原题链接：https://vjudge.net/problem/511814/origin
Description：
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.
Input
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.
Output
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.
Examples
Input
4
1 3 2 5
3 4 1 2
Output
5
4
3
0
Input
5
1 2 3 4 5
4 2 3 5 1
Output
6
5
5
1
0
Input
8
5 5 4 4 6 6 5 5
5 2 8 7 1 3 4 6
Output
18
16
11
8
8
6
6
0
Note
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.
题意 ：给你一串数组以及一串指令（每次破坏数组的第几个元素），输出每个指令后剩下的连通块的和最大是多少。
分析：有两种方法做，第一种用集合的插入和删除（强烈建议）；第二种用并查集：题目给的操作数从第 1 个到第 N 个数是删除原数组中的一个数, 那么反过来从后往前就是增加原数组中的一个数, 每增加一个值, 那么就有四种情况: 第一种和前后都不连续, 即自成一个集合; 第二种:和前面的数连续, 和前一个数在一个集合; 第三种和之后一个数连续, 和之后的一个数在一个集合; 第四种既和前面一个数连续也和后面一个数连续,那么说明前后两个集合被这个数合并成一个集合. 然后合并的时候维护每个集合的元素和 sum, 利用 max 记录当前集合 sum 和新增集合的 sum 中的最大值.
AC代码如下：
①使用集合插入和删除连通块的和。
#include <bits/stdc++.h>
using namespace std;
#define fast ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define ll long long
#define _for(i,a,b) for(int i = a;i < b;i++)
#define rep(i,a,b) for(int i = a;i <= b;i++)
#define all(s) s.begin(), s.end()
const int maxn = 100050;
int n,x;
ll a[maxn];
set<int>pos;
set<ll>sum;
map<ll, int>nsum;
int main()
{
pos.clear();
sum.clear();
nsum.clear();
scanf("%d", &n);
a[0] = 0;
rep(i, 1, n)
{
scanf("%d", &x);
a[i] = a[i - 1] + x;
}
a[n + 1] = a[n];
sum.insert(a[n]);
nsum[a[n]]++;
pos.insert(0), pos.insert(n + 1);
set<int>::iterator itt;
set<ll>::iterator it;
int r, l;
while (n--)
{
scanf("%d", &x);
itt = pos.lower_bound(x);
r = *itt; --itt; l = *itt;
pos.insert(x);
ll len = a[r - 1] - a[l];
if (nsum[len] > 1)nsum[len]--;
else
{
nsum[len]--;
sum.erase(len);
}
nsum[a[r - 1] - a[x]]++;
nsum[a[x - 1] - a[l]]++;
sum.insert(a[r - 1] - a[x]);
sum.insert(a[x - 1] - a[l]);
it = sum.end();
it--;
printf("%I64d
", *it);
}
return 0;
}②使用并查集：
#include <bits/stdc++.h>
using namespace std;
#define fast ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define ll long long
#define _for(i,a,b) for(int i = a;i < b;i++)
#define rep(i,a,b) for(int i = a;i <= b;i++)
#define all(s) s.begin(), s.end()
const int maxn = 1e5 + 10;
int n;
ll a[maxn], cmd[maxn], sum[maxn], par[maxn],ans[maxn];
bool vis[maxn];
int find_(ll x)
{
return (par[x] == x) ? x : par[x] = find_(par[x]);
}
void unite(ll x,ll y)
{
ll fx = find_(x);
ll fy = find_(y);
if (fx != fy) { par[fy] = fx; sum[fx] += sum[fy]; }//两连通块和相加
}
int main()
{
scanf("%d", &n);
rep(i, 1, n)
{
scanf("%lld", &a[i]);
par[i] = i; vis[i] = 0;
}
rep(i, 1, n)scanf("%lld", &cmd[i]);
ll rans = 0;
for (int i = n; i >= 1; i--)
{
ll t = cmd[i];
sum[t] = a[cmd[i]], vis[t] = 1, ans[i] = rans;
//新加入一个元素后的四种情况
if (t != 1 && vis[t - 1])unite(t, t - 1);
if (t != n && vis[t + 1])unite(t, t + 1);
rans = max(rans, sum[find_(t)]);
}
rep(i, 1, n)printf("%lld
", ans[i]);
return 0;
}