CF 390A:Inna and Alarm Clock
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简单的说100*100的矩阵,上面n个点中有闹钟,Inno只能选择以横向或纵向的方式关掉一行上任意多的闹钟,问需要多少步。
一开始题目没读懂,读懂了水题一道。
#include <cstdio>#include <iostream>using namespace std ;int xx[110] = {0} ;int yy[110] = {0} ;int Min(int x , int y) { return (x < y)? x : y ;}int main() { //freopen("in.txt" , "r" , stdin) ; int n , x = 0 , y = 0 ; int ans ; cin >> n ; for (int i = 1 ; i <= n ; i ++) { int a , b ; cin >> a >> b ; xx[a] ++ ; yy[b] ++ ; } for (int i = 0 ; i <= 100 ; i ++) { if (xx[i]) x ++ ; if (yy[i]) y ++ ; } ans = Min(x , y) ; cout << ans << endl ; return 0 ;}
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