牛客练习赛34-C little w and Segment Coverage

版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/C_13579/article/details/85109856

地址：https://ac.nowcoder.com/acm/contest/297/C

思路：先将所有点的覆盖次数求出来，再求出点覆盖一次的前缀和以及没有覆盖的点个数Sum，之后遍历线段找出最小的覆盖一次的值即可。求点的覆盖次数时code1为先将所有的点按照左端点由小到大排序，在遍历一遍求得，实在过于麻烦。code2则用扫描线求解

Code 1:

#include<iostream>
#include<algorithm>
using namespace std;

const int MAX_N=1e5+5;
const int MAX_M=1e5+5;
struct node{
	int l,r;
	int id;
	bool operator<(const node &p)const{
		return (l==p.l)?(r<p.r):(l<p.l);
	}
};
int n,m;
node a[MAX_M];
int d[MAX_N],pre[MAX_N];

int main()
{
	ios::sync_with_stdio(false);
    cin>>n>>m;
    for(int i=1;i<=m;a[i].id=i,++i)
		cin>>a[i].l>>a[i].r;
	sort(a+1,a+m+1);
	int l1=1,l2=1;
	for(int i=1;i<=m;++i)
	{
		l2=max(l2,a[i].l);
		while(l2<l1){
			if(d[l2])	d[l2]=2;
			++l2;
		}
		l1=max(l1,a[i].l);
		while(l1<=a[i].r){
			d[l1++]=1;
		}
	}
	int Sum=0;
	for(int i=1;i<=n;++i)
	{
		pre[i]=pre[i-1];
		if(d[i]==0)	++Sum;
		if(d[i]==1)	pre[i]=pre[i-1]+1;
	}
	int x,s1=0,s2=n+1;
	for(int i=1;i<=m;++i)
	{
		x=pre[a[i].r]-pre[a[i].l-1];
		if(s2>x){
			s2=x;	s1=a[i].id;
		}else	if(s2==x)	s1=max(s1,a[i].id);
	}
	s2+=Sum;
	cout<<s1<<" "<<s2<<endl;
	
    return 0;
}

Code 2:

#include<iostream>
using namespace std;
typedef pair<int,int> pr;

const int MAX_N=1e5+5;
const int MAX_M=1e5+5;
int n,m;
pr a[MAX_M];
int d[MAX_N];

int main()
{
	ios::sync_with_stdio(false);
	cin>>n>>m;
	for(int i=1;i<=m;++i)
	{
		cin>>a[i].first>>a[i].second;
		++d[a[i].first];
		--d[a[i].second+1];
	}
	for(int i=1;i<=n;++i)
		d[i]+=d[i-1];
	int Sum=0;
	for(int i=1;i<=n;++i)
	{
		if(!d[i])	++Sum;
		if(d[i]!=1)	d[i]=0;
		d[i]+=d[i-1];
	}
	pr ans=pr{0,n+1};
	int x;
	for(int i=1;i<=m;++i)
	{
		x=d[a[i].second]-d[a[i].first-1];
		if(ans.second>=x)	ans=pr{i,x};
	}
	ans.second+=Sum;
	cout<<ans.first<<" "<<ans.second<<endl;
	
	return 0;
}

/*
#include<iostream>
#include<algorithm>
using namespace std;
typedef pair<int,int> pr;

const int MAX_N=1e5+5;
const int MAX_M=1e5+5;
struct node{
	int x,w;
	bool operator<(const node &p)const{
		return (x==p.x)?(w<p.w):(x<p.x);
	}
};
int n,m;
pr b[MAX_M];
node a[MAX_M*2];
int d[MAX_N],pre[MAX_N];

int main()
{
	ios::sync_with_stdio(false);
	cin>>n>>m;
	int l,r,t=0;
	for(int i=1;i<=m;++i)
	{
		cin>>b[i].first>>b[i].second;
		a[t++]=node{b[i].first,1};
		a[t++]=node{b[i].second+1,-1};
	}
	m*=2;
	sort(a,a+m);
	l=t=0;
	for(int i=1;i<=n&&l<m;++i)
	{
		while(l<m&&a[l].x==i){
			t+=a[l++].w;
		}
		d[i]=t;
	}
	
	int Sum=0;
	for(int i=1;i<=n;++i)
	{
		pre[i]=pre[i-1];
		if(d[i]==0)	++Sum;
		if(d[i]==1)	pre[i]=pre[i-1]+1;
	}
	int x,s1=0,s2=n+1;
	m/=2;
	for(int i=1;i<=m;++i)
	{
		x=pre[b[i].second]-pre[b[i].first-1];
		if(s2>=x){
			s2=x;	s1=i;
		}
	}
	s2+=Sum;
	cout<<s1<<" "<<s2<<endl;
	
	return 0;
}
*/