编程之美——区间重合判断

将源区间按照x坐标大小排序，然后进行合并，使用二分查找目标区间x，y坐标在源区间中的分布，若两者分布于同一连续区间，则区间重合

代码：

 1 #include<iostream>
 2 #include<algorithm>
 3 using namespace std;
 4
 5 struct section
 6 {
 7     int low;
 8     int high;
 9     bool operator<(const section &l)const
10     {
11         return low<l.low;
12     }
13 };
14
15 int getIndex(int num);
16
17 const int N=1001;
18 section source[N];
19 const int K=101;
20 section des[K];
21 //源区间合并后段数
22 int ncnt=0;
23
24 int main()
25 {
26     //n:源区间个数 k：目标区间个数
27     int n,k;
28     cin>>n>>k;
29     for(int i=0;i<n;++i)
30         cin>>source[i].low>>source[i].high;
31     for(int i=0;i<k;++i)
32         cin>>des[i].low>>des[i].high;
33
34     sort(source,source+n);
35
36     int highnum=source[0].high;
37     for(int i=1;i<n;++i)
38         if(source[i].low<=highnum)
39         {
40             //处理[1,6] [2,5]查[2,6]这种情况
41             if(source[i].high<highnum)
42                 continue;
43             else
44                 highnum=source[i].high;
45         }
46         else
47         {
48             source[ncnt++].high=highnum;
49             source[ncnt].low=source[i].low;
50             highnum=source[i].high;
51         }
52     source[ncnt++].high=highnum;
53
54     for(int i=0;i<k;++i)
55     {
56         int indexlow=getIndex(des[i].low);
57         int indexhigh=getIndex(des[i].high);
58         if(indexlow==indexhigh&&indexlow!=-1&&des[i].high<source[indexhigh].high)
59             cout<<"the "<<i+1<<" section is match"<<endl;
60         else
61             cout<<"the "<<i+1<<" section is not match"<<endl;
62     }
63
64     return 0;
65 }
66
67 int getIndex(int num)
68 {
69     if(num<source[0].low||num>source[ncnt-1].high)
70         return -1;
71     int u=0;
72     int v=ncnt-1;
73     while(u<=v)
74     {
75         int m=(u+v)>>1;
76         if(num>=source[m].low)
77             u=m+1;
78         else
79             v=m-1;
80     }
81
82     return v;
83 }