1400 : Composition

• 時間限制:10000ms
• 單點時限:1000ms
• 內存限制: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子序列 然后與原始長度相減即可
假設valid(ch1, ch2)表示ch1帖池、ch2可以連在一起（很好實現其馏，二維2626的bool矩陣即可）
用一個數組dp[26]表示目前以'a'+i結尾的最長子序列的長度虎囚，dp初始化為全0湾揽。 然后線性掃描原始串，每次更新table表即可
復雜度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個字符結尾链快，最長的子序列
bool m[26][26];     //用來判斷兩個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;
}