UVA 10020 Minimal coverage(贪心 + 区间覆盖问题）

 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

2

1
-1 0
-5 -3
2 5
0 0

1
-1 0
0 1
0 0

Sample Output

0

1
0 1

题意：给定一个M，和一些区间[Li，Ri]。。要选出几个区间能完全覆盖住[0，M]区间。要求数量最少。。如果不能覆盖输出0.

思路：贪心的思想。。把区间按Ri从大到小排序。 然后遇到一个满足的[Li，Ri]，就更新缩小区间。。直到完全覆盖。

注意[Li，Ri]只有满足Li小于等于且Ri大于当前覆盖区间左端这个条件。才能选中。

代码：

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

int t;
int start, end, qn, outn;
struct M {
    int start;
    int end;
} q[100005], out[100005];

int cmp (M a, M b) {//按最大能覆盖到排序
    return a.end > b.end;
}
int main() {
    scanf("%d", &t);
    while (t --) {
	qn = 0; outn = 0; start = 0;
	scanf("%d", &end);
	while (~scanf("%d%d", &q[qn].start, &q[qn].end) && q[qn].start + q[qn].end) {
	    qn ++;
	}
	sort(q, q + qn, cmp);
	while (start < end) {
	    int i;
	    for (i = 0; i < qn; i ++) {
		if (q[i].start <= start && q[i].end > start) {
		    start = q[i].end;//更新区间
		    out[outn ++] = q[i];
		    break;
		}
	    }
	    if (i == qn) break;//如果没有一个满足条件的区间，直接结束。
	}
	if (start < end) printf("0\n");
	else {
	    printf("%d\n", outn);
	    for (int i = 0; i < outn; i ++)
		printf("%d %d\n", out[i].start, out[i].end);
	}
	if (t) printf("\n");
    }
    return 0;
}