POJ 2566 Bound Found(尺取法,前缀和)

Bound Found
Time Limit: 5000MS   Memory Limit: 65536K
Total Submissions: 5207   Accepted: 1667   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

Source

 

题意:给n和k,输入n个数和k个t,求一个连续子序列使得这个连续子序列的和最接近t,输出这个子序列的和和它的左右端点

解题思路:尺取法.要使用尺取法必须要保证数列的单调性,而这道题输入的n个数并不是单调的,然后借鉴了大佬的博客...这里用到了前缀和,预处理出前i个数的前缀和,和编号i一起放入pair中,然后根据前缀和从小到大排序,此时sum[r]-sum[l]就有了单调性。可以通过比较sum[r]-sum[l]与t的关系不断进行更新,如果sum[r]-sum[l]<t,说明和可以更大,所以r++;如果sum[r]-sum[l]>t,说明和可以更小,所以l++;如果sum[r]-sum[l]=t,必定是最小答案。由于序列不能为空,所以当l=r时,r++。更新答案的时候左右区间端点为乱序,输出的时候交换一下即可

 

因为前缀和不单调,所以需要先排序。在原数组开头添加0,求出前缀数组。题目即转化为在前缀数组中找pre[i],pre[j],两者之差最接近t,。对于每次找到的2个下标分别为i和j的2个数,所对应a的区间为[min(i, j) + 1, max(i, j)]。

那么前缀数组排序后,尺取法便可以求得最接近t的值。

 

 1 #include<cstdio>  
 2 #include<cstring>  
 3 #include<iostream>  
 4 #include<algorithm>  
 5 #define debu  
 6 using namespace std;
 7 const int maxn = 1e5 + 50;
 8 const int INF = 0x3f3f3f3f;
 9 struct Node
10 {
11     int id, tot;
12 };
13 Node a[maxn];
14 int n, q, ans, ansl, ansr;
15 int cmp(Node a, Node b)
16 {
17     if (a.tot == b.tot) return a.id<b.id;
18     else return a.tot<b.tot;
19 }
20 int main()
21 {
22     while (scanf("%d%d", &n, &q) != EOF && (n + q))
23     {
24         a[0].id = 0, a[0].tot = 0;
25         for (int i = 1; i <= n; i++)
26         {
27             int x;
28             scanf("%d", &x);
29             a[i].id = i;
30             a[i].tot = a[i - 1].tot + x;
31         }
32         sort(a, a + n + 1, cmp);
33         for (int i = 0; i<q; i++)
34         {
35             int t;
36             scanf("%d", &t);
37             int l = 0, r = 1, minx = INF;
38             while (l <= r && r <= n)
39             {
40                 int tmp = a[r].tot - a[l].tot;
41                 if (abs(tmp - t)<minx)
42                 {
43                     minx = abs(tmp - t);
44                     ans = tmp;
45                     ansl = a[l].id;
46                     ansr = a[r].id;
47                 }
48                 if (tmp<t) r++;
49                 else if (tmp>t) l++;
50                 else break;
51                 if (l == r) r++;
52             }
53             if (ansl>ansr) swap(ansl, ansr);
54             printf("%d %d %d\n", ans, ansl + 1, ansr);
55         }
56     }
57     return 0;
58 }

 

posted on 2018-02-21 23:58  蔡军帅  阅读(342)  评论(0编辑  收藏  举报