POJ 2566（尺取法+前缀和排序）

来源：互联网发布：怎么查看服务器端口编辑：程序博客网时间：2024/05/17 02:12

Bound Found

Time Limit: 5000MS Memory Limit: 65536KB 64bit IO Format: %lld & %llu

Submit Status

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 10310 2-9 8 -7 6 -5 4 -3 2 -1 05 1115 2-1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -115 1000 0

Sample Output

5 4 45 2 89 1 115 1 1515 1 15

题意：数列中有n个数字，给出k个数，做k次查询，求在数列中存在的区间[ left，right ]，使得区间中数的和的绝对值与t最接近。

题解：基本思路：统计出前缀和，然后找到两段前缀和差的绝对值最接近t。我们用尺取法，因为前缀和是无序，需要先将前缀和从小到大排好序，这样才能用尺取法推进查找记录最小值。在建立前缀和时，建立编号记录下标，通过编号还原位置。

在处理前缀和排序时的有很强的技巧性。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <climits>
#include <cmath>
using namespace std;
const int N = 100110;
const int inf = INT_MAX;
int v[N];
struct node
{
int id, sum;
} p[N];

int cmp(node a,node b)
{
return a.sum<b.sum;
}

int main()
{
int n, q;
while(scanf("%d %d", &n, &q),n||q)
{
for(int i=1; i<=n; i++)
{
scanf("%d", &v[i]);
}
int tmp;
p[0].sum=0;
p[0].id=0;
for(int i=1; i<=n; i++)
{
p[i].sum=p[i-1].sum+v[i];
p[i].id=i;
}
sort(p,p+n+1,cmp);
while(q--)
{
scanf("%d", &tmp);
int l=0, r=1, ans=inf, ansl=1, ansr=1;
while(r<=n)
{
int k1=min(p[l].id,p[r].id), k2=max(p[l].id,p[r].id);
int num=abs(p[r].sum-p[l].sum);
if(abs(ans-tmp)>=abs(num-tmp))
{
ansl=k1+1, ansr=k2;
ans=num;
}
if(num>tmp)
{
l++;
}
else if(num<tmp)
{
r++;
}
else if(num==tmp)
{
break;
}
if(r==l)
{
r++;
}
}
printf("%d %d %d\n",ans, ansl, ansr);
}
}
return 0;
}

0 0