Given set of line segments [L
i, R
i] with integer coordinates of their end points. Your task is to find the minimal subset of the given set which covers segment [0, M] completely (M is a positive integer).
Input
First line of the input contains an integer M (1 ≤ M ≤ 5000). Subsequent lines of input contain pairs of integers L
i
and R
i
(−50000 ≤ L
i
< R
i
≤ 50000). Each pair of coordinates is placed on separate line. Numbers in the pair are separated with space. Last line of input data contains a pair of zeroes. The set contains at least one and at most 99999 segments.
Output
Your program should print in the first line of output the power of minimal subset of segments which covers segment [0, M]. The list of segments of covering subset must follow. Format of the list must be the same as described in input with exception that ending pair of zeroes should not be printed. Segments should be printed in increasing order of their left end point coordinate.
If there is no covering subset then print “No solution” to output.
Samples
| input | output |
|---|---|
1 -1 0 -5 -3 2 5 0 0 |
No solution |
1 -1 0 0 1 0 0 |
1 0 1 |
**************************************************************************************
Greedy (reference to other people's code)
**************************************************************************************
1 #include<iostream> 2 #include<string> 3 #include<cstring> 4 #include<cstdio> 5 #include<cmath> 6 #include<algorithm> 7 using namespace std; 8 struct seg 9 { 10 int s; 11 int e; 12 seg &operator =(seg &b) 13 { 14 s=b.s; 15 e=b.e; 16 return *this; 17 } 18 19 }p[100010],ans[100010]; 20 int M,i,j,k,t; 21 int m,n; 22 int min; 23 int gs; 24 void init() 25 { 26 cin>>M; 27 i=0; 28 while(cin>>m>>n) 29 { 30 if(m==0&&n==0) 31 break; 32 p[i].s=m; 33 p[i++].e=n; 34 } 35 k=i; 36 } 37 bool cmp(seg a,seg b) 38 { 39 if(a.s<b.s)return true; 40 if(a.s>b.s)return false; 41 if(a.s==b.s) 42 { 43 if(a.e>b.e)return true; 44 else return false; 45 } 46 is } 47 int main () 48 { 49 the init (); 50 Sort (P, P + K, CMP); // first sort, followed by the end 51 is T = 0 ; 52 is for (I = . 1 ; I <K ; I ++ ) 53 is IF (P [I] .s> P [T] .s && P [I] .E> P [T] .E) P [T ++] = P [I]; // delete covered line 54 is K = T; 55 P [K + . 1 ] .s = M + . 1 ; // for comparison is not out of range 56 is P [K + . 1 ] = M + .E . 1 ; 57 is J =0 ; 58 int H = 0 ; 59 for (I = 0 ; I <= K; I ++ ) 60 IF (j <P [I + . 1 ] .s && P [I] .s <= j) // satisfies the condition that is guaranteed j This is not the next one line to avoid repetition 61 is { 62 is J = P [I] .E; 63 is ANS [++ H] = P [I]; // record; 64 IF (P [I] .E> = M) // satisfy the condition output 65 { 66 COUT H << << endl; 67 for ( int G = . 1 ; G <= H; G ++) 68 cout<<ans[g].s<<' '<<ans[g].e<<endl; 69 return 0; 70 } 71 } 72 cout<<"No solution"<<endl; 73 return 0; 74 75 76 77 }
Reproduced in: https: //www.cnblogs.com/sdau--codeants/p/3234803.html