LuoguP2765 魔术球问题

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;	
}