USACO Section 1.2 Milking Cows

大意：求最长连续区间，最长不连续区间。

/*
ID:g0feng1
LANG:C++
TASK:milk2
*/

#include <iostream>
#include <fstream>
#include <cstdlib>
#include <cstdio>
#include <string>
#include <cstring>
#include <cmath>
#include <map>
#include <algorithm>
using namespace std;

FILE *fin = fopen("milk2.in", "r");
FILE *fout = fopen("milk2.out", "w");

const int maxn = 5010;
const int INF = 0x3f3f3f3f;

int n;
int sa;

struct node
{
	int x, y;
}A[maxn];

int cmpx(node a, node b)
{
	return a.x < b.x;
}

int cmpy(node a, node b)
{
	return a.y < b.y;
}

void read_case()
{
	fscanf(fin, "%d", &n);
	for(int i = 0; i < n; i++) fscanf(fin, "%d%d", &A[i].x, &A[i].y);
	sort(A, A+n, cmpx);
	A[n].x = INF, A[n].y = INF;
}

int cal_max()
{
	int ans;
	int res = 0;
	for(int i = 0; i < n; i++)
	{
		int ans = A[i].y - A[i].x;
		int y = A[i].y;
		for(int j = i+1; j < n; j++)
		{
			if(y >= A[j].x)
			{
				if(y < A[j].y)
				{
					ans += A[j].y-y;
					y = A[j].y;
				}
			}
		}
		res = max(ans, res);
	}
	return res;
}

int cal_min()
{
	sort(A, A+n, cmpy);
	int ans = 0;
	int x = A[n-1].x;
	for(int i = n-2; i >= 0; i--)
	{
		if(x > A[i].y)
		{
			int t = x - A[i].y;
			ans = max(ans, t);
		}
		else if(A[i].x < x) x = A[i].x;
	}
	return ans;
}

void solve()
{
	read_case();
	int maxv = cal_max(), minv = cal_min();
	fprintf(fout, "%d %d\n", maxv, minv);
}

int main()
{
	solve();
	return 0;
}