离散化题目：人潮最多的時段（ Interval Partitioning Problem ）

题目：一群访客参加宴会，我们询问到每一位访客的进场时刻与出场时刻（即：已知访客的进场时刻与出场时刻），请问宴会现场挤进最多人的时段。

这个问题的关键不是访客，而是出入时刻。

struct Guest { int arrival; int leave; } g[7] = { { 10, 12 }, { 12, 14 }, { 12, 13 }, { 9, 15 }, {13, 15 }, { 14, 15 }, {15,15} };

bool cmp(const int& i, const int& j)
{
    return abs(i) < abs(j);
}

int maximum_guest()
{//返回宾客最多的人数
    
    vector<int> time;
    for (int i = 0; i<7; ++i)
    {
        time.push_back(+g[i].arrival);
        time.push_back(-g[i].leave);
    }
for (auto&u : time)
    {
        cout << u << " ";
    }
    sort(time.begin(), time.end(), cmp);
    cout << endl;
    for (auto&u : time)
    {
        cout << u << " ";
    }
    int n = 0, maximum = 0;
    int i;
    for ( i = 0; i<time.size(); ++i)
    {
        if (time[i] >= 0)
            n++;
        else
            n--;

        maximum = max(maximum, n);
    }

    cout<<endl << "人潮最多的時段有" << maximum << "人";

    return maximum;
}

void max_time()
{//问题主函数
    int maximum = maximum_guest();
    vector<int> time;
    for (int i = 0; i<7; ++i)
    {
        time.push_back(+g[i].arrival);
        time.push_back(-g[i].leave);
    }
    sort(time.begin(), time.end(), cmp);
    int n = 0;
    int flag = 1;
    int temp;
    for (int i = 0; i<time.size(); ++i)
    {
        if (time[i] >= 0)
            n++;
        else
            n--;

        if (n == maximum)
        {
            temp = i;
            flag = 0;
        }
        if (flag == 0 && n == maximum - 1)
        {
            cout << endl << "time:" << time[temp] << "to" << time[i];
        }
    }
}

int main()
{
    max_time();
    getchar();
    return 0;
}

转载请注明：http://www.cnblogs.com/firstcxj/p/4566385.html