魔术球问题

枚举放的球，先假设新建柱子，拆成两个点，第一个点连S，表示后面还可以放；第二个连T表示放到其他柱子上；再枚举放过的数和它是否组成完全平方数，枚举的数的第一个点向它的第二个点连边，表示这个球可以放到其他球上，容量都为一
每次跑最大流出来的表示会消掉的柱子个数，如果此时球-消去的比n大则break输出答案

# include <stdio.h>
# include <stdlib.h>
# include <iostream>
# include <string.h>
# include <math.h>
# define ll long long
# define RG register
# define IL inline
# define mem(a, b) memset(a, b, sizeof(a))
# define Max(a, b) (((a) > (b)) ? (a) : (b))
# define Min(a, b) (((a) < (b)) ? (a) : (b))
# define Sqr(a) ((a) * (a))
using namespace std;

IL int Get(){
    RG char c = '!'; RG int x = 0, z;
    for(; c > '9' || c < '0'; c = getchar()) z = (c == '-') ? -1 : 1;
    for(; c >= '0' && c <= '9'; c = getchar()) x = x * 10 + c - '0';
    return x * z;
}

const int MAXN = 10001, MAXM = 200001, INF = 2147483647;
int n, m, ft[MAXN], cnt, ans, level[MAXN], Q[MAXM], cur[MAXN], match[MAXN], vis[MAXN];
struct Edge{
    int to, nt, f;
} edge[MAXM];

IL void Add(RG int u, RG int v, RG int f){
    edge[cnt] = (Edge){v, ft[u], f}; ft[u] = cnt++;
}

IL bool Bfs(RG int S, RG int T){
    mem(level, 0);
    RG int head = 0, tail = 0;
    level[S] = 1; Q[0] = S;
    while(head <= tail){
        RG int u = Q[head++];
        for(RG int e = ft[u]; e != -1; e = edge[e].nt){
            RG int v = edge[e].to;
            if(!level[v] && edge[e].f){
                level[v] = level[u] + 1;
                Q[++tail] = v;
            }
        }
    }
    return level[T];
}

IL int Dfs(RG int u, RG int T, RG int maxf){
    if(u == T) return maxf;
    RG int ret = 0;
    for(RG int &e = cur[u]; e != -1; e = edge[e].nt){
        RG int v = edge[e].to;
        if(level[v] == level[u] + 1 && edge[e].f){
            RG int f = Dfs(v, T, Min(maxf - ret, edge[e].f));
            edge[e].f -= f; edge[e ^ 1].f += f;
            ret += f; match[u] = v;
            if(ret == maxf) break;
        }
    }
    if(!ret) level[u] = 0;
    return ret;
}

int main(){
    mem(ft, -1);
    n = Get();
    RG int S = 0, T = 4000, num = 0;
    while(++num){
        Add(S, num, 1); Add(num, S, 0);
        Add(num + 2000, T, 1); Add(T, num + 2000, 0);
        for(RG int i = 1; i < num; i++)
            if(Sqr((int)sqrt(i + num)) == i + num)
                Add(i, num + 2000, 1), Add(num + 2000, i, 0);
        while(Bfs(S, T)){
            for(RG int i = S; i <= T; i++)
                cur[i] = ft[i];
            ans += Dfs(S, T, INF);
        }
        if(num - ans > n){
            printf("%d
", num - 1);
            break;
        }
    }
    for(RG int i = 1; i < num; i++)
        if(!vis[i]){
            RG int j = i;
            while(j){
                if(j != 2000 && j != 4000) j > 2000 ? printf("%d ", j -= 2000) : printf("%d ", j);
                vis[j] = 1;
                j = match[j];
            }
            printf("
");
        }
    return 0;
}