Description

Minimal coverage

The Problem

Given several segments of line (int the X axis) with coordinates [L_i,R_i]. You are to choose the minimal amount of them, such they would completely cover the segment [0,M].

The Input

The first line is the number of test cases, followed by a blank line.

Each test case in the input should contains an integer M(1<=M<=5000), followed by pairs "L_i R_i"(|L_i|, |R_i|<=50000, i<=100000), each on a separate line. Each test case of input is terminated by pair "0 0".

Each test case will be separated by a single line.

The Output

For each test case, in the first line of output your programm should print the minimal number of line segments which can cover segment [0,M]. In the following lines, the coordinates of segments, sorted by their left end (L_i), should be printed in
the same format as in the input. Pair "0 0" should not be printed. If [0,M] can not be covered by given line segments, your programm should print "0"(without quotes).

Print a blank line between the outputs for two consecutive test cases.

Sample Input

2

1
-1 0
-5 -3
2 5
0 0

1
-1 0
0 1
0 0

Sample Output

0

1
0 1

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>

struct node
{
    int l;
    int r;
} q[100100];

int cmp(const void *a,const void *b)
{
    struct node *aa,*bb;
    aa = (struct node*)a;
    bb = (struct node*)b;
    return aa->l - bb->l;
}

int main()
{
    int T;
    int n;
    scanf("%d",&T);
    while(T--)
    {
        int m;
        scanf("%d",&m);
        n = 0;
        while(scanf("%d%d",&q[n].l,&q[n].r)!=EOF)
        {
            if(q[n].l == 0 && q[n].r == 0)
            {
                break;
            }
            n++;
        }
        qsort(q,n,sizeof(q[0]),cmp);
        int pmax = 0;
        int pi = 0;
        int ff = 0;
        for(int i=0; i<n; i++)
        {
            if(q[i].l<=0 && q[i].r>pmax)
            {
                ff = 1;
                pmax = q[i].r;
                pi = i;
            }
            else if(q[i].l>0)
            {
                break;
            }
        }
        if(ff == 0)
        {
            printf("0\n");
            continue;
        }
        int left = 0;
        int right = pmax;
        int maxx = 0;
        int a[10010];
        int k = 0;
        int p = 0;
        int h = 0;
        int flag = 0;
        a[p] = pi;
        p++;
        for(int i=0; i<n; i++)
        {

            if(q[i].l<=right && q[i].r>maxx)
            {
                if(flag == 0)
                {
                    right = q[i].r;
                    flag = 1;

                }

                maxx = q[i].r;
                h = i;
                if(maxx >=m)
                {
                    if(h!=a[p-1])
                    {
                        a[p] = h;
                        p++;
                    }
                    break;
                }
            }
            else if(q[i].l>right && right!=0)
            {
                right = q[h].r;
                if(h!=a[p-1])
                {

                    a[p] = h;
                    p++;
                }

                if(q[i].r>maxx)
                {
                    maxx = q[i].r;
                    h = i;
                    if(i == n-1)
                    {
                        a[p] = h;
                        p++;
                    }
                }
            }
        }

        if(p == 0)
        {
            printf("0\n");
        }
        else
        {
            printf("%d\n",p);
            for(int i=0; i<p; i++)
            {
                printf("%d %d\n",q[a[i]].l,q[a[i]].r);
            }
        }
        if(T)
        {
            printf("\n");
        }
    }
    return 0;
}