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Asteroids

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 24789		Accepted: 13439

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).

題意

一個n*n的網格，每個網格中有一些點，每次可以消滅一行或者一列的點，求最少的次數消滅所有的點。

分析

這是一個二分圖，行和列分別是兩個集合。求最少點覆蓋集。

結論：最小點覆蓋集=二分圖最大匹配數

建圖跑最大流即可。

建圖：S向行連流量為1的邊，列向T連流量為1的邊，對於每個點（u，v），由u向v連流量為1的邊

code

 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<cstring>
 4 
 5 using namespace std;
 6 const int N = 1010;
 7 const int INF = 1e9;
 8 struct Edge{
 9     int to,nxt,c;
10     Edge() {}
11     Edge(int x,int y,int z) {to = x,c = y,nxt = z;}
12 }e[1000100];
13 int q[1000100],L,R,S,T,tot = 1;
14 int dis[N],cur[N],head[N];
15 
16 inline char nc() {
17     static char buf[100000],*p1 = buf,*p2 = buf;
18     return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2) ? EOF :*p1++;
19 }
20 inline int read() {
21     int x = 0,f = 1;char ch=nc();
22     for (; ch<‘0‘||ch>‘9‘; ch=nc()) if(ch==‘-‘)f=-1;
23     for (; ch>=‘0‘&&ch<=‘9‘; ch=nc()) x=x*10+ch-‘0‘;
24     return x*f;
25 }
26 void add_edge(int u,int v,int c) {
27     e[++tot] = Edge(v,c,head[u]);head[u] = tot;
28     e[++tot] = Edge(u,0,head[v]);head[v] = tot;
29 }
30 bool bfs() {
31     for (int i=1; i<=T; ++i) cur[i] = head[i],dis[i] = -1;
32     L = 1,R = 0;
33     q[++R] = S;dis[S] = 1;
34     while (L <= R) {
35         int u = q[L++];
36         for (int i=head[u]; i; i=e[i].nxt) {
37             int v = e[i].to;
38             if (dis[v] == -1 && e[i].c > 0) {
39                 dis[v] = dis[u]+1;q[++R] = v;
40                 if (v==T) return true;
41             }
42         }
43     }
44     return false;
45 }
46 int dfs(int u,int flow) {
47     if (u==T) return flow;
48     int used = 0;
49     for (int &i=cur[u]; i; i=e[i].nxt) {
50         int v = e[i].to;
51         if (dis[v] == dis[u] + 1 && e[i].c > 0) {
52             int tmp = dfs(v,min(flow-used,e[i].c));
53             if (tmp > 0) {
54                 e[i].c -= tmp;e[i^1].c += tmp;
55                 used += tmp;
56                 if (used == flow) break;
57             }
58         }
59     }
60     if (used != flow) dis[u] = -1;
61     return used;
62 }
63 int dinic() {
64     int ret = 0;
65     while (bfs()) ret += dfs(S,INF);
66     return ret;
67 }
68 
69 int main() {
70     int n = read(),m = read();
71     S = n + n + 1,T = n + n + 2;
72     for (int i=1; i<=n; ++i) add_edge(S,i,1),add_edge(i+n,T,1);
73     for (int i=1; i<=m; ++i) {
74         int u = read(),v = read();
75         add_edge(u,v+n,1);
76     }
77     int ans = dinic();
78     printf("%d",ans);
79     return 0;
80 }

POJ 3041 Asteroids （二分圖最小點覆蓋集）