1400 : Composition

• 時(shí)間限制:10000ms
• 單點(diǎn)時(shí)限:1000ms
• 內(nèi)存限制:256MB

描述

Alice writes an English composition with a length of N characters. However, her teacher requires that M illegal pairs of characters cannot be adjacent, and if 'ab' cannot be adjacent, 'ba' cannot be adjacent either.
In order to meet the requirements, Alice needs to delete some characters.
Please work out the minimum number of characters that need to be deleted.

輸入

The first line contains the length of the composition N.
The second line contains N characters, which make up the composition. Each character belongs to 'a'..'z'.
The third line contains the number of illegal pairs M.
Each of the next M lines contains two characters ch1
and ch2
,which cannot be adjacent.
For 20% of the data: 1 ≤ N ≤ 10
For 50% of the data: 1 ≤ N ≤ 1000
For 100% of the data: 1 ≤ N ≤ 100000, M ≤ 200.

輸出

One line with an integer indicating the minimum number of characters that need to be deleted.

樣例提示

Delete 'a' and 'd'.

樣例輸入

5
abcde
3
ac
ab
de

樣例輸出

2

我思路是這樣的 求出滿足條件的最長(zhǎng)子序列 然后與原始長(zhǎng)度相減即可
假設(shè)valid(ch1, ch2)表示ch1、ch2可以連在一起（很好實(shí)現(xiàn)，二維2626的bool矩陣即可）
用一個(gè)數(shù)組dp[26]表示目前以'a'+i結(jié)尾的最長(zhǎng)子序列的長(zhǎng)度，dp初始化為全0。 然后線性掃描原始串，每次更新table表即可
復(fù)雜度O（N26）**

#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <functional>
#include <array>
#include <list>

using namespace std;

char table[100001];
int dp[26] = {0};   //以第'a'+i個(gè)字符結(jié)尾，最長(zhǎng)的子序列
bool m[26][26];     //用來(lái)判斷兩個(gè)char是否可以相連
int main(int argc, char **argv)
{
    int N = 0, M = 0;
    scanf("%d\n", &N);
    scanf("%s", table);
    //cin >> table;
    //cin >> M;
    scanf("%d\n", &M);
    /*for (int i = 0; i != 26; ++i)
    {
        for (int j=0; j!=26; ++j)
        {
            m[i][j] = false;
        }
    }*/
    for (int i=0; i!=M; ++i)
    {
        char ch1, ch2;
        //scanf("%c%c\n", &ch1, &ch2);
        cin >> ch1 >> ch2;
        m[ch1 - 'a'][ch2 - 'a'] = true;
        m[ch2 - 'a'][ch1 - 'a'] = true;
    }

    for (int i=0; i<N; ++i)
    {
        char ch1 = table[i];
        int temp = 1;
        for (int j=0; j!=26; ++j)
        {
            if (m[ch1 - 'a'][j])
            {
                continue;
            }
            else
            {
                temp = max(temp, dp[j] + 1);
            }
        }
        dp[ch1 - 'a'] = temp;
    }

    cout << N - *max_element(dp, dp+26) << endl;
    return 0;
}