uva 10020

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

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Alex Gevak
September 10, 2000 (Revised 2-10-00, Antonio Sanchez)

/* ********************************************************贪心：将线段按照 left 从小到大排序，r 从 0 增长到 M，每次取出left < r && right 最大的线段加入结果，然后 r 更新为 right题目中有一个要求："the coordinates of segments, sorted bytheir left end (Li), should be printed in the same formatas in the input"，如果输入为小数，必须输出和输入一样的数位，解决方法是在 Seg 结构体里增加保存输入的 string还有一点要注意的地方是，不论大小，如果 |a - b| < 1e-6，则可以认为 a == b******************************************************** */#include <cstdio>#include <cstdlib>#include <cassert>#include <cmath>#include <algorithm>#include <vector>#include <string>using namespace std;#define EPS 1e-10struct Seg{ string strL; string strR; double left; double right; Seg(const string &l = string("0"), const string &r = string("0")) : strL(l), strR(r), left(atof(l.c_str())), right(atof(r.c_str())) {}};// Sort by left pointstruct SegCmp{ bool operator()(const Seg &a, const Seg &b) { return a.left < b.left; }};int m; // Mvector <Seg> seg; // Segmentsint main(){ int cs; scanf("%d", &cs); while(cs-- > 0) { // Input scanf("%d", &m); seg.clear(); while(true) { char l[1024]; char r[1024]; scanf("%s%s", l, r); if(fabs(atof(l)) < EPS && fabs(atof(r)) < EPS) break; seg.push_back(Seg(l, r)); } sort(seg.begin(), seg.end(), SegCmp()); vector <bool> used(seg.size(), false); bool hasSolution = true; double r = 0; int i = 0; int n = 0; // Greedy while(r <= m - EPS) { // 找出 left < r && right 最大的线段，t 标记其下标 int t = i; while(i < seg.size() && (seg[i].left <= r || fabs(seg[i].left - r) < EPS)) { if(seg[i].right > seg[t].right) t = i; ++i; } if(t >= seg.size() || seg[t].left > r && fabs(seg[t].left -r) >= EPS) { hasSolution = false; break; } // 增加该线段到结果 used[t] = true; r = seg[t].right; ++n; } // Output if(! hasSolution) { printf("0/n"); } else { printf("%d/n", n); for(int i=0; i<used.size(); ++i) { if(used[i]) printf("%s %s/n", seg[i].strL.c_str(), seg[i].strR.c_str()); } } if(cs != 0) printf("/n"); } return 0;}