uva 10020 Minimal coverage(贪心-最小覆盖问题）

题目连接：10020 - Minimal coverage

题目大意：给出一个范围M,然后给出若干的区间，以0 0 终止， 要求用最少的区间将0 ~M 覆盖，输出最少个数以及方案。

解题思路：典型的区间覆盖问题，算法竞赛入门经典P154上有讲。

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
const int N = 1000005;

struct Seg {
    int x;
    int y;
}seg[N];
int n, cnt, order, vis[N];

bool cmp (const Seg& a, const Seg& b) {
    if (a.x != b.x)
	return a.x < b.x;
    else
	return a.y < b.y;
}

void input() {
    int p, q;
    n = cnt = 0;
    scanf("%d", &order);
    while (scanf("%d%d", &p, &q), p + q) {
	seg[n].x = p;
	seg[n].y = q;
	n++;
    }
}

int main() {
    int cas;
    scanf("%d", &cas);
    while (cas--) {
	memset(seg, 0, sizeof(seg));
	memset(vis, 0, sizeof(vis));
	input();

	sort(seg, seg + n, cmp);

	int begin = 0, Max, id;
	for (int i = 0; i < n; i++) {
	    if (seg[i].x > order)   break;

	    Max = 0, id;
	    int k;
	    for (k = 0; k < n - i; k++) {
		if (seg[i + k].x > begin) {
		  k--;
		  break;
		}
		if (seg[i + k].y > Max) {
		    Max = seg[i + k].y;
		    id = i + k;
		}
	    }
	    if (k > 0)
		i += k;
	    if (Max < begin)	break;

	    begin = Max; 
	    vis[cnt++] = id;

	    if (begin >= order)		break;
	}
	if (Max >= order) {
	    printf("%d\n", cnt);
	    for (int i = 0; i < cnt; i++)
		printf("%d %d\n", seg[vis[i]].x, seg[vis[i]].y);
	}
	else
	    printf("0\n");
	if (cas)	    printf("\n");
    }
    return 0;
}