POJ #3041 asteroids 30,412 min. overlay maximum matching Hungarian algorithm

Description

Bessie wants to navigate she spaceship through a dangerous asteroid field in the shape of an n x N grid (1 <= N <= 5 ). The grid contains K asteroids (1 <= k <=), which is conveniently located at the lattice points of the G Rid.
Fortunately, Bessie have a powerful weapon that can vaporize all the asteroids on any given row or column of the grid with A single shot. This weapon was quite expensive, so she wishes to use it sparingly. Given the location of the asteroids in the field, find the minimum number of shots Bessie needs to fire to eliminate a ll of the asteroids.

Input

* Line 1:two integers N and K, separated to a single space.
* Lines 2..k+1:each line contains, 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 41 11 32 23 2

Sample Output

2

Hint

INPUT DETAILS:
The following diagram represents the data, where "X" is a asteroid and "." is empty space:
x.x
. X.
. X.

OUTPUT DETAILS:
Bessie may fire across row 1 to destroy the asteroids at (all) and (1,3), and then she could fire down column 2 to destroy T He asteroids at (2,2) and (3,2).

Ideas

The modeling of this problem is very classical, we consider each row as an X node, each column as a Y node, each target one edge, then destroy all targets and the least number of launches means to select the fewest points, so that any one edge at least one endpoint is selected. And this corresponds to the definition of the concept of the minimum coverage of the dichotomy graph, which is solved by the Hungarian algorithm, by the theorem of the minimum point coverage number of the binary graph = maximum matching number.

The AC code is as follows:

#include <iostream>#include<algorithm>#include<cstdio>#include<cstring>Const intMAXN =510;Const intMAXK =10010;intn,k;intLINE[MAXN][MAXN];BOOLUSED[MAXN];intNEXT[MAXN];BOOLFind (intx) { for(inti =1; I <= N; i++) {        if(Line[x][i] &&!Used[i]) {Used[i]=true; if(!next[i] | |find (Next[i])) {Next[i]=x; return true; }        }    }    return false;}intmatch () {intsum =0;  for(inti =1; I <= N; i++) {memset (used,0,sizeof(used)); if(Find (i)) sum++; }    returnsum;}intMainvoid) {    intu,v;  while(~SCANF ("%d%d", &n, &k) {memset (line,0,sizeof(line)); Memset (Next,0,sizeof(next));  while(k--) {scanf ("%d%d", &u, &v); LINE[U][V]=1; } printf ("%d\n", Match ()); }    return 0;}

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POJ #3041 asteroids 30,412 min. overlay maximum matching Hungarian algorithm