UVa 10020 Minimal coverage（欲張り&区間カバー）

10020-Minimal coverage
Time limit：3.000 seconds 
http://uva.onlinejudge.org/index.php?option=com_オンラインjudge&Itemid=8&category=113&page=show_problem&problem=961
The Problem
Given several segments of line(int the X axis)with coordinans[Li，Ri].You are to chose the minimal amount of them，such the y would copletely 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 shout contains an integer M（1==M<=5000）、フォロワーby pairs“Li Ri”（124 Li|、124; Ri（=50000、i==100000）、each on a separate line.Eacterminates”
Each test case will be separated by a single line.
The Output
For each test case，in the first line of output your progrm shuld prininininininiml number of line segments ininininininininttttttttcan cover segment[0，M].In the follwing inininininininininininininininininininininininininininininininininininininininininininininininininininininininininininininininininininttttttttttttttdedededededededededededededests、thes、thes、thes、the the.If[0,M]can not be covered by given line segments、your program shoud print“0”。
Print a blank line between the outputts 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

『入門経典』P 154の経典欲張り。
問題のデータに欠陥があります。 if（nowL＜M）cnt=0；この言葉はACにもなります。
完全コード:

/*0.065s*/

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 100005;

struct seg
{
	int l, r;
	bool operator < (const seg a) const
	{
		return l < a.l;
	}
} p[N], ans[N];

int main()
{
	int T, n, M, i, nowL, maxR, cnt, templ, tempr;
	bool found;
	scanf("%d", &T);
	while (T--)
	{
		scanf("%d", &M);
		n = 0;
		while (scanf("%d%d", &templ, &tempr), templ || tempr)
            if (templ < M && tempr) p[n++] = (seg){templ, tempr};///              
        sort(p, p + n);
		for (i = nowL = cnt = 0; nowL < M; ++cnt)
		{
			found = false;
			maxR = 0;
			for (; i < n && p[i].l <= nowL; ++i)
				if (p[i].r > maxR)
				{
					found = true;
					maxR = p[i].r;///                       
					ans[cnt] = p[i];
				}
			if (!found) break;
			nowL = maxR;///     
		}
		if (nowL < M) cnt = 0;///    ~
		printf("%d
", cnt);
		for (i = 0; i < cnt; i++)
			printf("%d %d
", ans[i].l, ans[i].r);
		if (T) putchar(10);
	}
	return 0;
}