[Noi2016]区间

Posted 2020-09-03 CHADLZX

 

 这道题不愧是第一题,真的挺水的.

 看到要搞使区间的最大长度和最少长度之差最小,先按长度排序.

 要统计某个地方的出现次数,有这么大的数据范围,离散坐标少不了.

 排序完之后,选择的区间肯定是连续的一段了,不如二分枚举再加上线段树判定吧...

 当然不是这么搞的,搞两个指针前后移动维护最大值>=m的性质即可.

 丑丑的代码,不过懒得改了.

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<ctime>
#include<string>
#include<iomanip>
#include<algorithm>
#include<map>
using namespace std;
#define LL long long
#define FILE "dealing"
#define up(i,j,n) for(int i=j;i<=n;++i)
#define db double
#define ull unsigned long long
#define eps 1e-10
#define pii pair<int,int>
int read(){
	int x=0,f=1,ch=getchar();
	while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
	while(ch>=‘0‘&&ch<=‘9‘){x=(x<<1)+(x<<3)+ch-‘0‘;ch=getchar();}
	return f*x;
}
const int maxn=4002200,maxm=20000,mod=(int)(1e9+7+0.1),inf=(int)(2e9);
template<class T>bool cmax(T& a,T b){return a<b?a=b,true:false;}
template<class T>bool cmin(T& a,T b){return a>b?a=b,true:false;}
template<class T>T min(T& a,T& b){return a<b?a:b;}
template<class T>T max(T& a,T& b){return a>b?a:b;}
int delet[maxn],Max[maxn];
void add(int x,int d){
	delet[x]+=d;
	Max[x]+=d;
}
void pushdown(int x){
	if(delet[x]){
		add(x<<1,delet[x]);
		add(x<<1|1,delet[x]);
		delet[x]=0;
	}
}
void updata(int x){
	Max[x]=max(Max[x<<1],Max[x<<1|1]);
}
int L,R,key;
void change(int x,int l,int r){
	if(l>R||r<L)return ;
	if(l>=L&&r<=R){
		add(x,key);
		return ;
	}
	int mid=(l+r)>>1;
	pushdown(x);
	change(x<<1,l,mid);
	change(x<<1|1,mid+1,r);
	updata(x);
}
int n,m;
struct node{int x,y,id;}e[maxn],w[maxn];
bool cmp(node a,node b){return (a.y-a.x)<(b.y-b.x);}
pii a[maxn];int val[maxn],N,M,q[maxn][2];
int main(){
	freopen(FILE".in","r",stdin);
	freopen(FILE".out","w",stdout);
	n=read(),m=read();
	up(i,1,n)a[++N].first=e[i].x=read(),
		a[N].second=i,a[++N].first=e[i].y=read(),
		a[N].second=i,e[i].id=i;
	sort(e+1,e+n+1,cmp);
	sort(a+1,a+N+1);
	a[0].first=-1;
	up(i,1,N){
		if(a[i].first!=a[i-1].first)
			val[++M]=a[i].first;//真正权值
		if(q[a[i].second][0])q[a[i].second][1]=M;
		else q[a[i].second][0]=M;
	}
	up(i,1,n)w[i].x=q[e[i].id][0],w[i].y=q[e[i].id][1];//排序后hash权值
	int left=1,ans=inf;
	for(int i=1;i<=n;i++){
		if(Max[1]<m){
			L=w[i].x,R=w[i].y;key=1;
			change(1,1,M);
		}
		while(Max[1]>=m){
			cmin(ans,(val[w[i].y]-val[w[i].x])-(val[w[left].y]-val[w[left].x]));
			L=w[left].x,R=w[left].y,key=-1;
			change(1,1,M);
			left++;
		}
	}
	if(ans==inf)printf("-1\n");
	else printf("%d\n",ans);
	return 0;
}