Codeforces Round #229 (Div. 二) A. Inna and Alarm Clock
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; }
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