洛谷 P2765 魔术球问题 （最小路径覆盖 or 贪心）

题目

（戳我）[https://www.luogu.com.cn/problem/P2765]

思路

两种思路，第一种贪心很明显了，我们直接从第一个开始放球，可以放就放，不能放就新开一个。第二种的话建图，首先我们应该发现这个东西的特点，一个柱子可以放多个球，有多个柱子，但是每个柱子的球是不可能一样的，这和图论里面二分图的最小路径覆盖很像，因此我们选择拆点建图，然后对这个图不断加点，当我的顶点数-最大匹配大于最小路径的时候，就是答案了。贪心大法好！！！

代码实现

#include<cstdio>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>
#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;
#define rep(i,f_start,f_end) for (int i=f_start;i<=f_end;++i)
#define per(i,n,a) for (int i=n;i>=a;i--)
#define MT(x,i) memset(x,i,sizeof(x) )
#define rev(i,start,end) for (int i=start;i<end;i++)
#define inf 0x3f3f3f3f
#define mp(x,y) make_pair(x,y)
#define lowbit(x) (x&-x)
#define MOD 1000000007
#define exp 1e-8
#define N 1000005 
#define fi first 
#define se second
#define pb push_back
typedef long long ll;
typedef vector <int> VI;
typedef pair<int ,int> PII;
typedef pair<int ,PII> PIII;
ll gcd (ll a,ll b) {return b?gcd (b,a%b):a; }
inline int read() {
    char ch=getchar(); int x=0, f=1;
    while(ch<'0'||ch>'9') {
        if(ch=='-') f=-1;
        ch=getchar();
    } while('0'<=ch&&ch<='9') {
        x=x*10+ch-'0';
        ch=getchar();
    } return x*f;
}
const int maxn=1e4+10;
const int M=5e5+10;
struct edge {
    int u,v,flow,next;
}e[M];
int head[maxn],cnt=-1;

inline void add (int u,int v,int flow) {
    e[++cnt]=(edge) {u,v,flow,head[u]};
    head[u]=cnt;
}
inline void add_edge (int u,int v,int flow) {
    add (u,v,flow);
    add (v,u,0);
}

int dis[maxn],cur[maxn];
int ss,s,t,n,m,x,y,ans;
int bfs () {
    MT (dis,0);
    queue<int > q;
    q.push (ss);
    dis[ss]=1;
    while (q.size ()) {
        int x=q.front (); q.pop ();
        for (int i=head[x];i!=-1;i=e[i].next) {
            if (dis[e[i].v]==0&&e[i].flow) {
                dis[e[i].v]=dis[x]+1;
                q.push (e[i].v);
            }
        }
    }
    return dis[t];
}

int dfs (int now,int nowflow) {
    if (now==t) return nowflow;
    for (int &i=cur[now];~i;i=e[i].next) {
        if (dis[e[i].v]==dis[now]+1&&e[i].flow) {
            int canflow=dfs (e[i].v,min (nowflow,e[i].flow));
            if (canflow>0) {
                e[i].flow-=canflow;
                e[i^1].flow+=canflow;
                return canflow;
            }
        }
    }
    return 0;
}

inline int Dinic () {
	int ans=0;
    while (bfs ()) {
       memcpy (cur,head,sizeof (head));
       while (int val=dfs(ss,inf)) ans+=val;
    }
	return ans;
}

int pre[maxn],vis[maxn];

int main () {
	MT (head,-1);
    scanf ("%d",&n);
	t=10000;
	while (1) {
       ans++,s++;
	   rev (i,1,s) if (sqrt (i+s)==(int) (sqrt (i+s))) {
		   add_edge (i,s+5000,1);
	   }
	   add_edge (0,s,1);
	   add_edge (s+5000,t,1);
	   int st=Dinic ();
	   ans-=st;
	   if (ans>n) break;
	}
	printf ("%d
",s-1);

	rev (i,1,s) {
		int k=head[i];
		while (k) {
			if (!e[k].flow) {
				pre[i]=e[k].v-5000;
				break;
			}
			k=e[k].next;
		}
	}
	rev (i,1,s) {
		if (vis[i]) continue;
		int p=i;
		while (p!=-5000) {
			vis[p]=1;
			printf ("%d ",p);
			p=pre[p];
		}
		cout<<endl;

	}
	return 0;
}

#include<cstdio>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>
#include<set>
#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;
#define rep(i,f_start,f_end) for (int i=f_start;i<=f_end;++i)
#define per(i,n,a) for (int i=n;i>=a;i--)
#define MT(x,i) memset(x,i,sizeof(x) )
#define rev(i,start,end) for (int i=start;i<end;i++)
#define inf 0x3f3f3f3f
#define mp(x,y) make_pair(x,y)
#define lowbit(x) (x&-x)
#define exp 1e-8
#define N 1000005 
#define fi first 
#define se second
#define pb push_back
const int mod=1e9+7;
typedef long long ll;
typedef vector <int> VI;
typedef pair<int ,int> PII;
typedef pair<int ,PII> PIII;
ll gcd (ll a,ll b) {return b?gcd (b,a%b):a; }
inline int read() {
    char ch=getchar(); int x=0, f=1;
    while(ch<'0'||ch>'9') {
        if(ch=='-') f=-1;
        ch=getchar();
    } while('0'<=ch&&ch<='9') {
        x=x*10+ch-'0';
        ch=getchar();
    } return x*f;
}

const int maxn=60;
VI v[maxn];
set <int> s;
int n;

void prework () {
    rep (i,1,maxn) {
        s.insert (i*i);
    }
}

int main () {
    cin>>n;
    int cnt=0,ans=1;
    prework ();
    while (true) {
        rep (i,1,cnt) {
            int x=v[i].back ()+ans;
            if (s.count (x)) {
             v[i].pb (ans);
             ans++;
             i=0;
             continue;
            }
        }
        if (cnt<n) v[++cnt].pb (ans++);
        else break;
    }
    cout<<ans-1<<endl;
    rep (i,1,n) {
        for (auto it:v[i]) {
            printf ("%d ",it);
        }
        cout<<endl;
    }
    return 0;
}