Ultraviolet A 10020-minimal coverage

Link:

Http://uva.onlinejudge.org/index.php? Option = com_onlinejudge & Itemid = 8 & category = 113 & page = show_problem & problem = 961

Type: greedy

Original question:

The Problem

Given several segments of line (int the X axis) with coordinates [Li, Ri]. you are to choose the minimal amount of them, such they wocould 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 shoshould contains an integer m (1 <= m <= 5000), followed by pairs "Li Ri" (| Li |, | Ri | <= 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 shocould 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 (LI), shocould be printed in the same format as in the input. pair "0 0" shocould not be printed. if [0, m] can not be covered by given line segments, your programm shold 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

Analysis and Summary:

Greedy for the minimum coverage of classic intervals. Lrj algorithm entry classic p154.

For the range [left, right], sort by left from small to large, and set the start point to S, then select the biggest right in left <= s as the new start point S, so the loop goes on, until the starting point S is greater than the target length.

/*  *   UVa: 10020 - Minimal coverage *   Result: Accept *   Time: 0.080s *   Author: D_Double                                                                  */#include<iostream>#include<cstdio>#include<algorithm>#include<vector>#define MAXN 100005using namespace std;int M;struct Segment{int left;int right;friend bool operator < (const Segment&a, const Segment&b){if(a.left!=b.left) return a.left<b.left;return a.right>b.left;}}arr[MAXN];int nIndex;int ans[MAXN], N;int main(){freopen("input.txt","r",stdin);int T, i;scanf("%d",&T);while(T--){nIndex=0;scanf("%d",&M);while(~scanf("%d %d",&arr[nIndex].left, &arr[nIndex].right)){if(!arr[nIndex].left && !arr[nIndex].right) break;if(arr[nIndex].right>0) ++nIndex;}sort(arr, arr+nIndex);int cur = 0;N=-1;bool flag=false;if(arr[0].left > 0){printf("0\n");if(T) printf("\n");continue;}for(i=0; i<nIndex; ++i){if(arr[i].left <= cur){if(N==-1) ans[++N]=i;else if(arr[i].right > arr[ans[N]].right)ans[N]=i;}else{cur=arr[ans[N]].right;ans[++N]=i;}if(arr[ans[N]].right>=M){flag=true; break;}} if(flag){printf("%d\n", N+1);for(i=0; i<=N; ++i)printf("%d %d\n", arr[ans[i]].left, arr[ans[i]].right);}elseprintf("0\n");if(T) printf("\n");}return 0;}

-- The meaning of life is to give it meaning.
 

  Original
  Http://blog.csdn.net/shuangde800
  , By d_double (reprinted, please mark)