给定一个源区间[x,y]和N个无序的目标区间[x1,y1] [x2,y2] ... [xn,yn],判断源区间[x,y]是不是在目标区间内
解法1:
先用区间的左边界值对目标区间进行排序O(nlogn),对排好序的区间进行合并O(n),对每次待查找的源区间,二分法查出其左右两边界点分别处于合并后的哪个源区间中O(logn),若属于同一个源区间则说明其在目标区间中,否则就说明不在。
- struct region
- {
- int start;
- int over;
- bool operator<(const region& r) const
- {return start<r.start;}
- };
- //binary-research by merged[].start to find the most nearly merged[].start
- int check(region m[],int length,int testId)
- {
- int low=0,high=length+1;
- int mid;
- while(low<=high)
- {
- mid=(low+((high-low)>>1));
- if(m[mid].start <= testId) low=mid+1;
- else
- high=mid-1;
- }
- return high;//返回的是high值,此时high小于等于mid,小于low;
- }
- int main()
- {
- region r[30],merged[30],test;
- int n;//count of array
- int m;//conut of merged array
- cin>>n;
- for(int i=0; i<n; i++)
- cin>>r[i].start>>r[i].over;
- cin>>test.start>>test.over;
- sort(r,r+n);//sort by start time
- cout<<endl;
- cout<<"Sorted Region: "<<endl;
- for(int i=0;i<n;i++) cout<<r[i].start<<" "<<r[i].over<<endl;
- cout<<endl;
- //mergeRegions;
- m=0;
- int lasthigh = r[0].over;
- merged[0].start = r[0].start;
- merged[0].over = r[0].over;
- for (int i=1; i<n; i++)
- {
- if (lasthigh >= r[i].start)//注意:>= 合并操作
- {
- lasthigh = lasthigh>r[i].over?lasthigh:r[i].over;//lasthigh等于较大值
- merged[m].over = lasthigh;
- }
- else //扩展一个新的区间
- {
- m++;
- merged[m].start = r[i].start;
- merged[m].over = r[i].over;
- lasthigh = r[i].over;
- }
- }
- cout<<"Merged Region: "<<endl;
- for(int i=0;i<=m;i++) cout<<merged[i].start<<" "<<merged[i].over<<endl;
- cout<<endl;
- //check the test line binary-research
- int startId = check(merged, m, test.start);
- int overId = check(merged, m, test.over);
- if(startId==overId && test.over<=merged[overId].over)
- cout<<"OK! The test line is in the set."<<endl;
- else
- cout<<"False! The test line is not in the set."<<endl;
- }
解法2:
利用并查集,首先初始化每个元素的代表节点father[i]=i等于其本身,count[i]=1;然后每输入一个区间,合并一次(遍历区间内的每一个元素,更新其代表);最后查询待查区间的首位两个元素是否在同一区间内。这种方法将区间转化为离散的集合,操作容易,但是浪费空间比较严重,对于大规模的区间不太实用。
- int father[SIZE];
- int count[SIZE];
- void initail(int num)
- {
- for (int i=0; i<num; i++)
- {
- father[i]=i;//每个集合的代表是自己
- count[i]=1; //代表一个元素
- }
- }
- void merge(int x, int y)
- {
- if(father[x]==father[y])
- return;
- else
- {
- if(count[x]>=count[y])
- {
- father[y]=father[x];
- }
- else
- father[x]=father[y];
- count[father[x]]++;
- }
- }
- int main()
- {
- memset(father,-1,sizeof(father));
- memset(count,1,sizeof(count));
- int n,t0,t1;
- cin>>n;
- initail(SIZE);//需要初始化整个可能的区间
- while(n--)
- {
- cin>>t0>>t1;
- if(t0>t1)
- swap(t0,t1);
- for(int i=t0+1; i<=t1; i++)
- merge(t0,i);
- }
- int test0,test1;
- cin>>test0>>test1;
- if(father[test0]==father[test1])
- cout<<"Yes"<<endl;
- else
- cout<<"No"<<endl;
- }