TOJ 5945: interval coverage problems

5945: interval coverage problems

With N closed interval [Ai, Bi] number line, the selection interval covered as little as possible to specify a segment [S, T].

 

 

Entry

 

 

The first line contains three integers N (1 <= N <= 10 ^ 5), S and T (-10 ^ 6 <= S <T <= 10 ^ 6) 

Next N lines of two integers Ai, Bi (-10 ^ 6 <= Ai <Bi <10 ^ 6)

 

 

Export

 

 

Output contains an integer representing the number of a minimum required interval. If no solution, output No Solution

 

 

Sample input

 

8 1 10
-3 -1
13 16
0 3
2 6
3 5
6 10
5 6
4 10

Sample Output

 3

 Data input in accordance with the left point of ordering initialize a reference point for the need to cover the left point range than the reference point to find the left and right end points a little larger than the reference point to find the right point of maximum range of updated reference point in these intervals inside

#include <bits/stdc++.h>
using namespace std;
struct jilu
{
    int l,r;
}a[100010];
bool cmp(jilu a,jilu b)
{
    return a.l<b.l;
}
int main()
{
    int n,s,t,sum=0,f=1;
    cin>>n>>s>>t;
    for(int i=0;i<n;i++)
    {
        cin>>a[i].l>>a[i].r;
    }
    sort(a,a+n-, CMP);
     int jidian = S, = maxrl - 0x3f3f3f3f3f3f , F1 = 0 ; // this maxrl very beginning of the pit is provided a 0WA twice .. 
    for ( int I = 0 ; I <n-; I ++ ) 
    { 
        IF (jidian> = T)
             BREAK ;
         IF (A [I] .L <= jidian && A [I] .r> jidian) 
        { 
            maxrl = max (maxrl, A [I] .r); 
            F1 = . 1 ; 
        } 
        IF ( A [I] .L> jidian && F1 == 0 ) 
        { 
            BREAK ; 
            F= 0 ; 
        } 
        the else  IF (A [I] .L> == jidian && F1 . 1 ) 
        { 
            SUM ++ ; 
            jidian = maxrl; 
            I - ; 
            maxrl = - 0x3f3f3f3f3f3f ; 
            F1 = 0 ; 
        } 
        the else  IF (I == N- 1 ) // need to update the exit 
        { 
            SUM ++ ; 
            jidian = maxrl; 
        } 
    } 
        IF(f!=0&&jidian>=t)
            cout<<sum<<endl;
        else
            cout<<"No Solution"<<endl;
    }