uva 10020 - Minimal coverage

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

21-1 0-5 -32 50 01-1 00 10 0

Sample Output

010 1贪心求最少线段区间覆盖

#include<stdio.h>#include<algorithm>#include<iostream>#include<cmath>using namespace std;struct node{ int x,y;}a[100001],b[100001];int cmp(node a,node b){ return a.x<b.x;}int main(){ int max,pos,i,begin,end,m,n,sum,f,posmax,t; scanf("%d",&t); while (t--)  { scanf("%d",&m); n=0; while (scanf("%d%d",&begin,&end)) {  if (begin==0&&end==0) break;  if (end>=0)  {   a[n].x=begin;   a[n].y=end;   ++n;  } } sort(a,a+n,cmp); if (a[0].x>0) printf("0\n");//开始漏了没判断第一条线段就不能覆盖到0，wrongY-Y； else { pos=0; max=0; sum=1; while (pos+1<n&&a[pos+1].x<=0)//pos+1<N防止数组越界，还真有所有数据都从小于0开始的，re Y-Y； {  ++pos;  if (a[pos].y>a[max].y) max=pos; } pos=max; b[1].x=a[pos].x; b[1].y=a[pos].y; while (pos<n && b[sum].y<m)  {  f=1; posmax=max;  while (pos+1<n && a[pos+1].x<=a[max].y)  {  f=0;   ++pos;   if (a[pos].y>a[posmax].y) posmax=pos;  }  if (f) break;  max=posmax;  ++sum;  b[sum].x=a[max].x;  b[sum].y=a[max].y; } if (b[sum].y>=m)  {  printf("%d\n",sum);  for (i=1;i<=sum;i++)  printf("%d %d\n",b[i].x,b[i].y); }else printf("0\n"); } if (t) printf("\n"); } return 0;}