poj 3041Asteroids+3692Kindergarten(二分图匹配+公式补充)

Asteroids Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 13780   Accepted: 7493 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). Source USACO 2005 November Gold

问题描述
有一个N*N的网格，放有k个星星。

Bessie有一个强大的武器，每次可以清扫任意一行或者任意一列上所有的星星。当时武器非常昂贵，给出K个星星的位置，用最少的武器清扫干净所有的星星。

解题思路：
最小顶点覆盖：
用最少的顶点覆盖所有的边
二分图最小顶点覆盖 = 最大匹配

采用匈牙利算法：

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<string>
#include<stack>
#include<queue>
#include<vector>
#include<algorithm>
#include<iostream>
using namespace std;
#ifdef __int64
typedef __int64 LL;
#else
typedef long long LL;
#endif
bool _map[520][520];
int vis[520];
int link[520];//记录n2中的点在n1中所匹配的点的编号，即匹配集合M
int n,k;
int n1,n2;//为二分图的两个顶点集
bool Find(int x)
{
    for(int i=0;i<n2;i++)
    {
        if(_map[x][i]&&!vis[i])//x和i有边，且节点i未被搜索
        {
            vis[i]=1;
            if(link[i]==-1||Find(link[i]))//如果i不属于前一个匹配M,或者i匹配的节点可以找到一条增广路
            {
                link[i]=x;//更新
                return true;
            }
        }
    }
    return false;
}
int mach()
{
    int ans=0;
    memset(link,-1,sizeof(link));
    for(int i=0;i<n1;i++)
    {
        memset(vis,0,sizeof(vis));
        if(Find(i))//如果从第i个点找到了增广路
            ans++;
    }
    return ans;
}
int main()
{
    while(~scanf("%d%d",&n,&k))
    {
        memset(_map,0,sizeof(_map));
        int r,c;
        for(int i=1;i<=k;i++)
        {
            scanf("%d%d",&r,&c);
            _map[r-1][c-1]=true;
        }
        n1=n2=n;
        printf("%d\n",mach());
    }
    return 0;
}

Kindergarten Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 4798   Accepted: 2334 Description In a kindergarten, there are a lot of kids. All girls of the kids know each other and all boys also know each other. In addition to that, some girls and boys know each other. Now the teachers want to pick some kids to play a game, which need that all players know each other. You are to help to find maximum number of kids the teacher can pick. Input The input consists of multiple test cases. Each test case starts with a line containing three integers G, B (1 ≤ G, B ≤ 200) and M (0 ≤ M ≤ G × B), which is the number of girls, the number of boys and the number of pairs of girl and boy who know each other, respectively. Each of the following M lines contains two integers X and Y (1 ≤ X≤ G,1 ≤ Y ≤ B), which indicates that girl X and boy Y know each other. The girls are numbered from 1 to G and the boys are numbered from 1 to B. The last test case is followed by a line containing three zeros. Output For each test case, print a line containing the test case number( beginning with 1) followed by a integer which is the maximum number of kids the teacher can pick. Sample Input 2 3 3 1 1 1 2 2 3 2 3 5 1 1 1 2 2 1 2 2 2 3 0 0 0 Sample Output Case 1: 3 Case 2: 4 Source 2008 Asia Hefei Regional Contest Online by USTC

问题描述:

在幼儿园中，有很多小孩，男孩之间互相认识，女孩之间也互相认识。而且，知道一些女孩男孩之间相互认识。现在老师要组织一个游戏，需要参加游戏的人之间相互认识。求老师可挑选参加游戏的最大人数。

题目分析：

这个题目相当于求最大团。下面我就来补充几个概念，和几个公式

1、独立集：任何两点都不相邻。

2、团：任何两点都相邻。

3、最大独立集 = 顶点数 - 最大匹配数

4、最小顶点覆盖 = 最大匹配数

5、最小路径覆盖 = 最大独立集

6、最大团 = 补图的最大独立集

下面就直接运用第六个公式，先构建一个补图，之后求最大匹配数。结果用顶点数一减就完事了。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<string>
#include<stack>
#include<queue>
#include<vector>
#include<algorithm>
#include<iostream>
using namespace std;
#ifdef __int64
typedef __int64 LL;
#else
typedef long long LL;
#endif
bool _map[300][300];
int vis[300];
int link[300];
int n1,n2;
int G,B,M;
bool Find(int x)
{
    for(int i=0;i<n2;i++)
    {
        if(_map[x][i]&&!vis[i])
        {
            vis[i]=1;
            if(link[i]==-1||Find(link[i]))
            {
                link[i]=x;
                return true;
            }
        }
    }
    return false;
}
int mach()
{
    int ans=0;
    memset(link,-1,sizeof(link));
    for(int i=0;i<n1;i++)
    {
        memset(vis,0,sizeof(vis));
        if(Find(i))
            ans++;
    }
    return ans;
}
int main()
{
    int kase=1;
    while(scanf("%d%d%d",&G,&B,&M)&&G&&B&&M)
    {
        for(int i=0;i<G;i++)//构建原二分图的补图
            for(int j=0;j<B;j++)
                _map[i][j]=true;
        int g,b;
        for(int i=0;i<M;i++)
        {
            scanf("%d%d",&g,&b);
            _map[g-1][b-1]=false;
        }
        n1=G; n2=B;
        printf("Case %d: %d\n",kase++,G+B-mach());
    }
    return 0;
}