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EOJ 2069 Asteroids

/*
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EOJ 2069 Asteroids
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----問(wèn)題描述：
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Bessie wants to navigate her spaceship through a dangerous asteroid field in the shape of an N x N grid (1 <= N <= 500).
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The grid contains K asteroids (1 <= K <= 10,000), which are conveniently located at the lattice points of the grid.
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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.
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This weapon is quite expensive, so she wishes to use it sparingly.
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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.
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----輸入：
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* Line 1: Two integers N and K, separated by a single space.
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* 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.
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----輸出：
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* Line 1: The integer representing the minimum number of times Bessie must shoot.
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----樣例輸入：
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3 4
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1 1
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1 3
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2 2
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3 2
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----樣例輸出：
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2
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----分析：
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建立二分圖模型，
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若第 i 行和第 j 列處存在一個(gè) asteroid ，則 x[i] 與 y[j] 連一條邊，
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求二分圖最大匹配，使用匈牙利算法。
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*/
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#include <stdio.h>
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#include <string.h>
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#define L 503
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int adj[ L ][ L ], n, state[ L ], result[ L ];
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int find( int i ) {
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int j, k;
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for ( j = adj[ i ][ 0 ]; j > 0; --j ) {
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k = adj[ i ][ j ];
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if ( state[ k ] == 0 ) {
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state[ k ] = 1;
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if ( ( result[ k ] == 0 ) || find( result[ k ] ) ) {
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result[ k ] = i;
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return 1;
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}
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}
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}
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return 0;
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}
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int maxMatch() {
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int ans = 0, i;
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for ( i = 1; i <= n; ++i ) {
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memset( state, 0, sizeof( state ) );
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if ( find( i ) )
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++ans;
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}
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return ans;
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}
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int main() {
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int i, j, k;
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memset( adj, 0, sizeof( adj ) );
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memset( result, 0, sizeof( result ) );
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scanf( "%d%d", &n, &k );
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while ( k-- ) {
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scanf( "%d%d", &i, &j );
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adj[ i ][ ++adj[ i ][ 0 ] ] = j;
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}
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printf( "%d\n", maxMatch() );
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return 0;
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}
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posted on 2012-03-30 22:18 coreBugZJ 閱讀(544) 評(píng)論(0) 編輯收藏引用所屬分類: ACM 、Algorithm 、課內(nèi)作業(yè)

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EOJ 2069 Asteroids