BZOJ5321 JXOI2017加法（二分答案+贪心+堆+树状数组）

　　二分答案后得到每个位置需要被加的次数。考虑贪心。从左到右考虑每个位置，将以该位置为左端点的区间按右端点从大到小加进堆。看该位置还需要被加多少次，如果不需要加了就不管，否则取堆顶区间将其选择，BIT实现区间覆盖。

#include<iostream>

#include<cstdio>

#include<cmath>

#include<cstdlib>

#include<cstring>

#include<algorithm>

#include<queue>

#include<vector>

using namespace std;

#define ll long long

#define N 200010

char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<''||c>'')) c=getchar();return c;}

int gcd(int n,int m){return m==?n:gcd(m,n%m);}

int read()

{

    int x=,f=;char c=getchar();

    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}

    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();

    return x*f;

}

int T,n,m,k,a[N],c[N],tree[N];

priority_queue<int> q;

vector<int> b[N];

void add(int k,int x){while (k<=n) tree[k]+=x,k+=k&-k;}

int query(int k){int s=;while (k) s+=tree[k],k-=k&-k;return s;}

bool check()

{

    for (int i=;i<=n;i++) tree[i]=;int cnt=;

    while (!q.empty()) q.pop();

    for (int i=;i<=n;i++) add(i,c[i]-c[i-]);

    for (int i=;i<=n;i++)

    {

        for (int j=;j<b[i].size();j++) q.push(b[i][j]);

        int x=query(i);

        for (;x>;x--)

        {

            if (q.empty()) return ;

            int y=q.top();q.pop();

            if (y<i) return ;

            add(i,-),add(y+,);

            if ((++cnt)>k) return ;

        }

    }

    return ;

}

int main()

{

#ifndef ONLINE_JUDGE

    freopen("bzoj5321.in","r",stdin);

    freopen("bzoj5321.out","w",stdout);

    const char LL[]="%I64d\n";

#else

    const char LL[]="%lld\n";

#endif

    T=read();

    while (T--)

    {

        n=read(),m=read(),k=read();int p=read();

        int l=,r,ans;

        for (int i=;i<=n;i++) l=min(l,a[i]=read()),b[i].clear();

        for (int i=;i<=m;i++)

        {

            int x=read(),y=read();

            b[x].push_back(y);

        }

        r=l+m*p;

        while (l<=r)

        {

            int mid=l+r>>;

            for (int i=;i<=n;i++) c[i]=mid<=a[i]?:(mid-a[i]-)/p+;

            if (check()) l=mid+,ans=mid;

            else r=mid-;

        }

        printf("%d\n",ans);

    }

    return ;

}