[网络流24题]魔术球问题

每个球向可以组成的完全平方数连边，会连出一个有向无环图

柱子上的球对应图上的路径,相当于最小路径点覆盖

求出拆点二分图，再用n-最大匹配数即可

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<cmath>
#define maxn 100000
#define maxm 100000
#define INF 0x7fffffff
using namespace std;
int n;
struct edge{
    int from;
    int to;
    int next;
}E[maxm*4+5];
long long flow[maxm*4+5];
int head[maxn*2+5];
int sz=1;
void add_edge(int u,int v,int w){
//	if(u<v)printf("%d->%d %d\n",u,v,w);
    sz++;
    E[sz].from=u;
    E[sz].to=v;
    E[sz].next=head[u];
    head[u]=sz;
    flow[sz]=w;
}

int deep[maxn*2+5];
int bfs(int s,int t){
    queue<int>q;
    memset(deep,0,sizeof(deep));
    deep[s]=1;
    q.push(s);
    while(!q.empty()){
        int x=q.front();
        q.pop();
        for(int i=head[x];i;i=E[i].next){
            int y=E[i].to;
            if(flow[i]&&!deep[y]){
                deep[y]=deep[x]+1;
                q.push(y);
                if(y==t) return 1;
            } 
        }
    }
    return 0;
}

int pre[maxn];
long long dfs(int x,int t,long long minf){
    if(x==t) return minf;
    long long rest=minf,k;
    for(int i=head[x];i;i=E[i].next){
        int y=E[i].to;
        if(flow[i]&&deep[y]==deep[x]+1){
            k=dfs(y,t,min(rest,flow[i]));
            if(k==0){
                deep[y]=0;
                continue; 
            }
            flow[i]-=k;
            flow[i^1]+=k;
            rest-=k;
            pre[x/2]=y/2;
        }
    }
    return minf-rest;
}

long long dinic(int s,int t){
    long long maxflow,nowflow;
    maxflow=0;
    while(bfs(s,t)){
        while(nowflow=dfs(s,t,INF)){
            maxflow+=nowflow;
        }
    }
    return maxflow;
}

int is_sqr(int x){
    if(sqrt(x)==(double)x) return 1;
    else return 0; 
}


int tmp[maxn*2+5];
int vis[maxn*2+5];
int s=0,t=maxn+1;
void print_ans(int x){
    for(int i=x;i!=0&&i!=t/2;i=pre[i]){
        vis[i]=1;
        printf("%d ",i);
    }
    printf("\n");
}
int main(){
    scanf("%d",&n);
    int pil=0,ball=0;

    while(pil<=n){
        ball++;
        add_edge(s,ball*2,1);
        add_edge(ball*2+1,s,0);
        add_edge(ball*2+1,t,1);
        add_edge(t,ball*2+1,0);
        for(int i=sqrt(ball)+1;i*i<(ball*2);i++){
            add_edge((i*i-ball)*2,ball*2+1,1);
            add_edge(ball*2+1,(i*i-ball)*2,0);
        }
        int ans=dinic(s,t);
        if(!ans){
            tmp[++pil]=ball;
        }
    } 
    printf("%d\n",ball-1);
    for(int i=1;i<=n;i++){
        print_ans(tmp[i]);
    }
    
}