POJ_2566_Bound_Found_(尺取法+前缀和)

描述

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http://poj.org/problem?id=2566

给出一个整数序列,并给出非负整数t,求数列中连续区间和的绝对值最接近k的区间左右端点以及这个区间和的绝对值.

Bound Found

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 2592		Accepted: 789		Special Judge

Description

Signals of most probably extra-terrestrial origin have been received and digitalized by The Aeronautic and Space Administration (that must be going through a defiant phase: "But I want to use feet, not meters!"). Each signal seems to come in two parts: a sequence of n integer values and a non-negative integer t. We'll not go into details, but researchers found out that a signal encodes two integer values. These can be found as the lower and upper bound of a subrange of the sequence whose absolute value of its sum is closest to t.

You are given the sequence of n integers and the non-negative target
t. You are to find a non-empty range of the sequence (i.e. a continuous
subsequence) and output its lower index l and its upper index u. The
absolute value of the sum of the values of the sequence from the l-th to
the u-th element (inclusive) must be at least as close to t as the
absolute value of the sum of any other non-empty range.

Input

The
input file contains several test cases. Each test case starts with two
numbers n and k. Input is terminated by n=k=0. Otherwise,
1<=n<=100000 and there follow n integers with absolute values
<=10000 which constitute the sequence. Then follow k queries for this
sequence. Each query is a target t with 0<=t<=1000000000.

Output

For
each query output 3 numbers on a line: some closest absolute sum and
the lower and upper indices of some range where this absolute sum is
achieved. Possible indices start with 1 and go up to n.

Sample Input

5 1
-10 -5 0 5 10
3
10 2
-9 8 -7 6 -5 4 -3 2 -1 0
5 11
15 2
-1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1
15 100
0 0

Sample Output

5 4 4
5 2 8
9 1 1
15 1 15
15 1 15

Source

Ulm Local 2001

分析

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尺取法.

一眼看过去就是用尺取法找和t绝对值相差最小的区间和,但是这道题里的序列并不是非负的,这意味着固定左端点,移动右端点时,区间和不是单调递增的.尺取法的模板题中区间和是单调的,所以找到大于某一值的区间右端点就可以固定.而这道题中因为存在负值,所以区间的变化不是单调的,不能按照普通的方法解决.我们考虑一段区间和,不仅可以认为是a[l]+a[l+1]+...+a[r],还可以看作是sum[r]-sum[l-1],而这道题中要求的区间和的绝对值,就可以看作是max(sum[r],sum[l-1])-min(sum[r],sum[l-1]).这样求一个前缀和,进行排序,双指针确定区间左右端点,这样sum[r]-sum[l](排序后的编号)就代表一个区间和的绝对值,这样的区间就是单调的.如此一来,确定左端点,移动右端点,sum[r]-sum[l]就是单增的,找到sum[r]-sum[l]>t的位置即可停止,然后l++,sum[r-1]-sum[l]比之前的sum[r-1]-sum[l]更小(减数增大),本来就比t小,现在小得更多了,对于任意r'<=r-1都是如此,就不必考虑,从sum[r]-sum[l]开始继续即可.

注意:

1.sum[r]-sum[l-1],因为1<=l<=r,所以l-1>=0,注意sum[0].s=0,sum[0].num=0,要参与排序,并且每一次都要重新赋值,因为上一组数据排序后sum[0].s不一定是0,这将影响到语句"sum[i]=point(sum[i-1].s,i)" .

2.对于区间右端点表示的状态,最好是用来表示当前右端点的位置,然后每一次更新状态时都要判断,与ans比较,看是否更新最优解.

3.由于r>=l所以r>l-1,也就是说两个区间端点不能是同一个点,当l追上r时,r++.

4.其实尺取法的写法可以优化,现在的写法是两层循环,l改变后进行一次操作,然后对于同一个l改变r,每次进行一次操作,这样在一个外层循环内部要写两遍操作.其实对于l和r的改变是一样,都是区间状态的改变,所以只要一层循环即可.

即

  while (l<=n&&r<=n&&ans!=) 

 #include<cstdio>
 #include<cstdlib>
 #include<algorithm>
 using std :: sort;
 using std :: min;
 using std :: max;

 const int maxn=,INF=0x7fffffff;

 int a[maxn],s[maxn];
 int n,q,t;

 struct point
 {
     int s,num;
     point() {}
     point(int a,int b) : s(a) , num(b) {}
 }sum[maxn];

 bool comp(point x,point y) { return x.s<y.s; }

 int value(int l,int r) { return abs(sum[r].s-sum[l].s); }

 int L(int l,int r) { return min(sum[l].num,sum[r].num)+; }

 int R(int l,int r) { return max(sum[l].num,sum[r].num); }

 void solve(int t)
 {
     int ans=INF,res,idxl,idxr;
     int r=;
     for(int l=;l<=n;l++)
     {
         if(l==r) r++;
         if(r>n) break;
         int now=value(l,r);
         int d=abs(now-t);
         if(d<=ans)
         {
             ans=d;
             res=now;
             idxl=L(l,r);
             idxr=R(l,r);
         }
         while(r<n&&now<t)
         {
             r++;
             now=value(l,r);
             int d=abs(now-t);
             if(d<=ans)
             {
                 ans=d;
                 res=now;
                 idxl=L(l,r);
                 idxr=R(l,r);
             }
         }
         if(now<t) break;
         if(now==t)
         {
             break;
         }
     }
     printf("%d %d %d\n",res,idxl,idxr);
 }

 void init()
 {
     while(scanf("%d%d",&n,&q)&&(n!=||q!=))
     {
         sum[].s=;
         sum[].num=;
         for(int i=;i<=n;i++)
         {
             scanf("%d",&a[i]);
             sum[i]=point(sum[i-].s+a[i],i);
         }
         sort(sum,sum+n+,comp);
         for(int i=;i<=q;i++)
         {
             scanf("%d",&t);
             solve(t);
         }
     }
 }

 int main()
 {
     freopen("bound.in","r",stdin);
     freopen("bound.out","w",stdout);
     init();
     fclose(stdin);
     fclose(stdout);
     return ;
 }