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这题真的很妙

可以想到二分最大值

问题就转化为如何判断是否合法

从左到右扫描区间的左端点,扫描到的就把右端点放入堆

每个点的权值可以用线段树树状数组维护

扫描时遇到点值不够时,就从优先队列中找到最大的右端点,区间加一遍

当优先队列为空或次数不足时就不符合要求

#include<bits/stdc++.h>
namespace ZDY{
    #pragma GCC optimize(3)
    #define il __inline__ __attribute__ ((always_inline))
    #define rg register
    #define ll long long
    #define ull unsigned long long
    #define db double
    #define sht short
    #define MB template <class T>il
    #define Fur(i,x,y) for(int i=x;i<=y;++i)
    #define Fdr(i,x,y) for(int i=x;i>=y;--i)
    #define fl(i,x) for(int i=head[x],to;to=e[i].to,i;i=e[i].nxt)
    #define clr(x,y) memset(x,y,sizeof(x))
    #define cpy(x,y) memcpy(x,y,sizeof(x))
    #define fin(s) freopen(s".in","r",stdin)
    #define fout(s) freopen(s".out","w",stdout)
    #define fcin ios::sync_with_stdio(false)
    #define l2(n) ((int)(log2(n)))
    #define inf 2122219134
    MB T ABS(T x){return x>0?x:-x;}
    MB T MAX(T x,T y){return x>y?x:y;}
    MB T MIN(T x,T y){return x<y?x:y;}
    MB T GCD(T x,T y){return y?GCD(y,x%y):x;}
    MB void SWAP(T &x,T &y){T t=x;x=y;y=t;}
}using namespace ZDY;using namespace std;
namespace IO{const char* ln="\n";const int str=1<<20;struct IN{char buf[str],*s,*t;bool _;IN():s(buf),t(buf),_(0){}il char gc(){return s==t&&((t=(s=buf)+fread(buf,1,str,stdin))==s)?EOF:(*s++);}IN&operator>>(char&ch){if(_)return *this;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)_=1;else ch=c;return *this;}IN& operator>>(char* ch){clr(ch,0);if(_)return *this;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)return _=1,*this;*ch=c;ch++;while((c=gc())!=EOF&&!isspace(c))*ch=c,ch++;if(c==EOF)_=1;return *this;}IN& operator>>(string& ch){if(_)return *this;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)return _=1,*this;ch+=c;while((c=gc())!=EOF&&!isspace(c))ch+=c;if(c==EOF)_=1;return *this;}template<typename T>IN&operator>>(T&x){if(_)return *this;char c=gc();bool ff=0;while(c!=EOF&&(c<'0'||c>'9'))ff^=(c=='-'),c=gc();if(c==EOF){_=1;return *this;}x=0;while(c!=EOF&&'0'<=c&&c<='9')x=(x<<3)+(x<<1)+c-48,c=gc();if(c==EOF)_=1;if(ff)x=-x;return *this;}}in;struct OUT{char buf[str],*s,*t;OUT():s(buf),t(buf+str){}~OUT(){fwrite(buf,1,s-buf,stdout);}void pt(char c){(s==t)?(fwrite(s=buf,1,str,stdout),*s++=c):(*s++=c);}OUT&operator<<(const char*s){while(*s)pt(*s++);return *this;}OUT&operator<<(char*s){while(*s)pt(*s++);return *this;}OUT&operator<<(string s){for(int i=0;s[i];i++)pt(s[i]);return *this;}template<typename T>OUT&operator<<(T x){if(!x)return pt('0'),*this;if(x<0)pt('-'),x=-x;char a[30],t=0;while(x)a[t++]=x%10,x/=10;while(t--)pt(a[t]+'0');return *this;}}out;}using namespace IO;
#define N 200011
int n,q,k,V,a[N],s[N];
struct node{
    int l,r;
    il bool operator<(node x){
        return l<x.l;
    }
}b[N];
il void add(int x,int v){
    while(x<=n)s[x]+=v,x+=(x&-x);
}
il int ask(int x){
    int ans=0;
    while(x)ans+=s[x],x-=(x&-x);
    return ans;
}
int que[N];
priority_queue<int>T;
bool chk(int w){
    clr(s,0);
    int tot=0;
    Fur(i,1,n)if(a[i]<w)que[++tot]=i;
    while(!T.empty())T.pop();
    int j=1,cnt=0;
    Fur(i,1,tot){
        while(j<=q&&b[j].l<=que[i])
            T.push(upper_bound(que+1,que+tot+1,b[j].r)-que-1),++j;
        while(a[que[i]]+ask(i)<w){
            ++cnt;
            if(cnt>k||T.empty())return 0;
            add(i,V);add(T.top()+1,-V);
            T.pop();
        }
    }
    return 1;
}
void work(){
    in>>n>>q>>k>>V;
    int mi=inf;
    Fur(i,1,n)in>>a[i],mi=MIN(mi,a[i]);
    Fur(i,1,q)in>>b[i].l>>b[i].r;
    sort(b+1,b+q+1);
    int l=mi,r=mi+k*V,ans;
    while(l<=r){
        int mid=(l+r)>>1;
        if(chk(mid))ans=mid,l=mid+1;
        else r=mid-1;
    }
    out<<ans<<ln;
}
int main(){
    fin("in");
    int tt;in>>tt;
    while(tt--)work();
}
LG 4064 [JXOI2017]加法
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