51nod-1732 51nod Marriage Agency

1732 51nod Marriage Agency 

Topic source: original

Baseline Time Limit: 1 second Space Limit: 131072 KB Score: 40  Difficulty: Level 4 Algorithm Questions

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In addition to doing OJ, 51nod has also started a lot of side businesses. A marriage agency is one of them.

For a customer, we can use a string to describe the characteristics of the customer.

Suppose now we have two customers A and B.

A's trait string is: abcdefg

B's trait string is: abcxyz

Then the matching degree f(A, B) of A and B is the length of the longest common prefix of A and B, that is, len('abc') = 3

Due to the tight budget of 51nod recently, the Jacket Master designed a compression algorithm to save memory.

All user trait strings are stored in a string S of length n. (n <= 1000) The user's trait is represented by an integer p, indicating that the user's trait string is S[p...n - 1].

Now given a string S, query <ai, bi> q times (ai, bi are valid user trait integers respectively). Please output the customer matching degree corresponding to q queries.

Input

Now given string length n, and string S. Next is the integer q, which means that there are q queries next.
The following line q has two integers ai, bi. Represents the matching degree of users whose query characteristics are ai and bi.

1 <= n <= 1000
1 <= q <= 10^6

All data entered is legal.

Output

Each line outputs a user matching degree integer.

Input example

12
loveornolove
5
3 7
0 0
9 1
3 1
9 5

Output example

0
12
3
0
0

题解：本来想暴力预处理看看T了几个样例的，结果竟然过了。。样例真水。本方法预处理出所有的情况结果，最坏情况O(n^3)。。。。

AC代码

#include <stdio.h>
#include <iostream>
#include <string>
#include <queue>
#include <map>
#include <vector>
#include <algorithm>
#include <string.h>
#include <cmath>
 
using namespace std;

const int maxn = 1111;
char a[1111];
int dp[maxn][maxn] = {0};

int get_ans(int l, int r, int n){
	for(int i = 0; i + r < n; i++)
		if(a[l + i] != a[r + i])
			return i;
	return n - r;
}

void solve(int n){
	for(int len = 0; len < n; len++){
		for(int i = 0; i + len < n; i++){
			dp[i][i + len] = get_ans(i, i + len, n);
		}
	}
}

int main(){
	int n, q, l, r;
	scanf("%d", &n);
	scanf("%s", a);
	solve(n);
	scanf("%d", &q);
	for(int i = 0; i < q; i++){
		scanf("%d%d", &l, &r);
		printf("%d\n", dp[min(l, r)][max(l, r)]);
	}
	return 0;
}