AtCoder Beginner Contest 103 D(贪心)

AtCoder Beginner Contest 103 D

题目大意:n个点,除第n个点外第i与第i+1个点有一条边,给定m个a[i],b[i],求最少去掉几条边能使所有a[i],b[i]不相连.

按右端点从小到大排序,如果当前选的去掉的边在区间内,那么符合条件,否则ans++,并贪心地把去掉的边指向右端点,因为前面的区间都满足条件了,所以要去掉的边要尽量向右移使其满足更多的区间。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <cstring>
 5 #include <cmath>
 6 #include <queue>
 7 #include <map>
 8 #include <stack>
 9 #define ll long long
10 #define out(a) printf("%d",a)
11 #define writeln printf("\n")
12 using namespace std;
13 const int N=1e5+50;
14 int n,m;
15 int now,ans;
16 struct node
17 {
18     int x,y;
19 }a[N];
20 int read()
21 {
22     int s=0,t=1; char c;
23     while (c<'0'||c>'9'){if (c=='-') t=-1; c=getchar();}
24     while (c>='0'&&c<='9'){s=s*10+c-'0'; c=getchar();}
25     return s*t;
26 }
27 ll readl()
28 {
29     ll s=0,t=1; char c;
30     while (c<'0'||c>'9'){if (c=='-') t=-1; c=getchar();}
31     while (c>='0'&&c<='9'){s=s*10+c-'0'; c=getchar();}
32     return s*t;
33 }
34 bool cmp(node a,node b)
35 {
36     return a.y==b.y?a.x<b.x:a.y<b.y;
37 }
38 int main()
39 {
40     n=read(),m=read(); now=0;
41     for (int i=1;i<=m;i++)
42       a[i].x=read(),a[i].y=read();
43     sort(a+1,a+m+1,cmp);
44     for (int i=1;i<=m;i++)
45       if (now>a[i].x&&now<=a[i].y) continue;
46       else ans++,now=a[i].y;
47     out(ans);
48     return 0;
49 }

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