Network flow problem 24 questions Magic Ball

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Is a magical figure caption built, built Figure \ (Van \) ♂ whole will not ah

We generally is a one to put the ball, to put each of a, we obtain the current path of the minimum number of covers (current number of vertices - maximum flow) until the minimum number of paths covering ({>} \) \ column number. At this time, the number of balls \ (--1 \) is the answer to the first question. The second question is to ask each path, pushed down along just fine

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#include<cmath>
using namespace std;
struct zzz{
    int t,len,nex;
}e[100010<<1]; int head[20010],tot=1;
void add(int x,int y,int z){
    e[++tot].t=y;
    e[tot].len=z;
    e[tot].nex=head[x];
    head[x]=tot;
}
int s=0,t=20000,vis[20010],pre[20010];
bool bfs(){
    memset(vis,0,sizeof(vis));
    queue <int> q; q.push(s); vis[s]=1;
    while(!q.empty()){
        int k=q.front(); q.pop();
        for(int i=head[k];i;i=e[i].nex){
            int to=e[i].t;
            if(vis[to]||!e[i].len) continue;
            q.push(to); vis[to]=vis[k]+1;
            if(to==t) return 1;
        }
    }
    return vis[t];
}
int dfs(int now,int flow){
    if(now==t||!flow) return flow;
    int rest=0,fl;
    for(int i=head[now];i;i=e[i].nex){
        int to=e[i].t;
        if(vis[to]==vis[now]+1&&(fl=dfs(to,min(flow,e[i].len)))){
            e[i].len-=fl, e[i^1].len+=fl, rest+=fl;
            if(rest==flow) return rest;
        }
    }
    if(rest<flow) vis[now]=0;
    return rest;
}
int dinic(){
    int ans=0;
    while(bfs()) ans+=dfs(s,2147483647);
    return ans;
}
int read(){
    int k=0; char c=getchar();
    for(;c<'0'||c>'9';) c=getchar();
    for(;c>='0'&&c<='9';c=getchar())
      k=(k<<3)+(k<<1)+c-48;
    return k;
}
bool jl[20010]; int next[20010];
int main(){
    int n=read(),cnt=0,num=0;
    while(cnt<=n){
        num++; cnt++;
        for(int i=1;i<num;i++)
          if(sqrt(i+num)==(int)sqrt(i+num))
            add(i,num+10000,1),add(num+10000,i,0);
        add(s,num,1); add(num,s,0);
        add(num+10000,t,1); add(t,num+10000,0);
        cnt-=dinic();
    }
    printf("%d\n",--num);
    for(int i=1;i<=num;i++){
        for(int j=head[i];j;j=e[j].nex){
            if(!e[j].len){
                next[i]=e[j].t-10000;
                break;
            }
        }
    }
    for(int i=1;i<=num;i++){
        if(jl[i]) continue;
        int x=i;
        while(x!=-10000){
            jl[x]=1; printf("%d ",x);
            x=next[x];
        }
        printf("\n");
    }
    return 0;
}