poj 2566 Bound Found（取尺法，思路）

Bound Found

Time Limit: 5000MS Memory Limit: 65536KTotal Submissions: 2719 Accepted: 826 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 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：题意：给个t，求一个子区间与t的差的绝对值最小。

思路：n达到了100000，此题就不能用O（n²）枚举区间来做了。考虑连续区间上的xx问题，我们可以用尺取法。可以尺取法要求区间具有单调性，这里有正有负，怎么办？

很简单，我们要求任意区间的和就需要求出前缀数组，然后我们对前缀数组排序就可以了。  排完序后你找的任意两个点都会对应原序列的一个区间（因为会取绝对值，后面减去前面也没关系） 然后用尺取法推进，找出最小的一个即可。

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;#define N 100050#define INF 2000000010int a[N];struct Node{    int v,id;}sum[N];bool cmp(Node a,Node b){    return a.v<b.v;}int main(){    int n,k,t;    while(~scanf("%d %d",&n,&k)&&(n+k))    {        sum[0].v=0;sum[0].id=0;        for(int i=1;i<=n;i++)        {            scanf("%d",&a[i]);            sum[i].v=sum[i-1].v+a[i];            sum[i].id=i;        }        sort(sum,sum+1+n,cmp);        int ans;        while(k--)        {            scanf("%d",&t);            ans=INF;            int l=0,r=1;            int anss,ansl,ansr;            while(l<=n&&r<=n)            {                int w=abs(sum[r].v-sum[l].v);                int q=abs(w-t);                if(q<ans)                {                    anss=w;                    ans=q;                    ansl=sum[l].id;                    ansr=sum[r].id;                }                if(w>t) l++;                else if(w<t) r++;                else break;                if(l==r) r++;            }            if(ansl>ansr) swap(ansl,ansr);            printf("%d %d %d\n",anss,ansl+1,ansr);        }    }    return 0;}