USACO Milking Cows

题意：给出n个区间(a, b) (n<=5000, a,b<=1000000)

将n个区间合并，求最长连续区间和最长区间间隔

先按x升序和y升序排序

tempe保存目前连续区间的右边界

1、i与当前连续区间连续的话必须满足seg[i].e <= tempe，然后更新tempe和最长连续区间值

2、不满足条件，则新开一个连续区间，更新tempe和最长区间间隔值

attention！

1、three bugs= =

2、一开始想的思路错了，往离散化线段树方向想

/*
ID: byluolu1
PROG: milk2
LANG: C++
*/
#include <cstdio>
#include <cstdlib>
#define N 5005

struct Seg
{
    int s, e;
}seg[N];

int cmp(const void *a, const void *b)
{
    struct Seg *x = (struct Seg *)a;
    struct Seg *y = (struct Seg *)b;
    if(x->s == y->s)
        return x->e - y->e;
    return x->s - y->s;
}

int main()
{
    FILE *fin = fopen("milk2.in", "r");
    FILE *fout = fopen("milk2.out", "w");
    int n;
    fscanf(fin, "%d",&n);
    for(int i=0; i<n; i++)
        fscanf(fin, "%d%d",&seg[i].s, &seg[i].e);
    qsort(seg, n, sizeof(struct Seg), cmp);

    int lm = seg[0].e - seg[0].s, lnm = 0, maxlm = 0, maxlnm = 0, tempe = seg[0].e;
    maxlm = lm;// - bug1
    for(int i=1; i<n; i++)
    {
        if(seg[i].s <= tempe) // - bug2
        {
            if(seg[i].e > tempe)
            {
                lm += seg[i].e - tempe;
                tempe = seg[i].e;
            }
            maxlm = lm > maxlm ? lm : maxlm;
        }
        else
        {
            lnm = seg[i].s - tempe;// - bug3
            maxlnm = lnm > maxlnm ? lnm : maxlnm;
            lm = seg[i].e - seg[i].s;
            maxlm = lm > maxlm ? lm : maxlm;
            tempe = seg[i].e;
        }
    }
    fprintf(fout, "%d %d
",maxlm, maxlnm);
    return 0;
}