Asteroids
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 6748 | Accepted: 3530 |
Description
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
* 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
Sample Input
3 4
1 1
1 3
2 2
3 2
Sample Output
2
Hint
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).
解题思路:
求最小的边覆盖所有点,讲边和点对换所得的图G‘中,即求G’的点最小覆盖。
G中edge[r,c]=G'的点r,c。可看为二分图。
在二分图中,最小覆盖集=最大匹配数
匈牙利算法简单解决。
#include<iostream> #include<cstring> using namespace std; bool visit[501]; bool map[501][501]; int result[501]; int N; bool find(int startNodes) { for(int i=1; i<=N; i++) { if(map[startNodes][i] && !visit[i]) //startNode unvisited neighbor nodes { visit[i]=true;//try to in augment path if(result[i]==0 || find(result[i])) //i is not in matched graph || has a augment path { result[i]=startNodes; //matched return true; } } } return false; } int main() { int K; int r,c; int edge; memset(map,0,sizeof(map)); memset(result,0,sizeof(result)); edge=0; cin>>N>>K; for(int i=1; i<=K; i++) { cin>>r>>c; map[r][c]=true; } for(int i=1; i<=N; i++) { memset(visit,0,sizeof(visit)); if(find(i)) edge++; } cout<<edge<<endl; return 0; }