LuoguP2765 魔术球问题

首先,很难看出来这是一道网络流题.但是因为在网络流24题中，所以还是用网络流的思路

首先考虑完全平方数的限制。

如果\(i,j\)满足\(i < j\) 且 $i + j \(为完全平方数我们就在\)i - j $连一条有向边

练完之后我们会得到这样一个图(图来自luogu题解)

发现这是一个DAG，而且我们将柱子的限制转化为路径条数。问题就转化成了

LuoguP2764 最小路径覆盖问题

然后，我们就一直加球，加到需要的路径条数大于给定的柱子数为止

#include<cstdio>
#include<cctype>
#include<cstring>
#include<queue>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<vector>
using namespace std;
const int N = 2e5 + 3;
const int M = 2e6 + 3;
const int INF = 2e9;
int n,s,t,tot = 1,top;
vector <int> G[N];
int pre[N],head[N],cur[N],high[N],first[N];
bool vis[N];
struct edge{
    int from;
    int to;
    int nxt;
    int flow;   
}e[M];
inline void add(int x,int y,int z){
    e[++tot].to = y;
    e[tot].flow = z;
    e[tot].from = x;
    e[tot].nxt = head[x];   
    head[x] = tot;
    e[++tot].to = x;
    e[tot].flow = 0;
    e[tot].from = y;
    e[tot].nxt = head[y];
    head[y] = tot; 
}
inline int read(){
    int v = 0,c = 1;char ch = getchar();
    while(!isdigit(ch)){
        if(ch == '-') c = -1;
        ch = getchar(); 
    }
    while(isdigit(ch)){
        v = v * 10 + ch - 48;
        ch = getchar(); 
    }
    return v * c;
}
inline bool bfs(){
    queue <int> q;
    for(int i = 0;i <= t;++i) high[i] = 0;
    q.push(s);high[s] = 1;
    while(!q.empty()){
        int k = q.front();q.pop();
        for(int i = head[k];i;i = e[i].nxt){
            int y = e[i].to;
            if(!high[y] && e[i].flow > 0)
            high[y] = high[k] + 1,q.push(y);
        }
    }
    return high[t] != 0;
}
inline int dfs(int x,int dis){
    if(x == t) return dis;
    for(int &i = cur[x];i;i = e[i].nxt){
        int y = e[i].to;
        if(high[y] == high[x] + 1 && e[i].flow > 0){
            int flow = dfs(y,min(dis,e[i].flow));
            if(flow > 0){
                e[i].flow -= flow;
                e[i ^ 1].flow += flow;
                pre[x >> 1] = y >> 1;
                return flow;    
            }
        }
    }
    return 0;
}
inline int dinic(){
    int res = 0;
    while(bfs()){
        for(int i = 0;i <= t;++i) cur[i] = head[i];
        while(int now = dfs(s,INF)) res += now;
    }
    return res;
}
int main(){
    n = read();
    int sum = 0,now = 0;
    s = 100000 + 2,t = s + 1;
    while(now <= n){
        sum++;
        add(s,sum << 1,1);
        add(sum << 1 | 1,t,1);
        for(int i = sqrt(sum) + 1;i * i < sum * 2;++i)
            add((i * i - sum) << 1,sum << 1 | 1,1); 
        int f = dinic();
        if(!f) first[++now] = sum;
    }
    printf("%d\n",sum - 1);
    for(int i = 1;i <= n;++i){
        if(vis[first[i]]) continue;
        for(int j = first[i];j != 0 && j != (t >> 1);j = pre[j])
        vis[j] = 1,printf("%d ",j);
        puts("");
    }
    return 0;   
}