coreBugZJ

此 blog 已棄。

EOJ 2069 Asteroids

/*
2

EOJ 2069 Asteroids
3

----問題描述：
6

Bessie wants to navigate her spaceship through a dangerous asteroid field in the shape of an N x N grid (1 <= N <= 500).
8

The grid contains K asteroids (1 <= K <= 10,000), which are conveniently located at the lattice points of the grid.
9

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.
11

This weapon is quite expensive, so she wishes to use it sparingly.
12

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.
13

----輸入：
16

* Line 1: Two integers N and K, separated by a single space.
18

* 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.
19

----輸出：
22

* Line 1: The integer representing the minimum number of times Bessie must shoot.
23

----樣例輸入：
26

3 4
28

1 1
29

1 3
30

2 2
31

3 2
32

----樣例輸出：
35

2
37

----分析：
40

建立二分圖模型，
42

若第 i 行和第 j 列處存在一個 asteroid ，則 x[i] 與 y[j] 連一條邊，
43

求二分圖最大匹配，使用匈牙利算法。
44

*/
46

#include <stdio.h>
49

#include <string.h>
50

#define L 503
52

int adj[ L ][ L ], n, state[ L ], result[ L ];
54

int find( int i ) {
56

int j, k;
57

for ( j = adj[ i ][ 0 ]; j > 0; --j ) {
58

k = adj[ i ][ j ];
59

if ( state[ k ] == 0 ) {
60

state[ k ] = 1;
61

if ( ( result[ k ] == 0 ) || find( result[ k ] ) ) {
62

result[ k ] = i;
63

return 1;
64

}
65

}
66

}
67

return 0;
68

}
69

int maxMatch() {
71

int ans = 0, i;
72

for ( i = 1; i <= n; ++i ) {
73

memset( state, 0, sizeof( state ) );
74

if ( find( i ) )
75

++ans;
76

}
77

return ans;
78

}
79

int main() {
81

int i, j, k;
82

memset( adj, 0, sizeof( adj ) );
83

memset( result, 0, sizeof( result ) );
84

scanf( "%d%d", &n, &k );
85

while ( k-- ) {
86

scanf( "%d%d", &i, &j );
87

adj[ i ][ ++adj[ i ][ 0 ] ] = j;
88

}
89

printf( "%d\n", maxMatch() );
90

return 0;
91

}
92

posted on 2012-03-30 22:18 coreBugZJ 閱讀(544) 評論(0) 編輯收藏引用所屬分類: ACM 、Algorithm 、課內作業

只有注冊用戶登錄后才能發表評論。


相關文章: 微軟2014實習生及秋令營技術類職位在線測試 TopCoder SRM 593 DIV2 第三題生成全排列的非回溯方法（TopCoder SRM 591 DIV 2） A* 算法求解八數碼問題，POJ 1077 Eight POJ 1067 取石子游戲 POJ 2068 Nim POJ 2975 Nim POJ 3696 The Luckiest number POJ 3604 Professor Ben EOJ 1117 剩余定理

網站導航: 博客園 IT新聞 BlogJava 博問 Chat2DB 管理

青青草原综合久久大伊人导航_色综合久久天天综合_日日噜噜夜夜狠狠久久丁香五月_热久久这里只有精品

coreBugZJ

My Links

Blog Stats

常用鏈接

留言簿(10)

隨筆分類(458)

隨筆檔案(268)

相冊

ACM

AI

LaTeX

安全

編程語言

好有道理

技術

開源

科學

數學

圖形圖像

文化

問題（練習＆有趣）

資源

最新隨筆

搜索

最新評論

閱讀排行榜

評論排行榜

EOJ 2069 Asteroids