poj 3041 Asteroids

最新推荐文章于 2022-08-16 21:41:12 发布

转载最新推荐文章于 2022-08-16 21:41:12 发布 · 61 阅读

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原文链接：http://www.cnblogs.com/lsyyy/p/11291409.html

Bessie需要穿越一个充满小行星的N x N网格，使用武器清除每一行或列的小行星，目标是最少射击次数。通过建立图模型并求解最大流问题，找到最优解。

Description

Bessie wants to navigate her spaceship through a dangerous asteroid field in the shape of an N x N grid (1 <= N <= 500). The grid contains K asteroids (1 <= K <= 10,000), which are conveniently located at the lattice points of the grid.

Fortunately, Bessie has a powerful weapon that can vaporize all the asteroids in any given row or column of the grid with a single shot.This weapon is quite expensive, so she wishes to use it sparingly.Given the location of all the asteroids in the field, find the minimum number of shots Bessie needs to fire to eliminate all of the asteroids.

Input

* Line 1: Two integers N and K, separated by a single space.
* Lines 2..K+1: Each line contains two space-separated integers R and C (1 <= R, C <= N) denoting the row and column coordinates of an asteroid, respectively.

Output

* Line 1: The integer representing the minimum number of times Bessie must shoot.

Sample Input

Sample Output

你要很多个点
每个点在方格里
你每次消除可以是一行或是一列
求最小消除次数

将每个点对应的行列连边，求最大流

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<vector>
#include<queue>
#include<algorithm>
#define re return
#define inc(i,l,r) for(int i=l;i<=r;++i)
const int maxn=1505,maxm=20005;
using namespace std;
template<typename T>inline void rd(T&x)
{
	char c;bool f=0;
	while((c=getchar())<'0'||c>'9')if(c=='-')f=1;
	x=c^48;
	while((c=getchar())>='0'&&c<='9')x=x*10+(c^48);
	if(f)x=-x;
} 

int n,m,s,t,deep[maxn],k=1,hd[maxn],cur[maxn];
struct node{
	int to,nt,w;
}e[maxm];

inline void add(int x,int y,int z)
{
	e[++k].to=y;e[k].nt=hd[x];hd[x]=k;e[k].w=z;
	e[++k].to=x;e[k].nt=hd[y];hd[y]=k;e[k].w=0;
}

inline bool  bfs()
{
	queue<int>q;
	inc(i,1,t)deep[i]=0;
	deep[s]=1;
	q.push(s);
	while(!q.empty())
	{
		int u=q.front();
		q.pop();
		for(int i=hd[u];i;i=e[i].nt)
		{
			int v=e[i].to;
			if(!deep[v]&&e[i].w)
			{
				deep[v]=deep[u]+1;
				if(v==t)re 1;
				q.push(v);
			} 
		}
	}
	re 0;
}

inline int dfs(int u,int flow)
{
	if(u==t) re flow;
	int delta=flow;
	for(int &i=cur[u];i;i=e[i].nt)
	{
		int v=e[i].to;
		if(deep[v]==deep[u]+1&&e[i].w)
		{
			int d=dfs(v,min(delta,e[i].w));
			e[i].w-=d;e[i^1].w+=d;
			delta-=d;
			if(!delta)re flow;
		}
	}
	re flow-delta;
}

int main()
{
	rd(n),rd(m);
	s=(n<<1)+1;
	t=s+1;
	int x,y;
	inc(i,1,m)
	{
		rd(x),rd(y);
		add(x,y+n,1);
	}
	inc(i,1,n){
		add(s,i,1);
		add(i+n,t,1);
	}
	int ans=0;
	while(bfs())
	{
		inc(i,1,t)cur[i]=hd[i];
		ans+=dfs(s,n);
	} 
	printf("%d",ans);
	re 0;
}

转载于:https://www.cnblogs.com/lsyyy/p/11291409.html