【二分+線段樹】BZOJ4552 [Tjoi2016&Heoi2016]排序

題面在這裏

首先想到二分

然後就可以把整個序列轉化成01序列（0比mid小，1比mid大）

這樣排序的操作就可以用線段樹區間覆蓋來實現

最後判斷K 這個位置是0還是1，就完成了二分的驗證

竟然1A了，好高興

示例程序：

#include<cstdio>
inline char nc(){
    static char buf[100000],*p1=buf,*p2=buf;
    return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;
}
inline int red(){
    int res=0,f=1;char ch=nc();
    while (ch<'0'||'9'<ch) {if (ch=='-') f=-f;ch=nc();}
    while ('0'<=ch&&ch<='9') res=res*10+ch-48,ch=nc();
    return res*f;
}

const int maxn=100005,maxs=4*maxn;
int n,q,a[maxn],op[maxn],l[maxn],r[maxn],K;
struct node{
    node *l,*r;
    int s,tag;
    node () {}
    node (node *_l,node *_r,int _s):l(_l),r(_r),s(_s),tag(-1) {}
    inline void pushup() {s=l->s+r->s;}
    inline void pushdown(int L,int R){
        if (L==R||tag<0) return;
        int mid=L+R>>1;
        l->s=(mid-L+1)*tag;l->tag=tag;
        r->s=(R-mid)*tag;r->tag=tag;
        tag=-1;
    }
}base[maxs],nil;
typedef node* P_node;
P_node null,Rot,len;
void init(){
    nil=node(null,null,0);null=&nil;len=base;
}
P_node build(int L,int R){
    *len=node(null,null,0);
    P_node x=len++;
    if (L==R) return x;
    int mid=L+R>>1;
    x->l=build(L,mid);x->r=build(mid+1,R);
    x->pushup(); return x;
}
void insert(P_node x,int L,int R,int QL,int QR,int w){
    if (QR<L||R<QL) return;
    if (QL<=L&&R<=QR) {x->s=(R-L+1)*w;x->tag=w;return;}
    x->pushdown(L,R);
    int mid=L+R>>1;
    insert(x->l,L,mid,QL,QR,w);insert(x->r,mid+1,R,QL,QR,w);
    x->pushup();
}
int query(P_node x,int L,int R,int QL,int QR){
    if (QR<L||R<QL) return 0;
    if (QL<=L&&R<=QR) return x->s;
    x->pushdown(L,R);
    int mid=L+R>>1;
    return query(x->l,L,mid,QL,QR)+query(x->r,mid+1,R,QL,QR);
}
bool check(int mid){
    for (int i=1;i<=n;i++)
     insert(Rot,1,n,i,i,a[i]>=mid);
    for (int i=1;i<=q;i++)
     if (op[i]){
        int x=query(Rot,1,n,l[i],r[i]);
        insert(Rot,1,n,l[i],l[i]+x-1,1); insert(Rot,1,n,l[i]+x,r[i],0);
     }else{
        int x=query(Rot,1,n,l[i],r[i]);
        insert(Rot,1,n,r[i]-x+1,r[i],1); insert(Rot,1,n,l[i],r[i]-x,0);
     }
    return query(Rot,1,n,K,K);
}
int main(){
    n=red(),q=red();
    for (int i=1;i<=n;i++) a[i]=red();
    for (int i=1;i<=q;i++) op[i]=red(),l[i]=red(),r[i]=red();
    K=red();
    init();Rot=build(1,n);
    int L=1,R=n,ans=1;
    while (L<=R){
        int mid=L+R>>1;
        if (check(mid)) ans=mid,L=mid+1;else R=mid-1;
    }
    printf("%d",ans);
    return 0;
}