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  • poj 2566 Bound Found(尺取法 好题)

    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

     
    尺取法,注意 inf 初始化
     
     1 #include<iostream>
     2 #include<cstdio>
     3 #include<cstring>
     4 #include<algorithm>
     5 #include<set>
     6 #include<map>
     7 using namespace std;
     8 #define N 106006
     9 #define inf 1<<30
    10 pair<int,int> g[N];
    11 int main()
    12 {
    13     int n,k;
    14     while(scanf("%d%d",&n,&k)==2)
    15     {
    16         if(n==0 && k==0) 
    17           break;
    18         int sum=0;
    19         g[0]=make_pair(0,0);
    20         for(int i=1;i<=n;i++){
    21             int x;
    22             scanf("%d",&x);
    23             sum=sum+x;
    24             g[i]=make_pair(sum,i);
    25         }
    26         sort(g,g+n+1);
    27         while(k--){
    28             
    29             int val;
    30             scanf("%d",&val);
    31             
    32             int minn=inf;
    33             int ans,ansl=1,ansr=1;
    34             int s=0,t=1;
    35             for(;;){
    36                 if(t>n)
    37                   break;
    38                    if(minn==0)
    39                      break;
    40                 int num=g[t].first-g[s].first;
    41                 if(abs(num-val)<minn){
    42                     minn=abs(num-val);
    43                     ans=num;
    44                     ansl=g[s].second;
    45                     ansr=g[t].second;
    46                 }
    47                 
    48                  if(num<val)
    49                    t++;
    50                 if(num>val)
    51                   s++;
    52                 if(s==t)
    53                   t++;
    54             }
    55             if(ansl>ansr)  
    56                 swap(ansl,ansr);  
    57             printf("%d %d %d
    ",ans,ansl+1,ansr); 
    58         }
    59         
    60     }
    61     return 0;
    62 }
    View Code
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  • 原文地址:https://www.cnblogs.com/UniqueColor/p/4765295.html
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