hdu6103（尺取法）

Kirinriki

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 456    Accepted Submission(s): 160

Problem Description

We define the distance of two strings A and B with same length n is
disA,B=∑i=0n−1|Ai−Bn−1−i|
The difference between the two characters is defined as the difference in ASCII.
You should find the maximum length of two non-overlapping substrings in given string S, and the distance between them are less then or equal to m.

 

Input

The first line of the input gives the number of test cases T; T test cases follow.
Each case begins with one line with one integers m : the limit distance of substring.
Then a string S follow.

Limits
T≤100
0≤m≤5000
Each character in the string is lowercase letter, 2≤|S|≤5000
∑|S|≤20000

 

Output

For each test case output one interge denotes the answer : the maximum length of the substring.

 

Sample Input

1 5 abcdefedcb

 

Sample Output

5

Hint

[0, 4] abcde [5, 9] fedcb The distance between them is abs('a' - 'b') + abs('b' - 'c') + abs('c' - 'd') + abs('d' - 'e') + abs('e' - 'f') = 5

 

Source

2017 Multi-University Training Contest - Team 6

題意：

給出字符串s，尋找其兩個長度相同且不重疊的子串，滿足其每位的ascil差值之和不大於m，且長度最長。

思路：

因爲字符串的長度爲5000，所以可以枚舉兩個子串的對稱軸，然後對其區間進行尺取法，這樣只需要o(n^2)的複雜度即可。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <algorithm>
#include <queue>
#include <map>
using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 5010;

char a[maxn], num[maxn];
int m;

int solve(int len)
{
    int s = 0, t = 0, ans = 0, sum = 0;
    while (1)
    {
        while (sum + num[t] <= m && t < len)
        {
            sum += num[t];
            ans = max(t - s + 1, ans);
            t++;
        }
        sum -= num[s++];
        if (s >= len) break;
    }
    return ans;
}

int main()
{
    int t;
    scanf("%d", &t);
    while (t--)
    {
        int len, res = 0;
        scanf("%d", &m);
        scanf("%s", a);
        len = strlen(a);
        for (int i = 0; i <= len; i++)   //枚舉折的中點
        {
            int cnt = 0;
            for (int j = 1; j + i < len && i - j >= 0; j++)   //枚舉奇數情況
                num[cnt++] = abs(a[j + i] - a[i - j]);
            res = max(res, solve(cnt));
        }
        for (int i = 0; i <= len; i++)   //枚舉折的中點
        {
            int cnt = 0;
            for (int j = 1; j + i - 1 < len && i - j >= 0; j++)   //枚舉偶數情況
                num[cnt++] = abs(a[j + i - 1] - a[i - j]);
            res = max(res, solve(cnt));
        }
        printf("%d\n", res);
    }
    return 0;
}

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <algorithm>
#include <queue>
#include <map>
using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 5010;

char a[maxn], num[maxn];
int m;

int solve(int len)
{
	int s = 0, t = 0, ans = 0, sum = 0;
	while (1)
	{
		while (sum + num[t] <= m && t < len)
		{
			sum += num[t];
			ans = max(t - s + 1, ans);
			t++;
		}
		sum -= num[s++];
		if (s >= len) break;
	}
	return ans;
}

int main()
{
	int t;
	scanf("%d", &t);
	while (t--)
	{
		int len, res = 0;
		scanf("%d", &m);
		scanf("%s", a);
		len = strlen(a);
		for (int i = 0; i <= len; i++)	//枚舉折的中點
		{
			int cnt = 0;
			for (int j = 1; j + i < len && i - j >= 0; j++)	//枚舉奇數情況
				num[cnt++] = abs(a[j + i] - a[i - j]);
			res = max(res, solve(cnt));
		}
		for (int i = 0; i <= len; i++)	//枚舉折的中點
		{
			int cnt = 0;
			for (int j = 1; j + i - 1 < len && i - j >= 0; j++)	//枚舉偶數情況
				num[cnt++] = abs(a[j + i - 1] - a[i - j]);
			res = max(res, solve(cnt));
		}
		printf("%d\n", res);
	}
	return 0;
}