poj 2566 Bound Found（尺取法 好题）
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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\n",ans,ansl+1,ansr); 58 } 59 60 } 61 return 0; 62 }
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