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。

 1 struct Guest { int arrival; int leave; } g[7] = { { 10, 12 }, { 12, 14 }, { 12, 13 }, { 9, 15 }, {13, 15 }, { 14, 15 }, {15,15} };
 2 
 3 bool cmp(const int& i, const int& j)
 4 {
 5     return abs(i) < abs(j);
 6 }
 7 
 8 int maximum_guest()
 9 {//返回宾客最多的人数
10     
11     vector<int> time;
12     for (int i = 0; i<7; ++i)
13     {
14         time.push_back(+g[i].arrival);
15         time.push_back(-g[i].leave);
16     }
17 for (auto&u : time)
18     {
19         cout << u << " ";
20     }
21     sort(time.begin(), time.end(), cmp);
22     cout << endl;
23     for (auto&u : time)
24     {
25         cout << u << " ";
26     }
27     int n = 0, maximum = 0;
28     int i;
29     for ( i = 0; i<time.size(); ++i)
30     {
31         if (time[i] >= 0)
32             n++;
33         else
34             n--;
35 
36         maximum = max(maximum, n);
37     }
38 
39     cout<<endl << "人潮最多的時段有" << maximum << "人";
40 
41     return maximum;
42 }
43 
44 void max_time()
45 {//问题主函数
46     int maximum = maximum_guest();
47     vector<int> time;
48     for (int i = 0; i<7; ++i)
49     {
50         time.push_back(+g[i].arrival);
51         time.push_back(-g[i].leave);
52     }
53     sort(time.begin(), time.end(), cmp);
54     int n = 0;
55     int flag = 1;
56     int temp;
57     for (int i = 0; i<time.size(); ++i)
58     {
59         if (time[i] >= 0)
60             n++;
61         else
62             n--;
63 
64         if (n == maximum)
65         {
66             temp = i;
67             flag = 0;
68         }
69         if (flag == 0 && n == maximum - 1)
70         {
71             cout << endl << "time:" << time[temp] << "to" << time[i];
72         }
73     }
74 }
75 
76 int main()
77 {
78     max_time();
79     getchar();
80     return 0;
81 }

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posted @ 2015-06-10 16:17  ChessChan  阅读(1059)  评论(0)  编辑  收藏  举报