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.
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.
* 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
3 4 1 1 1 3 2 2 3 2
Sample Output
2
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;
}