B - Bound Found POJ - 2566(尺取 + 对区间和的绝对值

B - Bound Found POJ - 2566

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

思路

• 题意：这一题就是给你一个有正有负数的序列ar[]，然后又给我们一个 值val
  
  问能否在这个ar序列中找到一个子串，是这个子串的数字和的绝对值 最接近 val 的那个子串
• 思路：这一题思路真实很神奇，也很有用，首先求这个序列的前缀和，并对这个前缀和（注意这个前缀合是由结构体组成、存有当前前缀和终点位置的id下标）按从小到大的顺序排序，这样我们就的得到一个有序的前缀和序列br,由于我们让求的是一个区间和的绝对值，所以我们让br中任意两个位置的前缀和相减（大位置 - 小位置的前缀和），我们就能得到任意的 子区间 和；之后要做的就是用 尺取 不断的维护最接近val 的子区间了，
  • 在尺取的时候有一点细节要明白，我们尺取某个区间的时候，我们假设这个区间前缀的差值为tem，，，当tem <= val 这个时候r（尺取是的有边界）要 ++，当tem > val 的时候就没有必要 r++了，因为在++，只会让tem更大、更不接近val，所这个时候我们要 l ++，去减小区间前缀和差值，

代码

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cmath>

using namespace std;

const int Len = 1e5+10;
struct Node
{
    int sum,id;
    bool operator < (const Node x) const
    {
        return sum < x.sum;
    }
} a[Len];

int main()
{
    //freopen("A.txt","r",stdin);
    int n,t;
    while(scanf("%d %d", &n, &t) && n + t)
    {
        a[0].id = 0, a[0].sum = 0;
        for(int i = 1; i <= n; i ++)
        {
            scanf("%d", &a[i].sum);
            a[i].sum += a[i - 1].sum;
            a[i].id = i;
        }
        sort(a, a + 1 + n);
        while(t --)
        {
            int val;
            scanf("%d", &val);
            int l = 0, r = 1, mn = 1e9 + 7, al, ar, sum;
            while(true)
            {
                while(r <= n)   //小于n之前的任意一对的值我们都要统计一下
                {
                    int  tem = a[r].sum - a[l].sum;     //获取区间和的差值
                    if(abs(tem - val) < mn)             //判断这个区间和的差值是否跟接近 val，通过小于原来的最优值值来判断
                    {
                        mn = abs(tem - val);
                        sum = tem;
                        al = min(a[r].id, a[l].id) + 1; //注意：左边界 +1
                        ar = max(a[r].id, a[l].id);
                    }

                    if(tem < val)                       //如果当前值还是小于要靠近的值，那么我们增大区间和的差值来 缩小与给定值的差值，并且继续循环看能否更新最优值
                        r ++;
                    else
                        break;
                }

                if(r > n) break;                        //到达边界结束

                l ++;
                if(l == r) r ++;                        //如果 l == r ，这种情况不符合讨论的情况 让 r ++
            }
            printf("%d %d %d\n", sum, al, ar);
        }

    }

    return 0;
}

总结

1.当让我们求某个区间的和的绝对值大小时，我们可以求 前缀，在对这个前缀可排序（这个前缀要保留 id位置的），这数据变的有序了我们更容易处理问题

2.确定好 我们尺取 维护的是 那个值、什么值