poj 2566 Bound Found 尺取法变形

时间：2016-01-14 14:10:26 阅读：168 评论：0 收藏：0 [点我收藏+]

标签：

Bound Found

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 2277		Accepted: 703		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

Source

Ulm Local 2001

题意：给你n个数字，这些数字可正可负，再给你个数字t,求在这个数列中一个连续的子序列，和的绝对值与t相差最小

错因分析：刚开始看到这道题目觉得很变态，尺取法怎么可能做得出来，这些数字是可正可负的，，后来看了题解，才恍然大悟，正因为是可正可负所以不能直接用尺取法去解题（尺取法必须是找到单调性），需要进行转化，找到单调性，但是a[i]数组是不能改变的，因为要求连续，那么就自然想到对a[i]求前缀和

先贴上别人好的代码，核心思想：对sum进行从小到大的排序，注意需要加入<0,0>这点（为了求得单独的sum，就是从第一个节点到sum[i]对应的i节点），然后尺取法设置左右两个端点，直接在sum数组上进行尺取法，需要注意的是对两个端点移动的判断，假设得出的int temp=p[r].first - p[l].first<k,说明temp偏小，那么r端点右移，否则>k的话，说明temp偏大，需要l端点右移，但是需要注意不能出现l==r的情况，因为序列不能为空，l==r，则序列为空;还有一点很重要的是因为求得是和的绝对值，即|sum[j]-sum[i]|，那么只需要对sum数组进行从小到大排序，用大的减去小的就可以了（这也是去绝对值的体现）

#include<cstdio>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include<map>
#include <algorithm>
#include <set>
using namespace std;
#define MM(a) memset(a,0,sizeof(a))
typedef long long LL;
typedef unsigned long long ULL;
const int mod = 1000000007;
const double eps = 1e-10;
const int inf = 0x3f3f3f3f;
pair<int, int > p[100005];
int n, m, k;
void solve(int k)
{
	int l = 0, r = 1, al, ar, av, minn = inf;
	  while (l<=n&&r<=n&&minn!=0)
		{
			int temp=p[r].first - p[l].first;
			if (abs(temp - k) < minn)
			{
				minn = abs(temp - k);
				ar = p[r].second;
				al = p[l].second;
				av = temp;
			}
			if (temp> k)
				l++;
			else if (temp < k)
				r++;
			else
				break;
			if (r == l)
				r++;
	   }
    if(al>ar)
        swap(al,ar);//因为al和ar大小没有必然关系（）取绝对值，所以//要交换
	printf("%d %d %d\n", av, al+1, ar);
}
int main()
{
	while (~scanf("%d %d", &n, &m))
	{
		if (!n&&!m) return 0;
		p[0] = make_pair(0, 0);
		for (int i = 1; i <= n; i++)
		{
			scanf("%d", &p[i].first);
			p[i].first += p[i - 1].first;
			p[i].second = i;
		}
		sort(p, p + n + 1);
		while (m--)
		{
			scanf("%d", &k);
			solve(k);
		}
	}
	return 0;
}

　　下面是自己的wa代码

好好找茬

#include<cstdio>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include<map>
#include <algorithm>
#include <set>
using namespace std;
#define MM(a) memset(a,0,sizeof(a))
typedef long long LL;
typedef unsigned long long ULL;
const int mod = 1000000007;
const double eps = 1e-10;
const int inf = 0x3f3f3f3f;
pair<int, int > p[100005];
int n, m, k;
void solve(int k)
{
	int l = 0, r = 1, al, ar, av, minn = inf;
	  while (l<=n&&r<=n)
		{
			int temp = p[r].first - p[l].first;
			if (abs(temp - k) < minn)
			{
				minn = abs(temp - k);
				ar = p[r].second;
				al = p[l].second;
				av = temp;
			}
			if (temp > k)
				l++;
			else if (temp < k)
				r++;
			else
				break;
			if (r == l)
				r++;
	}
	printf("%d %d %d\n", av, al+1, ar);
}
int main()
{
	while (~scanf("%d %d", &n, &m))
	{
		if (!n&&!m) return 0;
		p[0] = make_pair(0, 0);
		for (int i = 1; i <= n; i++)
		{
			scanf("%d", &p[i].first);
			p[i].first += p[i - 1].first;
			p[i].second = i;
		}
		sort(p + 1, p + n + 1);
		while (m--)
		{
			scanf("%d", &k);
			solve(k);
		}
	}
	return 0;
}

poj 2566 Bound Found 尺取法变形

标签：

原文地址：http://www.cnblogs.com/smilesundream/p/5129758.html

踩

(0)

评论一句话评论（0）

分享档案

更多>

2021年07月29日 (22)
2021年07月28日 (40)
2021年07月27日 (32)
2021年07月26日 (79)
2021年07月23日 (29)
2021年07月22日 (30)
2021年07月21日 (42)
2021年07月20日 (16)
2021年07月19日 (90)
2021年07月16日 (35)

周排行

poj 2566 Bound Found 尺取法 变形

poj 2566 Bound Found 尺取法变形