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• # Timus 1303 Minimal Coverage DP或贪心

## 1303. Minimal Coverage

Given set of line segments [Li, Ri] with integer coordinates of their end points. Your task is to find the minimal subset of the given set which covers segment [0, M] completely (M is a positive integer).

### Input

First line of the input contains an integer M (1 ≤ M ≤ 5000). Subsequent lines of input contain pairs of integers Li and Ri (−50000 ≤ Li < Ri ≤ 50000). Each pair of coordinates is placed on separate line. Numbers in the pair are separated with space. Last line of input data contains a pair of zeroes. The set contains at least one and at most 99999 segments.

### Output

Your program should print in the first line of output the power of minimal subset of segments which covers segment [0, M]. The list of segments of covering subset must follow. Format of the list must be the same as described in input with exception that ending pair of zeroes should not be printed. Segments should be printed in increasing order of their left end point coordinate.
If there is no covering subset then print “No solution” to output.

### Samples

inputoutput
```1
-1 0
-5 -3
2 5
0 0
```
```No solution
```
```1
-1 0
0 1
0 0
```
```1
0 1
```

输入m
给出一些线段，
问最少选择多少条可以覆盖区间[0,m],
输出条数
然后按左端点小到大输出这些线段。

维护当前的最右端点(刚开始是0)
然后找左端点小于这个点的且右端点最大的线段
更新最右端点

主要看注释

```  1 #include<cstdio>
2 #include<cstring>
3 #include<algorithm>
4
5 using namespace std;
6
7 const int maxn=100000+4;
8
9 struct Edge
10 {
11     int l,r;
12 }edge[maxn];
13
14 bool cmp(Edge a,Edge b)
15 {
16     if(a.l==b.l)
17         return a.r<b.r;
18     return a.l<b.l;
19 }
20
21 int ans[maxn];
22
23 int main()
24 {
25     int m;
26
27     while(scanf("%d",&m)!=EOF)
28     {
29
30         int tot=1;
31
32         scanf("%d%d",&edge[tot].l,&edge[tot].r);
33
34         if(edge[tot].l==0&&edge[tot].r==0)
35             break;
36
37         while(edge[tot].l||edge[tot].r)
38         {
39             tot++;
40             scanf("%d%d",&edge[tot].l,&edge[tot].r);
41         }
42
43         sort(edge+1,edge+tot,cmp);
44
45         int num=0;
46         int cur=0;
47         int cnt=0;
48         edge[0].l=edge[0].r=-55555;
49
50         bool flag=true;
51
52         while(true)
53         {
54             if(cur>=m)
55             {
56                 break;
57             }
58
59
60             //这里犯了一个极大的错误
61             //我刚开始的代码是这样的：
62             //while(cnt+1<tot&&edge[cnt+1].l<=cur&&edge[cnt+1].r>edge[cnt].r)
63             //    cnt++;
64             //我们是要找出满足左端点在cur左边的所有线段中，
65             //的右端点最远的点
66             //所以要用变量right表示当前最远的，
67             //然后跟right比较
68             //而不是跟前面的线段比较。
69
70
71             int right=cur;
72             for(int i=cnt;i+1<tot;i++)
73             {
74                 if(edge[i+1].l<=cur&&edge[i+1].r>right)
75                 {
76                     cnt=i+1;
77                     right=edge[cnt].r;
78                 }
79             }
80
81             if(edge[cnt].r<=cur)
82             {
83                 flag=false;
84                 break;
85             }
86
87             else
88             {
89                 ans[num++]=cnt;
90                 cur=edge[cnt].r;
91             }
92         }
93
94         if(!flag)
95         {
96             printf("No solution
");
97
98             continue;
99         }
100
101         printf("%d
",num);
102
103         for(int i=0;i<num;i++)
104         {
105             printf("%d %d
",edge[ans[i]].l,edge[ans[i]].r);
106         }
107
108     }
109
110     return 0;
111 }```
View Code

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• 原文地址：https://www.cnblogs.com/-maybe/p/4483609.html