for (int i = n; i >= 1; --i){int x = p[i];f[x] = x;sum[x] = a[x];if (x > 1 && f[x - 1]){int y = find(x - 1);if (x != y){sum[x] += sum[y];f[y] = x;}}if (x < n && f[x + 1]){int y = find(x + 1);if (x != y){sum[x] += sum[y];f[y] = x;}}gmax(ANS, sum[x]);ans[i - 1] = ANS;}for (int i = 1; i <= n; ++i)printf("%lld\n", ans[i]);}return 0;
}
/*
【题意】
给你n个数，我们按照一个全排列的次序全部删除，问你在每次做完删除操作之后，剩余的和最大的联通块的和是多少
定义联通块为没有删除的一段连续的数【类型】
并查集 正难则反【分析】
这道题是正难则反并查集的典型应用
倒着合并就可以了【时间复杂度&&优化】
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