Poj 2566 Bound Found（尺取法经典题目，前缀和排序）

题目链接：http://poj.org/problem?id=2566点击打开链接

Bound Found
Time Limit: 5000MS         Memory Limit: 65536K
Total Submissions: 5089         Accepted: 1622         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

题目大意：给定一个t，求一个子区间与t的差的绝对值最小。

最近刚学尺取法碰到这道经典题，看懂题解后以新手的角度解析这道题。

经典的尺取法。

首先我们看到这道题用尺取法O（n）复杂度简单方便，但是尺取法要单调序列才能用，但题目有正有负该怎么办呢？

这时候我们看到题目是一个子区间 ，而且是单调，这时候我们会想到前缀和数组。刚好满足尺取条件。果断构造前缀数组，然后排序（注意ad[0]=0不要忘，而且要加入排序。因为ad[j]-ad[0]相当去前j个序列和）最后输入一个t执行尺取法一次，再输出即可。

尺取法理解后十分简单，主要就是区间两端的指针l，r不断变换，保留一定条件，以达到o（n）复杂度。多刷几道相关简单题，明白思路后刷几道变形题，基本就会了。

#include
#include
#include
using namespace std;
#define INF 2000000010
#define maxn 100000+100
int su[maxn],qu[maxn];
struct point
{
	int s;
	int loc;
}ad[maxn];
bool cmp(point a,point b)
{
	return a.st)
		        l++;
		    else
		        break;//尺取爬行过程 
		    if(l==r)
		        r++; //注意不要忘了相等时右指针向右移 
		}
		if(pl>pr)
		swap(pl,pr);//由于abs(sum[r]-sum[l])==abs(sum[l]-sum[r])，排序不影响结果，但最后要把顺序正过来。 
		printf("%d %d %d\n",pans,pl+1,pr);
	}
}
	return 0;	
}