[Timue: 1019. Line Painting]

原题地址：[Timue: 1019. Line Painting]

不知道该怎么解比较好，于是选择去模拟这个过程：

搞了一个map去存储住这些区间，并且在添加和删除时把前后有相交的区间处理一下，合并还是切割还是怎么样。

#include <iostream>
#include <map>
#include <algorithm>

using namespace std;

int main()
{
    map<int, pair<int, bool>> A;
    A[0] = make_pair(999999999, false);
    int N;
    cin >> N;
    for (int i = 0; i < N; ++i)
    {
        int x, y;
        char z;
        cin >> x >> y >> z;
        --y;
        bool c = z == 'b';
        decltype(A)::iterator iter;
        while ((iter = A.lower_bound(x)) != A.end() && iter->first <= y)
        {
            if (iter->second.first > y)
            {
                if (iter->second.second == c)
                {
                    y = iter->second.first;
                }
                else
                {
                    A[y + 1] = make_pair(iter->second.first, !c);
                }
            }
            A.erase(iter);
        }
        iter = A.lower_bound(x);
        bool add = true;
        if (iter != A.begin())
        {
            --iter;
            if (iter->second.first >= x)
            {
                if (iter->second.second == c)
                {
                    iter->second.first = max(y, iter->second.first);
                    add = false;
                }
                else
                {
                    if (iter->second.first > y)
                    {
                        A[y + 1] = make_pair(iter->second.first, !c);
                    }
                    iter->second.first = x - 1;
                }
            }
        }
        if (add)
        {
            A[x] = make_pair(y, c);
        }
    }
    pair<int, int> ans(0, -1);
    pair<int, int> current;
    bool continued = false;
    for (auto &x : A)
    {
        if (x.second.second)
        {
            continued = false;
        }
        else
        {
            if (!continued)
            {
                current.first = x.first;
                current.second = x.second.first;
            }
            else
            {
                current.second = x.second.first;
            }
            if (current.second - current.first > ans.second - ans.first)
            {
                ans = current;
            }
            continued = true;
        }
    }
    cout << ans.first << " " << ans.second + 1<< endl;

    return 0;
}