BZOJ4653: [Noi2016]区间

Posted 2020-08-09

传送门

UOJ上卡掉一个点，COGS上卡掉两个点..弃疗，不改了，反正BZOJ上过啦hhh

 

先把区间按长度递增排序。然后每次用线段树维护区间最大覆盖次数，用一个指针随便扫扫就行了。

 

  1 //NOI 2016 D2T1
  2 //by Cydiater
  3 //2016.9.18
  4 #pragma GCC optimize("O2")
  5 #include <iostream>
  6 #include <cstdio>
  7 #include <cstdlib>
  8 #include <cstring>
  9 #include <string>
 10 #include <algorithm>
 11 #include <queue>
 12 #include <map>
 13 #include <ctime>
 14 #include <cmath>
 15 #include <iomanip>
 16 using namespace std;
 17 #define ll long long
 18 #define up(i,j,n)        for(INT i=j;i<=n;i++)
 19 #define down(i,j,n)        for(INT i=j;i>=n;i--)
 20 #define FILE "interval"
 21 #define INT unsigned int
 22 const INT MAXN=1e6+5;
 23 const INT oo=0x3f3f3f3f;
 24 map<INT,INT> lable;
 25 inline INT read(){
 26     char ch=getchar();INT x=0,f=1;
 27     while(ch>‘9‘||ch<‘0‘){if(ch==‘-‘)f=-1;ch=getchar();}
 28     while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
 29     return x*f;
 30 }
 31 INT num[MAXN],N,M,top=0,cnt=0,check=0,x,y,v,ans=oo;
 32 struct Query{
 33     INT st,nd,len;
 34 }q[MAXN];
 35 struct Tree{
 36     INT maxx,delta;
 37 }t[MAXN<<3];
 38 namespace solution{
 39     inline bool cmp(Query a,Query b){return a.len<b.len;}
 40     inline void downit(INT node){
 41         if(t[node].delta==0)return;
 42         t[node<<1].delta+=t[node].delta;t[node<<1|1].delta+=t[node].delta;
 43         t[node<<1].maxx+=t[node].delta;t[node<<1|1].maxx+=t[node].delta;
 44         t[node].delta=0;
 45     }
 46     inline void reload(INT node){t[node].maxx=max(t[node<<1].maxx,t[node<<1|1].maxx);}
 47     void updata(INT leftt,INT rightt,INT root){
 48         downit(root);
 49         if(leftt>y||rightt<x)        return;
 50         if(leftt>=x&&rightt<=y){
 51             t[root].maxx+=v;
 52             t[root].delta+=v;
 53             return;
 54         }
 55         INT mid=(leftt+rightt)>>1;
 56         updata(leftt,mid,root<<1);
 57         updata(mid+1,rightt,root<<1|1);
 58         reload(root);
 59     }
 60     void init(){
 61         N=read();M=read();
 62         up(i,1,N){
 63             q[i].st=read();q[i].nd=read();
 64             num[++top]=q[i].st;num[++top]=q[i].nd;
 65         }
 66         sort(num+1,num+top+1);
 67         up(i,1,top)if(!lable[num[i]])lable[num[i]]=++cnt;
 68         up(i,1,N){
 69             q[i].len=q[i].nd-q[i].st;
 70             q[i].st=lable[q[i].st];
 71             q[i].nd=lable[q[i].nd];
 72         }
 73         sort(q+1,q+N+1,cmp);
 74     }
 75     void slove(){
 76         up(i,1,N){
 77             while(t[1].maxx<M&&check<N){
 78                 check++;
 79                 x=q[check].st;y=q[check].nd;v=1;
 80                 updata(1,cnt,1);
 81             }
 82             if(t[1].maxx>=M)ans=min(ans,q[check].len-q[i].len);
 83             x=q[i].st;y=q[i].nd;v=-1;
 84             updata(1,cnt,1);
 85         }
 86     }
 87     void output(){
 88         if(ans==oo)puts("-1");
 89         else cout<<ans<<endl;
 90     }
 91 }
 92 int main(){
 93     //freopen(FILE".in","r",stdin);
 94     //freopen(FILE".out","w",stdout);
 95     //freopen("input.in","r",stdin);
 96     using namespace solution;
 97     init();
 98     slove();
 99     output();
100     return 0;
101 }

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