UVa OJ 1451 - Average
Problem
給定一個長度爲n的01串,選一個長度至少爲L的連續子串,使得子串中數字的平均值最 大。如果有多解,子串長度應儘量小;如果仍有多解,起點編號儘量小。序列中的字符編號 爲1~n,因此[1,n]就是完整的字符串。1≤n≤100000,1≤L≤1000。
例如,對於如下長度爲17的序列00101011011011010,如果L=7,最大平均值爲6/8(子 序列爲[7,14],其長度爲8);如果L=5,子序列[7,11]的平均值最大,爲4/5。
Input
Your program is to read from standard input. The input consists of T test cases. The number of test cases T is given in the first line of the input. Each test case starts with a line containing two integers n (1 ≤ n ≤ 100, 000) and L (1 ≤ L ≤ 1, 000) which are the length of a binary sequence and a length lower bound, respectively. In the next line, a string, binary sequence, of length n is given.
Output
Your program is to write to standard output. Print the starting and ending index of the subsequence.
Sample Input
2
17 5
00101011011011010
20 4
11100111100111110000
Sample Output
7 11
6 9
Solution
說來慚愧,這道題目我WA了快一面了。一開始看錯題目,提交了好多次都錯了,自己還沒反應過來,一個勁傻傻地改。後來發現的時候,已經沒有做下去的興致了。
不過最終還是把題目完成了的。總的來說,這道題目還是讓我有些收穫的。題目的解題思路是通過將數列轉化成座標軸上的圖像,平均值這時候也就變成了斜率了。然後一個個點去維護下凸函數的單調數列,找到最優解。
一開始我的cntAverage()這個函數只有三個參數,使用的是除法,提交之後花了0.1s,自己感覺慢了。於是把除法改成了現在的乘法,時間縮短一半,變成了0.05s。可見乘法和除法在計算的效率上還是相差很多的。
#include <iostream>
#include <string>
using namespace std;
const int maxn = 100005;
int n, L, start, ending,temp;
double maxd;
int DNA[maxn],cav[maxn];
string str;
inline int cntAverage(int L, int r, int LL, int rr, int DNA[]){
return (DNA[r]-DNA[L])*(rr-LL) - (DNA[rr]-DNA[LL])*(r-L);
}
inline void changePoint(int pits,int bump){
double temp;
temp = cntAverage(pits, bump, start-1, ending, DNA);
if (temp < 0) return;
if (temp || (bump-pits) < (ending-start+1)){
maxd = temp;
start = pits + 1;
ending = bump;
}
}
int main(){
ios::sync_with_stdio(false);
cin.tie(0);
int cas;
cin >> cas;
while (cas--){
int find = 0, cor = -1;
cin >> n >> L; cin.get();
getline(cin, str);
for (int i = 1; i <= n; ++i)
DNA[i] = DNA[i-1] + (str[i-1] == '1');
maxd = DNA[L] / L, start = 1, ending = L;
for (int i = L; i <= n; ++i){
temp = i - L;
while (find < cor && cntAverage(cav[cor], temp, cav[cor - 1], cav[cor], DNA) <= 0)
--cor;
cav[++cor] = temp;
while (find < cor && cntAverage(cav[find], i, cav[find + 1], i, DNA) <= 0)
++find;
changePoint(cav[find],i);
}
cout << start << ' ' << ending << '\n';
}
return 0;
}