loj6062(hall定理+线段树)

题目链接

https://loj.ac/problem/6062

题解

这个题可以看成二分图完美匹配的判定,可以用 hall 定理求解,当固定子集大小,会发现其实最差的情况是选取最小的几个数,那么我们不妨对 $B$ 排序,然后看其前缀的相邻点数。维护相邻点数时,可以用线段树维护




代码

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/*
*         ┏┓    ┏┓
*         ┏┛┗━━━━━━━┛┗━━━┓
*         ┃       ┃  
*         ┃   ━    ┃
*         ┃ >   < ┃
*         ┃       ┃
*         ┃... ⌒ ...  ┃
*         ┃ ┃
*         ┗━┓ ┏━┛
*          ┃ ┃ Code is far away from bug with the animal protecting          
*          ┃ ┃ 神兽保佑,代码无bug
*          ┃ ┃           
*          ┃ ┃       
*          ┃ ┃
*          ┃ ┃           
*          ┃ ┗━━━┓
*          ┃ ┣┓
*          ┃ ┏┛
*          ┗┓┓┏━━━━━━━━┳┓┏┛
*           ┃┫┫ ┃┫┫
*           ┗┻┛ ┗┻┛
*/

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<queue>
#include<map>
#include<stack>
#include<cmath>
#include<set>
#include<bitset>
#include<assert.h>
#define inc(i,l,r) for(int i=l;i<=r;i++)
#define dec(i,l,r) for(int i=l;i>=r;i--)
#define link(x) for(edge *j=h[x];j;j=j->next)
#define mem(a) memset(a,0,sizeof(a))
#define ll long long
#define eps 1e-8
#define succ(x) (1<<x)
#define lowbit(x) (x&(-x))
#define sqr(x) ((x)*(x))
#define mid (x+y>>1)
#define NM 150005
#define nm 200005
#define pi 3.1415926535897931
#define mp(x,y) make_pair(x,y)
using namespace std;
const int inf=1e9;
ll read(){
ll x=0,f=1;char ch=getchar();
while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
return f*x;
}



int n,m,p,_x,_t;
int a[NM],b[NM],ans;

struct node{
int x,y,s,tag;
node*l,*r;
node(int x,int y,node*l=0,node*r=0):x(x),y(y),l(l),r(r),s(0){if(l)upd();else s=-x;}
void upd(){s=min(l->s,r->s);}
void push(){
if(tag){
l->s+=tag;l->tag+=tag;
r->s+=tag;r->tag+=tag;
tag=0;
}
}
void mod(){
if(_x<=x){s+=_t;tag+=_t;return;}
push();if(_x<=mid)l->mod();r->mod();upd();
}
}*root;
node*build(int x,int y){return x==y?new node(x,y):new node(x,y,build(x,mid),build(mid+1,y));}

void ins(int x){
_x=lower_bound(b+1,b+1+m,p-x)-b;
if(_x<=m)root->mod();
}

int main(){
n=read();m=read();p=read();
inc(i,1,m)b[i]=read();
inc(i,1,n)a[i]=read();
sort(b+1,b+1+m);
root=build(1,m);
inc(i,1,m)_t=1,ins(a[i]);
if(root->s>=0)ans++;
inc(i,m+1,n){
_t=1;ins(a[i]);
_t=-1;ins(a[i-m]);
if(root->s>=0)ans++;
}
printf("%d\n",ans);
return 0;
}