UVa 10020 Minimal coverage (贪心&区间覆盖)

UVA 10020 Minimal Coverage解析

最新推荐文章于 2019-01-27 03:28:51 发布

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本文解析了UVA 10020 Minimal Coverage问题，这是一个经典的贪心算法题目。文章提供了完整的代码实现，并针对输入数据的特点进行了讨论。

10020 - Minimal coverage

Time limit: 3.000 seconds

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

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

Sample Output

《入门经典》P154上的经典贪心。

注意，题目的数据有缺陷，即使你不加 if (nowL < M) cnt = 0; 这一句也能AC。

完整代码：

/*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};///根本不能用的区间就不存进去了
        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;///在区间左端点符合要求的情况下所能覆盖到的最右端
					ans[cnt] = p[i];
				}
			if (!found) break;
			nowL = maxR;///更新左端点
		}
		if (nowL < M) cnt = 0;///任务失败~
		printf("%d\n", cnt);
		for (i = 0; i < cnt; i++)
			printf("%d %d\n", ans[i].l, ans[i].r);
		if (T) putchar(10);
	}
	return 0;
}