LibreOJ 6003 魔术球 (最大流)

题解:每次加入两个点,对于和为平方数的两个值所对应的点建边,反正网络流可以跑残量网络,所以就没有什么关系了……

代码如下:

#include<cmath>
#include<queue>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define inf 0x3f3f3f3f
using namespace std;
int head[],next[],w[],v[],deep[];
int s,t,cnt;
void init()
{
    cnt=-;
    memset(head,-,sizeof(head));
    memset(next,-,sizeof(next));
}
void add(int from,int to,int cost)
{
    cnt++;
    next[cnt]=head[from];
    w[cnt]=cost;
    v[cnt]=to;
    head[from]=cnt;
}
void add_edge(int from,int to,int cost)
{
    add(from,to,cost);
    add(to,from,);
}
int bfs(int s,int t)
{
    queue<int> q;
    memset(deep,,sizeof(deep));
    deep[s]=;
    q.push(s);
    while(!q.empty())
    {
        int u=q.front();
        q.pop();
        for(int i=head[u]; i!=-; i=next[i])
        {
            if(!deep[v[i]]&&w[i]>)
            {
                deep[v[i]]=deep[u]+;
                q.push(v[i]);
            }
        }
    }
    if(!deep[t])
    {
        return ;
    }
    return ;
}
int dfs(int u,int t,int dist)
{
    if(u==t)
    {
        return dist;
    }
    for(int i=head[u]; i!=-; i=next[i])
    {
        if(w[i]&&(deep[u]+==deep[v[i]]))
        {
            int di=dfs(v[i],t,min(w[i],dist));
            if(di>)
            {
                w[i]-=di;
                w[i^]=di;
                return di;
            }
        }
    }
    return ;
}
int dinic(int s,int t)
{
    int ans=;
    while(bfs(s,t))
    {
        while(int d=dfs(s,t,inf))
        {
            ans+=d;
        }
    }
    return ans;                //
}
int id[],wid[],vis[];
void solve(int x,int &f)
{
    int loc=wid[x];
    vis[x]=;
    for(int i=head[loc]; i!=-; i=next[i])
    {
        if(w[i]==&&v[i]!=t)
        {
            solve(v[i]/,f);
        }
    }
    if(f==)
    {
        f=;
    }
    else
    {
        putchar(' ');
    }
    printf("%d",x);
}
int main()
{
    init();
    int n;
    scanf("%d",&n);
    s=;
    t=;
    int i,tmp=,ans=,tmp2=;
    for(i=; i-ans<=n+; i++)
    {
        id[i]=tmp++;
        wid[i]=tmp++;
        add_edge(s,id[i],);
        add_edge(wid[i],t,);
        for(int j=; j<i; j++)
        {
            int tmp1=sqrt(i+j);
            if(tmp1*tmp1==i+j)
            {
                add_edge(id[j],wid[i],);
            }
        }
        ans+=dinic(s,t);
    }
    tmp2=i-;
    printf("%d\n",tmp2);
    for(int i=head[t]; i!=-; i=next[i])
    {
        if(w[i]==&&!vis[v[i]/])
        {
            int f=;
            solve(v[i]/,f);
            puts("");
        }
    }
    for(int i=; i<=tmp2; i++)
    {
        if(!vis[i])
        {
            printf("%d\n",i);
        }
    }
}