CS Academy Array Removal

题目链接：https://csacademy.com/contest/archive/task/array-removal/statement/

题意：给一个元素个数为n的数组，给出一个n的排列，从前到后按照排列的元素从数组中删去一个数，每次删除之前输出数组的最大子数组值。最大子数组值为所有子数组的和的最大值。

思路：由于关心的是子数组的值，而随着操作的执行原数组所有元素都会被删除，因此可以从后往前考虑，并且使用 并查集 快速获取子数组的和。

从后往前，对于排列中的每个元素，如果当前位置两边有已经恢复过（从前到后删除，从后到前即为恢复）的数据，那么将当前元素作为父亲，两边作为儿子合并，并且更新当前元素值以及临时ans值为父亲+儿子的值。

如果当前位置两边都没回复，则当前位置临时ans值为自己

注意：由于每次都要输出最大值，因此可以用栈保存ans，并且每次确定一个临时ans， 取max(ans, stack.top())作为这一次的值。

代码：

 1 const int inf = 0x3f3f3f3f;
 2 const int maxn = 1e5 + 5;
 3 
 4 ll a[maxn];
 5 int n, perm[maxn];
 6 int fa[maxn], vis[maxn];
 7 stack<ll> ans;
 8 
 9 void init(){
10     while(!ans.empty()) ans.pop();
11     memset(a, 0, sizeof(a));
12     memset(perm, 0, sizeof(perm));
13     memset(vis, 0, sizeof(vis));
14     for(int i = 1; i <= n; i++) fa[i] = i;
15 }
16 int find(int x){
17     return (fa[x] == x? fa[x]: find(fa[x]));
18 }
19 ll unite(int x, int y){
20     int fx = find(x), fy = find(y);
21     fa[fy] = fx;
22     return a[fx] = a[fx] + a[fy];
23 }
24 
25 int main(){
26     scanf("%d", &n);
27     init();
28     for(int i = 1; i <= n; i++){
29         scanf("%d", &a[i]);
30     }
31     for(int i = 0; i < n; i++){
32         scanf("%d", &perm[i]);
33     }
34     int i = n / 2, j = n / 2;
35     if(n % 2 == 0) i--;
36     for( i, j; i >= 0 && j < n; i--, j++){
37         swap(perm[i], perm[j]);
38     }
39     for(int i = 0; i < n; i++){
40         int tmp = perm[i];
41         vis[tmp] = 1;
42         if(tmp > 1 && tmp < n && vis[tmp -1] && vis[tmp + 1]){
43             ll tmans = unite(tmp, tmp - 1);
44             tmans = unite(tmp, tmp + 1);
45             if(!ans.empty()) tmans = max(tmans, ans.top());
46             ans.push(tmans);
47         }
48         else if(tmp < n && vis[tmp + 1]){
49             ll tmans = unite(tmp, tmp + 1);
50             if(!ans.empty()) tmans = max(tmans, ans.top());
51             ans.push(tmans);
52         }
53         else if(tmp > 1 && vis[tmp - 1]){
54             ll tmans = unite(tmp, tmp - 1);
55             if(!ans.empty()) tmans = max(tmans, ans.top());
56             ans.push(tmans);
57         }
58         else {
59             ll tmans = 0;
60             if(!ans.empty())  tmans = ans.top();
61             tmans = max(tmans, a[tmp]);
62             ans.push(tmans);
63         }
64     }
65     while(!ans.empty()){
66         ll tmans = ans.top();
67         ans.pop();
68         printf("%lld\n", tmans);
69     }
70 }

题目：

Array Removal

Time limit: 1000 ms
Memory limit: 128 MB

 

Alex has an array of NN integers. On this array he can perform the following operation: choose an element that was not previously chosen and mark it as unavailable. Alex wants to perform exactly NN operations, until all the elements are marked.

Alex defines the cost of a subarray as the sum of all the elements in the subarray. Before performing an operation, Alex wants to know the maximum cost of a subarray that doesn't contain any unavailable elements.

Standard input

The first line contains a single integer NN, the length of the array.

The second line contains the NN values of the array.

The third line contains a permutation of size NN, representing the indices of the elements chosen for the operations, in order.

Standard output

The output should contain NN lines. On each line output a single integer, the answer before each operation.

Constraints and notes

  • 1 \leq N \leq 10^51≤N≤10​5​​
  • The values of the array are integers between 00 and 10^910​9​​
InputOutputExplanation
 
5
6 1 2 3 2
2 5 1 4 3

14
7
6
5
2

Available elements - Max sum - Max sum subarray:

11111 - 14 - [1, 5]11111−14−[1,5]

10111 - 7- [3, 5]10111−7−[3,5]

10110 - 6 - [1, 1]10110−6−[1,1]

00110 - 5 - [2, 3]00110−5−[2,3]

00100 - 2 - [3, 3]00100−2−[3,3]

 
6
8 3 4 1 5 10
1 4 6 3 2 5

31
23
15
7
5
5

111111 - 31 - [1, 6]111111−31−[1,6]

011111 - 23 - [2, 6]011111−23−[2,6]

011011 - 15 - [5, 6]011011−15−[5,6]

011010 - 7 - [2, 3]011010−7−[2,3]

010010 - 5 - [5, 5]010010−5−[5,5]

000010 - 5 - [5, 5]000010−5−[5,5]

 

posted @ 2017-08-02 11:21  EricJeffrey  阅读(246)  评论(0)  编辑  收藏  举报