【区间覆盖问题】uva 10020 - Minimal coverage

可以说是区间覆盖问题的例题...

Note:

区间包含+排序扫描；

　　要求覆盖区间[s, t];

　　1、把各区间按照Left从小到大排序，如果区间1的起点大于s，则无解（因为其他区间的左起点更大）；否则选择起点在s的最长区间;

　　2、选择区间[li, ri]后，新的起点应更新为ri，并且忽略所有区间在ri之前的部分；

 

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



代码如下

 1 #include<iostream>

 2 #include<cstdio>

 3 #include<cstdlib>

 4 #include<cstring>

 5 #include<algorithm>

 6 using namespace std;

 7 const int maxn = 100010;

 8 struct Seg

 9 {

10     int L, R;

11     bool operator < (const Seg &a) const

12     {

13         if(L != a.L) return L < a.L;

14         else return R > a.R;

15     }

16 } seg[maxn], ans[maxn];

17 void solve(int n, int m)

18 {

19     int cnt = 0, l = 0, r = 0;//起点l，起点l最长区间的右端点r

20     if(seg[0].L > l)

21     {

22         printf("0\n");

23         return ;

24     }//无解

25     bool ok = false;

26     while(1)

27     {

28         if(l >= m)    break;

29         int i, pos = 0;

30         ok = false;//判断变量ok，重要！

31         for(i = 0; i < n; i++)

32         {

33             if(seg[i].L <= l)

34             {

35                 if(seg[i].R > r)

36                 {

37                     pos = i;

38                     r = seg[pos].R;

39                     ok = true;

40                 }//寻找起点在l的最长区间

41             }

42         }

43         if(ok)

44         {

45             l = r;

46             ans[cnt].L = seg[pos].L;

47             ans[cnt++].R = seg[pos].R;

48         }//更新起点l

49         else break;

50     }

51     if(ok)

52     {

53         printf("%d\n", cnt);

54         for(int i = 0; i < cnt; i++)

55             printf("%d %d\n", ans[i].L, ans[i].R);

56     }

57     else printf("0\n");

58 }

59 int main()

60 {

61     int T;

62     scanf("%d", &T);

63     for(int kase = 0; kase < T; kase++)

64     {

65         if(kase) printf("\n");

66         memset(seg, 0, sizeof(seg));

67         memset(ans, 0, sizeof(ans));

68         int m; scanf("%d", &m);

69         int i = 0, L, R;

70         while(scanf("%d%d", &L, &R))

71         {

72             if(!L && !R)  break;

73             seg[i].L = L; seg[i].R = R;

74             i++;

75         }

76         sort(seg, seg+i);//排序

77         solve(i, m);

78     }

79     return 0;

80 }

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