POJ 3041 Asteroids 最小顶点覆盖

原创 于 2015-05-01 17:22:07 发布 · 412 阅读 · 0 ·
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该博客介绍了一个经典的算法问题，即如何使用最少次数的武器射击来清除网格中的所有小行星。提供了两种解决方案，一种是使用二分图的最大匹配算法（匈牙利算法），另一种是采用网络流算法（Dinic算法）。通过这两种方法，可以有效地找到解决问题的最佳策略。

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Asteroids
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 16509 Accepted: 9004

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).



两个做法 二分图最大匹配或网络流

匈牙利：

/** Author: ☆·aosaki(*’(OO)’*)  niconiconi★ **/
//#pragma comment(linker, "/STACK:1024000000,1024000000")
//#include<bits/stdc++.h>
#include <iostream>
#include <sstream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <functional>
#include <cmath>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <list>
#include <stack>
//#include <tuple>
#define ALL(v) (v).begin(),(v).end()
#define foreach(i,v) for (__typeof((v).begin())i=(v).begin();i!=(v).end();i++)
#define SIZE(v) ((int)(v).size())
#define mem(a) memset(a,0,sizeof(a))
#define mem1(a) memset(a,-1,sizeof(a))
#define lp(k,a) for(int k=1;k<=a;k++)
#define lp0(k,a) for(int k=0;k<a;k++)
#define lpn(k,n,a) for(int k=n;k<=a;k++)
#define lpd(k,n,a) for(int k=n;k>=a;k--)
#define sc(a) scanf("%d",&a)
#define sc2(a,b) scanf("%d %d",&a,&b)
#define lowbit(x) (x&(-x))
#define ll long long
#define pi pair<int,int>
#define vi vector<int>
#define PI acos(-1.0)
#define pb(a) push_back(a)
#define mp(a,b) make_pair(a,b)
#define TT cout<<"*****"<<endl;
#define TTT cout<<"********"<<endl;
inline int gcd(int a,int b)
{
    return a==0?b:gcd(b%a,a);
}

#define INF 1e9
#define eps 1e-8
#define mod 10007
#define MAX 10010

using namespace std;


const int MAXN=550;
int uN,vN;
int g[MAXN][MAXN];
int linker[MAXN];
bool used[MAXN];

int n,m;
bool dfs(int u)
{
    int v;
    for(v=0;v<vN;v++)
        if(g[u][v]&&!used[v])
        {
            used[v]=true;
            if(linker[v]==-1||dfs(linker[v]))
            {
                linker[v]=u;
                return true;
            }
        }
    return false;
}

int hungary()
{
    int re=0;
    int u;
    memset(linker,-1,sizeof(linker));
    for(u=0;u<uN;u++)
    {
        memset(used,0,sizeof(used));
        if(dfs(u))  re++;
    }
    return re;
}

int main()
{
    //freopen("in.txt","r",stdin);
    int xx,yy;
    while(~sc2(n,m))
    {
        mem(g);
        uN=n;
        vN=n;
        while(m--)
        {

            sc2(xx,yy);
            g[xx-1][yy-1]=1;

        }
     printf("%d\n",hungary());
    }
    return 0;
}


dinic:

/** Author: ☆·aosaki(*’(OO)’*)  niconiconi★ **/
//#pragma comment(linker, "/STACK:1024000000,1024000000")
//#include<bits/stdc++.h>
#include <iostream>
#include <sstream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <functional>
#include <cmath>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <list>
#include <stack>
//#include <tuple>
#define ALL(v) (v).begin(),(v).end()
#define foreach(i,v) for (__typeof((v).begin())i=(v).begin();i!=(v).end();i++)
#define SIZE(v) ((int)(v).size())
#define mem(a) memset(a,0,sizeof(a))
#define mem1(a) memset(a,-1,sizeof(a))
#define lp(k,a) for(int k=1;k<=a;k++)
#define lp0(k,a) for(int k=0;k<a;k++)
#define lpn(k,n,a) for(int k=n;k<=a;k++)
#define lpd(k,n,a) for(int k=n;k>=a;k--)
#define sc(a) scanf("%d",&a)
#define sc2(a,b) scanf("%d %d",&a,&b)
#define lowbit(x) (x&(-x))
#define ll long long
#define pi pair<int,int>
#define vi vector<int>
#define PI acos(-1.0)
#define pb(a) push_back(a)
#define mp(a,b) make_pair(a,b)
#define TT cout<<"*****"<<endl;
#define TTT cout<<"********"<<endl;
inline int gcd(int a,int b)
{
    return a==0?b:gcd(b%a,a);
}

#define inf 1e9
#define eps 1e-8
#define mod 10007
#define maxn 550*550

using namespace std;

int n,m,tot,S,T,f2,d2;
int pre[maxn],cpyh[maxn],q[maxn],d[maxn];

struct edge
{
    int from,to,next,val;
} ji[2*maxn];

void add(int x,int y,int w)
{
    ji[tot].from=x;
    ji[tot].to=y;
    ji[tot].val=w;
    ji[tot].next=pre[x];
    pre[x]=tot++;
    ji[tot].from=y;
    ji[tot].to=x;
    ji[tot].val=0;
    ji[tot].next=pre[y];
    pre[y]=tot++;
}

int dinic()
{
    int re=0;
    while(1)
    {
        mem1(d);
        int head=0,tail=0;
        q[tail++]=S;
        d[S]=0;
        while(head<tail)
        {
            int t=q[head++];
            for(int k=pre[t];k;k=ji[k].next)
            {
                int toit=ji[k].to;
                int val=ji[k].val;
                if(val&&d[toit]==-1)
                {
                    d[toit]=d[t]+1;
                    q[tail++]=toit;
                    if(toit==T)
                    {
                        head=tail;
                        break;
                    }
                }
            }
        }
        if(d[T]==-1) break;
        memcpy(cpyh,pre,(T+1)*sizeof(int));
        for(int i=S,top=0;;)
        {
            if(i==T)
            {
                int flow=inf,st;
                lp0(k,top)
                {
                    if(ji[q[k]].val<flow)
                    {
                       st=k;
                       flow=ji[q[st]].val;
                    }
                }
                lp0(k,top)
                {
                    ji[q[k]].val-=flow;
                    ji[q[k]^1].val+=flow;
                }
                re+=flow;
                top=st;
                i=ji[q[top]].from;
            }
            for(int j=cpyh[i];j;j=cpyh[i]=ji[j].next)
                if(ji[j].val>0&&d[i]+1==d[ji[j].to])
                    break;
            if(cpyh[i])
            {
                q[top++]=cpyh[i];
                i=ji[cpyh[i]].to;
            }
            else
            {
                if(top==0) break;
                d[i]=-1;
                i=ji[q[--top]].from;
            }
        }
    }
    return re;
}


void init()
{
    S=0;
    tot=2;
    T=2*n+1;
    mem(pre);
}

int main()
{
    //freopen("in.txt","r",stdin);
    int xx,yy;
    while(~sc2(n,m))
    {
        init();
        lp(i,m)
        {
            sc2(xx,yy);
            add(xx,yy+n,1);

        }
        lp(i,n)
        {
            add(0,i,1);
            add(i+n,T,1);
        }
        printf("%d\n",dinic());

    }

    return 0;
}




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