About analog network flow

The meaning of problems

In fact, CF 724 E

This is in fact the brothers adapted the title QMQ, the true scope should be \ (n <= 10000 \)

Thinking

This question may flow obviously do with the network:

But the scope of the direct T.

Then we can use the greedy

Then we found this map looks like a magical nature, using the maximum flow = minimum cut, we can enumerate minimum cut.

Minimum cut of nature that every point is either located in the \ (\ st) set or \ (ed \) collection, so we will be able to enumerate the case:

We found this in the middle of the edge we need to address, so the \ (st, ed \) we have to deal with it, but we want to do with minimal cut it?

We set \ (f [i] [j ] \) indicated that the first \ (i \) points, there \ (j \) a return to \ (st \) collection.

Then no count \ (C \) effects, it is not difficult listed:
\ (F [I] [J] = min (F [I-. 1] [J-. 1] + A [I], F [I-. 1 ] [J] + B [I]) \) .

但是如何处理\(C\)呢，不难发现对于\(i,j(i<j)\)如果\(i\)选了\(ed\)，\(j\)选了\(st\)，那么就会有个\(C\)。

这个该归到\(st\)还是\(ed\)呢，很明显为了DP没有后效性，我们就归到\(ed\)吧。

那么DP转移方程变成了：\(f[i][j]=min(f[i-1][j-1]+a[i],f[i-1][j]+b[i]+j*C)\)。

就可以转移了，时间复杂度\(O(n^2)\)，当然，这个又叫模拟网络流。

我的思路

我是采用贪心，这道题目\(n\)才\(2000\)，还开\(2s\)，那么我们不难想到一种贪心思路，就是使得所有卖完粮食后剩余粮食大于\(C\)的位置的粮食尽量的平衡。

也就是尽可能的榨干这个\(C\)，使得后面每个位置都可以堆满\(C\)。

As for the balance of values, we use binary search, then the time complexity \ (O (n ^ 2log range) \) .

But also optimized, \ (1S \) had lost inside.

#include<cstdio>
#include<cstring>
#include<queue>
#define  N  3100
using  namespace  std;
typedef  long  long  LL;
template  <class  T>
inline  T  mymin(T  x,T  y){return  x<y?x:y;}
int  a[N],b[N];
priority_queue<int>q;//储存粮食信息，当然是大于C的才给放
int  sta[N],top;//表示可以移动的粮食 
int  list[N],tail;
int  n,m;
LL  zans;
bool  check(int  x,int  k/*原本的数字*/)
{
    for(int  i=1;i<=top;i++)
    {
        if(sta[i]<=x)return  false;
        k+=mymin(sta[i]-x,m);//可以给你多少的数字 
        if(k>=x)return  true;
    }
    return  false;
}
void  work(int  x,int  y)
{
    top=0;
    while(!q.empty())sta[++top]=q.top(),q.pop();
    int  k=x-y;//可能是负数
    if(top)//有大于C的粮食位置
    {
        int  l=k+1,r=mymin((LL)sta[1],(LL)k+(LL)top*(LL)m),mid,ans=k/*就是不变*/;//表示范围，而且能防止m过小时时间过大
        while(l<=r)
        {
            mid=(l+r)/2;
            if(check(mid,k)==true)ans=mid,l=mid+1;
            else  r=mid-1;
        }
        for(int  i=1;i<=top;i++)
        {
            if(k==ans)break;
            int  zjj=mymin(mymin(ans-k,m),sta[i]-ans);sta[i]-=zjj;k+=zjj;
        }
    }
    if(k<0)
    {
        while(k<0  &&  tail)k+=list[tail--];
    }
    zans+=mymin(y,k+y);
    if(k>0)
    {
        if(k<=m)list[++tail]=k;
        else  q.push(k);
    }
    for(int  i=1;i<=top;i++)
    {
        if(sta[i]<=m)list[++tail]=sta[i];
        else  q.push(sta[i]);
    }
}
int  main()
{
//  freopen("c.in","r",stdin);
//  freopen("c.out","w",stdout);
    scanf("%d%d",&n,&m);
    for(int  i=1;i<=n;i++)scanf("%d",&a[i]);
    for(int  i=1;i<=n;i++)scanf("%d",&b[i]);
    for(int  i=1;i<=n;i++)
    {
        work(a[i],b[i]);
    }
    printf("%lld\n",zans);
    return  0;
}