HDU 6103 Kirinriki(思维尺取)

题意:在一个字符串中,选取两个长度相同且不重合的子串,计算dis值,定义为abcde efcgh 的a~h+b~g+c~c+d~f..即计算对应位置的差值和.

给定m,找到满足dis值小于等于m的最大子串对长度.

关键:枚举对称轴.进行尺取

复杂度 O(m^2).m<=5000

事实上,在见到小于等于m的时候,要想到尺取,对应位置差值和,应想到枚举对称轴.处理时注意可能对称到空处,*2后分奇偶讨论.

/*  xzppp  */
#include <iostream>
#include <vector>
#include <cstdio>
#include <string.h>
#include <algorithm>
#include <queue>
#include <map>
#include <string>
#include <cmath>
#include <bitset>
#include <iomanip>
using namespace std;
#define FFF freopen("in.txt","r",stdin);freopen("out.txt","w",stdout);
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define MP make_pair
#define PB push_back
#define _ %MOD
typedef long long  LL;
typedef unsigned long long ULL;
typedef pair<int,int > pii;
typedef pair<LL,LL> pll;
typedef pair<double,double > pdd;
typedef pair<double,int > pdi;
const int MAXM = 2e3+17;
const int MAXV = 2*1e3+17;
const int BIT = 15+3;
const int INF = 0x7fffffff;
const LL INFF = 0x3f3f3f3f3f3f3f3f;
const int MOD = 1e9+7;
const int MAXN = 1e5+17;
string str;
int main()
{
#ifdef GoodbyeMonkeyKing
FFF
#endif
int t,x;
cin>>t;
x = t;
while(t--)
{
int m;
cin>>m>>str;
int ans = 0;
for (int i = 1; i < 2*str.length(); ++i)
{
int l = i/2-1,r=i/2+1;
if(i&1) l++;
int temp = 0,lth = 0;
queue<int > q;
while(l>-1&&r<str.length())
{
q.push(abs(str[r]-str[l]));
temp += abs(str[r]-str[l]);
lth++;
if(temp>m&&!q.empty())
{
temp -= q.front(); q.pop();
lth--;
}
ans = max(lth,ans);
l--;r++;
}
}
cout<<ans<<endl;
}
return 0;
}