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poj 1258 -- Asteroids

2024-07-11 20:19:50   220人阅读

Asteroids
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 14426 Accepted: 7851

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 41 11 32 23 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).
 
第一道最大匹配题。行与列匹配。其实匹配到的点就是要消除的那个点的所在行或列。 用到匈牙利算法。
 
 1 /*====================================================================== 2  *           Author :   kevin 3  *         Filename :   Asteroids.cpp 4  *       Creat time :   2014-07-13 08:51 5  *      Description : 6  ========================================================================*/ 7 #include <iostream> 8 #include <algorithm> 9 #include <cstdio>10 #include <cstring>11 #include <queue>12 #include <cmath>13 #define clr(a,b) memset(a,b,sizeof(a))14 #define M 55015 16 using namespace std;17 18 int g[M][M];19 int n,k,used[M],linker[M];20 bool DFS(int u)21 {22     for(int v = 0; v < n; v++){23         if(g[u][v] && !used[v]){24             used[v] = 1;25             if(linker[v] == -1 || DFS(linker[v])){26                 linker[v] = u;27                 return true;28             }29         }30     }31     return false;32 }33 int MaxMatch()34 {35     int res = 0;36     clr(linker,-1);37     for(int u = 0; u < n; u++){38         clr(used,0);39         if(DFS(u)) res++;40     }41     return res;42 }43 int main(int argc,char *argv[])44 {45     while(scanf("%d%d",&n,&k)!=EOF){46         int a,b;47         clr(g,0);48         for(int i = 0; i < k; i++){49             scanf("%d%d",&a,&b);50             g[a-1][b-1] = 1;51         }52         int ans = MaxMatch();53         printf("%d\n",ans);54     }55     return 0;56 }
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