POJ 2566 Bound Found(最经典尺取法)

最新推荐文章于 2020-10-07 20:48:42 发布

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本文介绍了一种算法，用于解决从给定整数序列中找到一个连续子序列的问题，该子序列的绝对和最接近给定的目标值。通过使用前缀和排序并进行两指针搜索的方法来高效地解决此问题。

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Bound Found

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions:5177		Accepted:1657		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

15 1 15

求前缀和，对前缀和排序，对前缀和作差找出最接近t的区间
代码：

#include<cstdio>
#include<algorithm>
#include<math.h>
using namespace std;
struct node
{
    int id,cost;
} c[100004];
bool cmp(const node& a,const node& b)
{
    return a.cost<b.cost;
}
int main()
{
    int n,K;
    while(~scanf("%d%d",&n,&K))
    {
        if(!(n+K))break;
        int i,j;
        c[0].id=0,c[0].cost=0;
        for(i=1; i<=n; i++)
        {
            scanf("%d",&c[i].cost);
            c[i].cost+=c[i-1].cost;
            c[i].id=i;
        }
        sort(c,c+n+1,cmp);
        while(K--)
        {
            int st=0,en=1,t,res=1e9+7,q,r,mc;
            scanf("%d",&t);
            while(en<=n)
            {
                int s=c[en].cost-c[st].cost;
                if(abs(s-t)<res)
                {
                    res=abs(s-t);
                    mc=s;
                    q=c[st].id;r=c[en].id;
                }
                if(s<t)en++;
                else st++;
                if(st==en)en++;
            }
            printf("%d %d %d\n",mc,min(q,r)+1,max(q,r));
        }
    }
    return 0;
}