Codeforces Round #229 (Div. 2) A. Inna and Alarm Clock

题目链接：http://codeforces.com/contest/390/problem/A

题意：设置n个闹钟，给你每个闹钟的位置，选择竖列消除或者横行消除（只能用一种），问你关闭所有闹钟最小的操作数；

思路：横着扫一遍，竖着扫一遍，求较小值即可。

代码如下：

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;const int N = 1e5+10;struct node{int x, y;}p[N];int n;bool cmp1(node b, node c){if(b.x != c.x) return b.x < c.x;}bool cmp2(node b, node c){if(b.y != c.y) return b.y < c.y;}int main(){scanf("%d", &n);for(int i = 0; i < n; i++){scanf("%d%d", &p[i].x, &p[i].y);}int ans1 = 1, ans2 = 1;sort(p, p+n, cmp1);for(int i = 1; i < n; i++){if(p[i].x == p[i-1].x)continue;else ans1++;}sort(p, p+n, cmp2);for(int i = 1; i < n; i++){if(p[i].y == p[i-1].y) continue;else ans2++;}int ans = min(ans1, ans2);printf("%d\n", ans);return 0;}