Codeforces Round #331 (Div. 2)

水 A - Wilbur and Swimming Pool

自从打完北京区域赛,对矩形有种莫名的恐惧..

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;

int main(void)	{
	int n, x[4], y[4];
	cin >> n;
	for (int i=0; i<n; ++i)	{
		cin >> x[i] >> y[i];
	}
	if (n == 1)	puts ("-1");
	else if (n == 2)	{
		if (x[0] == x[1] || y[0] == y[1])	puts ("-1");
		else	{
			int ans = abs (x[0] - x[1]) * abs (y[0] - y[1]);
			cout << ans << endl;
		}
	}
	else if (n == 3)	{
		int ans = -1;
		if (x[0] == x[1])	{
			ans = abs (y[0] - y[1]) * abs (x[2] - x[0]);
		}
		else if (x[0] == x[2])	{
			ans = abs (y[0] - y[2]) * abs (x[1] - x[0]);
		}
		else if (x[1] == x[2])	{
			ans = abs (y[1] - y[2]) * abs (x[0] - x[1]);
		}
		cout << ans << endl;
	}
	else	{
		int ans = -1;
		if (x[0] == x[1])	{
			ans = abs (y[0] - y[1]) * abs (x[2] - x[0]);
		}
		else if (x[0] == x[2])	{
			ans = abs (y[0] - y[2]) * abs (x[1] - x[0]);
		}
		else if (x[0] == x[3])	{
			ans = abs (y[0] - y[3]) * abs (x[1] - x[0]);
		}
		cout << ans << endl;
	}

	return 0;
}

贪心(水) B - Wilbur and Array

直接扫一边,记录后面的值.忘开long long

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
const int N = 2e5 + 10;
const int INF = 0x3f3f3f3f;
ll a[N];

int main(void)	{
	int n;	cin >> n;
	for (int i=1; i<=n; ++i)	{
		cin >> a[i];
	}
	ll ans = 0;
	ll now = 0, oth = 0;
	for (int i=1; i<=n; ++i)	{
		now = oth;
		if (now == a[i])	continue;
		else if (now < a[i])	{
			ll tmp = a[i] - now;
			oth += tmp;
			ans += tmp;
		}
		else	{
			ll tmp = now - a[i];
			oth -= tmp;
			ans += tmp;
		}
	}
	cout << ans << endl;

	return 0;
}

贪心+排序 C - Wilbur and Points

题意:给出n个点以及n个w[i],问一个点的序列,使得w[i] = p[j].y - p[j].x,并且保证不出现p[i].x <= p[j].x && p[i].y <= p[j].y (i > j)

分析:其实就是到水题,map映射下,用set + pair来存储点,自动从小到大排序,最后再判序列是否满足题意.tourist的判断代码好神奇,一直怀疑自己读错题,应该是能查看数据的吧.

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
map<int, set<pair<int, int> > >mp;
int w[N];
pair<int, int> ans[N];

int main(void)	{
	int n;	scanf ("%d", &n);
	for (int x, y, i=1; i<=n; ++i)	{
		scanf ("%d%d", &x, &y);
		mp[y-x].insert (make_pair (x, y));
	}
	for (int i=1; i<=n; ++i)	{
		scanf ("%d", &w[i]);
	}
	for (int i=1; i<=n; ++i)	{
		if (mp[w[i]].empty ())	{
			puts ("NO");	return 0;
		}
		ans[i] = *(mp[w[i]].begin ());
		mp[w[i]].erase (mp[w[i]].begin ());
	}
	for (int i=2; i<=n; ++i)	{
		int x1 = ans[i].first, x2 = ans[i-1].first;
		int y1 = ans[i].second, y2 = ans[i-1].second;
		if (x1 <= x2 && y1 <= y2)	{
			puts ("NO");	return 0;
		}
	}
	puts ("YES");
	for (int i=1; i<=n; ++i)	{
		int x = ans[i].first;
		int y = ans[i].second;
		printf ("%d %d\n", x, y);
	}

	return 0;
}