POJ 3041 Asteroids 二分图

原题连接：http://poj.org/problem?id=3041

Asteroids

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 17985		Accepted: 9798

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

题意：

给你一个$N*N$的棋盘，其中某些地方有棋子，你可以每次消除一列或者一行，问你至少需要多少次操作，把棋子消除完。

题解：

这是一道二分匹配的经典问题。将每个棋子的横纵坐标连边，那么就会构成一个二分图，问题就转化为求解这个二分图的最小点覆盖。由最小点覆盖就是最大匹配，跑一发dinic就好。

代码：

#include<iostream>
#include<stack>
#include<vector>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<cstdio>
#include<queue>
#define MAX_S (1<<10)+10
#define MAX_V 1234
#define MAX_N MAX_V
#define INF 1000009
using namespace std;

struct edge {
    int to, cap, rev;
    bool isRev;

    edge(int t, int c, int r, bool i)
            : to(t), cap(c), rev(r), isRev(i) { }

    edge() { }
};

template <class T>
inline bool scan_d(T &ret)
{
    char c;
    int sgn;
    if(c=getchar(),c==EOF) return 0; //EOF
    while(c!=' -' &&(c<'0' ||c>'9' )) c=getchar();
    sgn=(c==' -' )?-1:1;
    ret=(c==' -' )?0:(c-'0' );
    while(c=getchar(),c>='0' &&c<='9' ) ret=ret*10+(c-'0' );
    ret*=sgn;
    return 1;
}

vector<edge> G[MAX_N];
int level[MAX_V];
int iter[MAX_V];

void init(int totNode) {
    for (int i = 0; i <= totNode; i++)
        G[i].clear();
    memset(level, 0, sizeof(level));
    memset(iter, 0, sizeof(iter));
}

void add_edge(int from,int to,int cap) {
    G[from].push_back(edge (to, cap, G[to].size(),0));
    G[to].push_back(edge (from, 0, G[from].size() - 1,1));
}

void bfs(int s) {
    queue<int> que;
    memset(level, -1, sizeof(level));
    level[s] = 0;
    que.push(s);
    while (!que.empty()) {
        int v = que.front();
        que.pop();
        for (int i = 0; i < G[v].size(); i++) {
            edge &e = G[v][i];
            if (e.cap > 0 && level[e.to] < 0) {
                level[e.to] = level[v] + 1;
                que.push(e.to);
            }
        }
    }
}

int dfs(int v,int t,int f) {
    if (v == t)return f;
    for (int &i = iter[v]; i < G[v].size(); i++) {
        edge &e = G[v][i];
        if (e.cap > 0 && level[v] < level[e.to]) {
            int d = dfs(e.to, t, min(f, e.cap));
            if (d > 0) {
                e.cap -= d;
                G[e.to][e.rev].cap += d;
                return d;
            }
        }
    }
    return 0;
}

int max_flow(int s,int t) {
    int flow = 0;
    for (; ;) {
        bfs(s);
        if (level[t] < 0)return flow;
        memset(iter, 0, sizeof(iter));
        int f;
        while ((f = dfs(s, t, INF)) > 0) {
            flow += f;
        }
    }
}

int n,m;
int S=1233,T=1212;

int main() {
    cin.sync_with_stdio(false);
    cin >> n >> m;
    for (int i = 0; i < m; i++) {
        int u, v;
        cin >> u >> v;
        add_edge(u, v+n, 1);
    }
    for (int i = 1; i <= n; i++) {
        add_edge(S, i, 1);
        add_edge(i+n, T, 1);
    }
    cout << max_flow(S, T) << endl;
    return 0;
}