[Virtual Judge]URAL1303:Minimal Coverage

点击打开题目链接

1303. Minimal Coverage Time limit: 1.0 second Memory limit: 64 MB Given set of line segments [L i, R i] with integer coordinates of their end points. Your task is to find the minimal subset of the given set which covers segment [0, M] completely (M is a positive integer). Input First line of the input contains an integer M (1 ≤ M ≤ 5000). Subsequent lines of input contain pairs of integers L i and R i (−50000 ≤ L i < R i ≤ 50000). Each pair of coordinates is placed on separate line. Numbers in the pair are separated with space. Last line of input data contains a pair of zeroes. The set contains at least one and at most 99999 segments. Output Your program should print in the first line of output the power of minimal subset of segments which covers segment [0, M]. The list of segments of covering subset must follow. Format of the list must be the same as described in input with exception that ending pair of zeroes should not be printed. Segments should be printed in increasing order of their left end point coordinate. If there is no covering subset then print “No solution” to output. Samples input output 1 -1 0 -5 -3 2 5 0 0 No solution 1 -1 0 0 1 0 0 1 0 1

=====================================题目大意=====================================

使用最少的提供的区间覆盖指定区间。

=====================================算法分析=====================================

贪心算法（此题即刘汝佳白书P154：区间覆盖问题，但是下面代码中我使用的算法与白书不同）。

=======================================代码=======================================

#include<cstdio>
#include<algorithm>

using namespace std;

int M,InterSum,AnsID[100005],CntAns;
                                   
struct Interval { int L,R; } Inter[100005];

bool cmp(Interval& I1,Interval& I2) 
{
	return I1.L<I2.L;
}

bool Greedy()
{
	CntAns=0;	
	sort(Inter,Inter+InterSum,cmp);
	int i=0,CurCoverEnd=0;
	while(i<InterSum&&CurCoverEnd<M)
	{
		int Best;
		for(Best=-1;i<InterSum;++i)
		{
			if(Inter[i].R<CurCoverEnd) { continue; }
			if(Inter[i].L>CurCoverEnd) { break; }
			if(Best==-1||Inter[Best].R<Inter[i].R) { Best=i; }
		}
		if(Best==-1) { break; }
		CurCoverEnd=Inter[Best].R;  
		AnsID[CntAns++]=Best;  
	}
	return M<=CurCoverEnd;
}

int main()
{
	while(scanf("%d",&M)==1)
	{
		InterSum=0;
		while(1)
		{
			scanf("%d%d",&Inter[InterSum].L,&Inter[InterSum].R);
			if(Inter[InterSum].L==0&&Inter[InterSum].R==0) { break; }
			++InterSum;
		}
		if(Greedy()==true)
		{
			printf("%d\n",CntAns);
			for(int i=0;i<CntAns;++i)
			{
				printf("%d %d\n",Inter[AnsID[i]].L,Inter[AnsID[i]].R);
			}
		}
		else
		{         
			printf("No solution\n");
		}
	}
	return 0;
}