poj 3041(最小点覆盖及二分图的最大匹配)

Asteroids

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 14127		Accepted: 7685

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

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).

题目意思就是要求一行或者一列里一炮最多能打掉多少个点

简单的求二分图的最大匹配(还不知道二分图匹配算法的可以百度一下二分图最大匹配先学学，这题就很容易了)，下面分别附上数组实现和vector实现的两种AC代码

16ms数组代码：

#include<cstdio> 
#include<cstring>
using namespace std;
int map[505][505];
int match[505];
int lk[505];
int vis[505];
int n;
int dfs(int u){
    for(int i=1;i<=n;i++){
        if(map[u][i]&&!vis[i]){
            vis[i]=1;
            if(!match[i]||dfs(match[i])){
                match[i]=u;
                lk[u]=i;
                return 1;
            }
        }
    }
    return 0;
}
int main(){
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int k;
    while(scanf("%d %d",&n,&k)!=EOF){
        memset(map,0,sizeof(map));
        for(int i=1;i<=k;i++){
            int x,y;
            scanf("%d %d",&x,&y);
            map[x][y]=1;
        }
        int sum=0;
        memset(match,0,sizeof(match));
        memset(lk,0,sizeof(lk));
        for(int i=1;i<=n;i++){
            if(!lk[i]){
                memset(vis,0,sizeof(vis));
                if(dfs(i))
                    sum++;
            }
        }
        printf("%d\n",sum);
    }
    return 0;
}

0ms vector实现代码：

#include<cstdio> 
#include<cstring>
#include<vector>
using namespace std;
vector <int> map[505];
int match[505];
int lk[505];
int vis[505];
int n;
int dfs(int u){
    for(int i=0;i<map[u].size();i++){
        if(!vis[map[u][i]]){
            vis[map[u][i]]=1;
            if(!match[map[u][i]]||dfs(match[map[u][i]])){
                match[map[u][i]]=u;
                lk[u]=map[u][i];
                return 1;
            }
        }
    }
    return 0;
}
int main(){
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int k;
    while(scanf("%d %d",&n,&k)!=EOF){
        for(int i=1;i<=n;i++)
            if(!map[i].empty())
                map[i].clear();
        for(int i=1;i<=k;i++){
            int x,y;
            scanf("%d %d",&x,&y);
            map[x].push_back(y);
        }
        int sum=0;
        memset(match,0,sizeof(match));
        memset(lk,0,sizeof(lk));
        for(int i=1;i<=n;i++){
            if(!lk[i]){
                memset(vis,0,sizeof(vis));
                if(dfs(i))
                    sum++;
            }
        }
        printf("%d\n",sum);
    }
    return 0;
}