UVA - 10020 Minimal coverage（区间覆盖问题）

最新推荐文章于 2019-02-23 15:50:45 发布

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#uva #10020 #Minimal coverage

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本文介绍了一个经典的区间覆盖问题，目标是最小化所需区间的数量来完全覆盖指定区间[0, m]。通过排序和贪心算法策略，文章详细解析了如何选择最佳区间组合，并提供了完整的代码实现。

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

Sample Output

题目大意：
输入一个数字m，再输入一组区间，问：能否用最少的区间覆盖[0,m]的区间。

解析：
典型的贪心题，思路按照刘汝佳白书P154页上的解析。
在这里再解释一下
(1)先把区间按照起点从小到大排序，先用一个变量left指向0
|----------------------|
0(left) m

(2)然后遍历所有的线段，取出起点小于left，且终点尽量最大的区间，记录下该区间
(3)然后让left指向该区间的终点，重复(2)的操作，直到left >= m，或者没有线段可以更新为止，结束操作。
|----------|-----------|
0 left m

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cstdlib>
using namespace std;
typedef long long ll;
const int INF = 0x3f3f3f3f;
const int N = 100005;
struct Segment {
	int l ,r;
}s[N], path[N];
int n, m, cnt;
bool cmp(Segment a, Segment b) {
	return a.l < b.l;
}
void input() {
	scanf("%d", &m);
	for(n = 0;; n++) {
		scanf("%d%d", &s[n].l, &s[n].r);
		if(!s[n].l && !s[n].r) break;
	}
}
bool solve() {
	int left, Max, pos;

	cnt = left = Max = pos = 0;
	for(int i = 0; i < n;) {
		int j = i;
		while(j < n && s[j].l <= left) {
			if(s[j].r > Max) {
				Max = s[j].r;
				pos = j;
			}
			j++;
		}
		if(i == j) break;

		path[cnt].l = s[pos].l;
		path[cnt++].r = s[pos].r;

		left = Max, i = j;

		if(left >= m) return true;
	}
	return false;
}
int main() {
	int T;
	scanf("%d", &T);
	while(T--) {
		input();
		sort(s,s + n, cmp);
		if(s[0].l > 0)
			printf("0\n");
		else if(solve()) {  
			printf("%d\n",cnt);  
			for(int i = 0; i < cnt; i++)  
				printf("%d %d\n",path[i].l,path[i].r);  
		}else
			printf("0\n");  
		
		if(T) puts("");
	}
	return 0;
}