Educational Codeforces Round 69 (Rated for Div. 2)

直接贴原网站题解了  代码是自己的

1197C - Array Splitting

Idea: Roms

Tutorial

1197C - Array Splitting

code:

#include<bits/stdc++.h>
using namespace std;

const int maxn=3e5+5;

int n,k;int a[maxn];


priority_queue<int,vector<int>,greater<int> >pq;

int main()
{
	ios::sync_with_stdio(0);cin.tie(0);
	
	cin>>n>>k;
	for(int i=1;i<=n;i++)
	{
		cin>>a[i];
	}
	for(int i=1;i<n;i++)
	{
		pq.push(a[i]-a[i+1]);
	}
	
	int ans=0;ans+=a[n]-a[1];
	for(int i=1;i<k;i++)
	{
		ans+=pq.top();pq.pop();
	}
	cout<<ans<<'\n';
	
	return 0;
}

1197E - Culture Code

Idea: adedalic

Tutorial

1197E - Culture Code

线段树维护时的pushup还是有一些技巧的，可以减少很多代码量

（题解代码线段树写法有点看不懂。。。）

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int>pii;
const int mod=1e9+7,maxn=2e5+5,inf=2e9;
const ll linf=1e18;

int modulo(int a)
{
	if(a>=mod)a-=mod;if(a<0)a+=mod;return a;
}

int cmp(const pii&a,const pii&b)//sort by in
{
	if(a.second!=b.second)return a.second<b.second;
	return a.first<b.first;
}

int n;
pii p[maxn];

struct node{int val;int cnt;};
struct segt
{
	node t[maxn<<2];
	node pushup(node a,node b)
	{node ret;
		if(a.val<b.val)return a;
		else if(a.val>b.val)return b;
		else{ret.val=a.val,ret.cnt=modulo(a.cnt+b.cnt);return ret;}
	}
	void build(int x,int l,int r)
	{
		if(l==r){t[x].val=inf;t[x].cnt=0;return ;}
		int m=l+r>>1;build(x<<1,l,m);build(x<<1|1,m+1,r);
		t[x]=pushup(t[x<<1],t[x<<1|1]);
	}
	node query(int ql,int qr,int l,int r,int x)
	{
		node ret={inf,0};
		if(ql<=l&&r<=qr)return t[x];
		int m=l+r>>1;
		if(ql<=m)ret=pushup(ret,query(ql,qr,l,m,x<<1));
		if(m<qr) ret=pushup(ret,query(ql,qr,m+1,r,x<<1|1));
		return ret;
	}
	
	void spmodify(int l,int r,int x,int tar,node modify)
	{
		if(l==r)
		{
			t[x]=modify;return ;
		}
		int m=l+r>>1;
		if(tar<=m)spmodify(l,m,x<<1,tar,modify);
		if(m<tar)spmodify(m+1,r,x<<1|1,tar,modify);
		t[x]=pushup(t[x<<1],t[x<<1|1]);
	}
}T;
// DP with d[i]=(x,y) — the minimum extra space x and number of such subsequences y among all nested subsets, where the i-th doll is minimal.
void solve()
{
	sort(p+1,p+n+1,cmp);//sort by in
	T.build(1,1,n);
	
	for(int i=n;i>=1;i--)
	{
		int pos=lower_bound(p+1,p+n+1,make_pair(0,p[i].first),cmp)-p;//search for in < out
		if(pos>n){T.spmodify(1,n,1,i,node{p[i].second,1});continue;}
		node ret=T.query(pos,n,1,n,1);
//		cerr<<"-- "<<ret.val<<' '<<ret.cnt<<'\n';
		T.spmodify(1,n,1,i,node{ret.val-p[i].first+p[i].second,ret.cnt});
	}
	printf("%d\n",T.query(1,n,1,n,1).cnt);
}

int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
		scanf("%d%d",&p[i].first,&p[i].second);//out in
	}
	
	solve();
	
	return 0;
}