codeforces-772C-Destroying Array
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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.
41 3 2 53 4 1 2
5430
51 2 3 4 54 2 3 5 1
65510
85 5 4 4 6 6 5 55 2 8 7 1 3 4 6
18161188660
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:给你一个数组,每次删除一个数,求每一次操作后的区间最大和(注意:这里删除一个数后,该数所在的位置不清零,即该树的前一个数和后一个数
不连续)
题解:很容易想到用线段树写,原谅本弱不会。。。这里采用更新区间左右端点。
逆向思维,把每次删除一个数想象成每次添加一个数,这样在你每次操作后更新一波左右端点,使其保存该区间和(中间点是无用点),
最后在于maxs比较即可。
代码如下:
#include<stdio.h>#include<string.h>#include<stdlib.h>#include<limits.h>#include<iostream>#include<algorithm>#include<queue>#include<stack>#include<vector>#include<math.h>#include<map>using namespace std;#define maxn 100005typedef __int64 ll;struct node{ll l,r,sum;}ans[maxn];bool flag[maxn];ll a[maxn],b[maxn],c[maxn];int main(){ll i,j,n,m,maxs=0,x;scanf("%I64d",&n);for(i=1;i<=n;i++){ scanf("%I64d",&a[i]); ans[i].l=i; ans[i].r=i; ans[i].sum=a[i]; }for(i=1;i<=n;i++) scanf("%I64d",&b[i]);for(i=n;i>0;i--){x=b[i];flag[x]=1;if(flag[x-1]){ans[x].l=ans[x-1].l;ans[ans[x-1].l].sum=ans[ans[x-1].l].sum+ans[ans[x].r].sum;ans[x].sum=ans[ans[x].l].sum;if(flag[x+1])ans[ans[x].l].r=ans[x+1].r;else ans[ans[x].l].r=ans[x].r;}if(flag[x+1]){ans[x].r=ans[x+1].r;ans[ans[x].r].l=ans[x].l;ans[ans[x].r].sum=ans[ans[x].r].sum+ans[ans[x].l].sum;ans[ans[x].l].sum=ans[ans[x].r].sum;if(flag[x-1])ans[ans[x].r].l=ans[x].l;}maxs=max(maxs,ans[ans[x].l].sum);maxs=max(maxs,ans[ans[x].r].sum);c[i]=maxs;}for(i=2;i<=n;i++)printf("%I64d\n",c[i]);printf("0\n");}
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