One-by-one to solve the Sequence Problem (1)_solve it one by one-优快云博客

本文探讨了如何通过删除最少数量的字符来解决序列问题，确保非法字符对不相邻。使用C++实现了一个算法，该算法考虑了不同数据规模的要求，并提供了一个具体的代码示例。

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One-by-one to solve the Sequence Problem (1)

1400 : Composition （from hihocoder）

时间限制:10000ms
单点时限:1000ms
内存限制: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.
输入
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.
输出
One line with an integer indicating the minimum number of characters that need to be deleted.

样例提示
Delete ‘a’ and ‘d’.

样例输入

5
abcde
3
ac
ab
de

样例输出

Code

#include <iostream>
#include <list>
#include <vector>
#include <string>
#include <unordered_map>
#include <unordered_set>
#include <utility>
using namespace std;
typedef unsigned int unint;
typedef unordered_set<string> setstr;
typedef unordered_map<char, unint> mapchar;
typedef pair<char, unint> pcu;
int main() {
    int N = 0;
    while (cin >> N) {
        string ts;
        cin >> ts;
        list<char> mstr(ts.begin(), ts.end());
        int m = 0;
        cin >> m;
        setstr mp;
        for (int i = 0; i < m; ++i)
        {
            string s1;
            cin >> s1;
            mp.insert(s1);
            char tc = s1[0];
            s1[0] = s1[1];
            s1[1] = tc;
            mp.insert(s1);

        }
        mapchar mmp;
        mmp.insert(pair<char,unint>(ts[0], 0));
        for (int i =1;i<N; ++i) {
            char c = ts[i];
            mapchar tmmp;
            string s11(2,'0');
            auto resf0 = mmp.find(c);
            tmmp.insert(pcu(c, i));
            s11[1] = c;
            for (auto it : mmp) {
                // delete
                it.second += 1;
                auto resf = tmmp.find(it.first);
                if (resf == tmmp.end()) {
                    tmmp.insert(it);
                }
                else {
                    if (resf->second > it.second)
                        resf->second = it.second;
                }
                // keep it
                it.second -= 1;
                s11[0] = it.first;
                auto resf1 = mp.find(s11);
                if (resf1 == mp.end()) {
                    // insert it
                    auto resf11 = tmmp.find(c);
                    if (resf11 == tmmp.end()) 
                    {
                        tmmp.insert(pcu(c, it.second));
                    }
                    else {
                        if (resf11->second > it.second)
                            resf11->second = it.second;
                    }
                }
            }// for it : mmp
            mmp = tmmp;
        }
        unint res1 = 0x0fffffff;
        for (auto i : mmp) {
            if (res1 > i.second)
                res1 = i.second;
        }
        cout << res1 << endl;
    }// end of while
    return 0;
}