poj 3041 (水题)

https://vjudge.net/problem/POJ-3041

 

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

3 4
1 1
1 3
2 2
3 2

Sample Output

2

Hint

INPUT DETAILS:
The following diagram represents the data, where "X" is an asteroid and "." is empty space:
X.X
.X.
.X.

OUTPUT DETAILS:
Bessie may fire across row 1 to destroy the asteroids at (1,1) and (1,3), and then she may fire down column 2 to destroy the asteroids at (2,2) and (3,2).

 

 

题意：

            行与行之间相互独立，一个行可以就炸掉很多列。（列的道理一样），如果替换一些字。

               点与点之间相互独立，一个点就可以炸掉很多边。

 

 

 

 

思路就是上图

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
#include<set>
#include<algorithm>
#include<map>
#define maxn 2005
typedef long long ll;
#define inf 10000009
using namespace std;
int n,k;
int zzz,x,y;
bool can[maxn][maxn];//can[][]表示计算机i能够处理的任务
struct edge{
   int to,cap,rev;    //边的终点，容量，反向边
};
vector<edge>mp[maxn];  // 邻接图
bool vis[maxn];
void add_edge(int from,int to,int cap)   //建图
{
    mp[from].push_back((edge){to,cap,mp[to].size()});
     mp[to].push_back((edge){from,0,mp[from].size()-1});  //反向弧
}
int dfs(int v,int t,int f)   //找增广路 v，t是最终点  用了f的流量
{
   if(v==t)return f;
   vis[v]=true;
   for(int i=0;i<mp[v].size();i++)
   {
       edge &e=mp[v][i];
       if(!vis[e.to]&&e.cap>0)
       {
           int d=dfs(e.to,t,min(f,e.cap));  //遍历所有的路径
           if(d>0)
           {
               e.cap-=d;                    //求增加的流量
               mp[e.to][e.rev].cap+=d;
               return d;
           }
       }
   }
   return 0;
}
int max_flow(int s,int t)
{
    int flow=0;
    for(;;)
    {
        memset(vis,0,sizeof(vis));   //初始化
        int f=dfs(s,t,inf);
        if(f==0)return flow;
        flow+=f;
    }
}
void solve()
{
    int s=n+n+1;
    int t=s+1;
    for(int i=0;i<n;i++)
    {
        add_edge(s,i,1);
    }
    for(int i=0;i<n;i++)
    {
        add_edge(n+i,t,1);
    }
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<n;j++)
        {
            if(can[i][j]==1)
            {
                //cout<<i<<" "<<j;
               add_edge(i,n+j,1);
            }
        }
    }
    cout<<max_flow(s,t)<<endl;
}
int main()
{
    while(cin>>n>>k)
    {
        memset(can,0,sizeof(can));
        for(int i=0;i<k;i++)
        {
            cin>>x>>y;
            x--;
            y--;
          can[x][y]=1;
        }
        solve();
    }
    return 0;
}