求矩形并的周长 poj 1177 picture

#include
#include
#include

using namespace std;
struct node
{
    int st, ed, m, lbd, rbd;
    int sequence_line, count;
} ST[40005];
void build(int st, int ed, int v)
{
    ST[v].st = st;
    ST[v].ed = ed;
    ST[v].lbd = ST[v].rbd = ST[v].m = 0;
    ST[v].sequence_line = ST[v].count = 0;
    if(ed - st > 1)
    {
        int mid = (st+ed)/2;
        build(st, mid, 2*v+1);
        build(mid, ed, 2*v+2);
    }
}
inline void UpData(int v)
{
    if(ST[v].count>0)
    {
        ST[v].m = ST[v].ed - ST[v].st;
        ST[v].lbd = ST[v].rbd = 1;
        ST[v].sequence_line = 1;
        return ;
    }
    if(ST[v].ed - ST[v].st == 1)
    {
        ST[v].m = 0;
        ST[v].lbd = ST[v].rbd = 0;
        ST[v].sequence_line = 0;
    }
    else
    {
        int left = 2*v+1, right = 2*v+2;
        ST[v].m = ST[left].m + ST[right].m;
        ST[v].sequence_line = ST[left].sequence_line + ST[right].sequence_line - (ST[left].rbd&ST[right].lbd);
        ST[v].lbd = ST[left].lbd;
        ST[v].rbd = ST[right].rbd;
    }
}
void insert(int st, int ed, int v)
{
    if(st <= ST[v].st && ed >= ST[v].ed)
    {
        ST[v].count++;
        UpData(v);
        return ;
    }
    int mid = (ST[v].st + ST[v].ed)/2;
    if(st < mid)insert(st, ed, 2*v+1);
    if(ed > mid)insert(st, ed, 2*v+2);
    UpData(v);
}
void Delete(int st, int ed, int v)
{
    if(st <= ST[v].st && ed >= ST[v].ed)
    {
        ST[v].count--;
        UpData(v);
        return ;
    }
    int mid = (ST[v].st + ST[v].ed)/2;
    if(st < mid) Delete(st, ed, 2*v+1);
    if(ed > mid)Delete(st, ed, 2*v+2);
    UpData(v);
}
struct line
{
    int x, y1, y2;//y1 < y2
    bool d;//d = true 表示线段为 矩形的左边， 反之为 右边.
}a[10003];
bool cmp(line t1, line t2)
{
    return t1.x }
void cal_C(int n);
int main()
{
    freopen("input.txt", "r", stdin);
   int n, x1, x2, y1, y2, suby, upy, i, j;
   while(scanf("%d", &n) != EOF)
   {
       j = 0;
       suby = 10000;
       upy = -10000;
       j = 0;
       for(i = 0; i < n; i++)
       {
           scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
           a[j].x = x1; a[j].y1 = y1; a[j].y2 = y2;
           a[j].d = 1;
           j++;
           a[j].x = x2; a[j].y1 = y1; a[j].y2 = y2;
           a[j].d = 0;
           j++;
           if(suby > y1)  suby = y1;
           if(upy < y2)   upy = y2;
       }
       sort(a, a+j, cmp);
       build(suby, upy, 0);
       cal_C(j);
   }
    return 0;
}
void cal_C(int n)
{
    int i, t2, sum = 0;
    t2 = 0;
    a[n] = a[n-1];
    for(i = 0; i < n; i++)
    {
        if(a[i].d)
            insert(a[i].y1, a[i].y2, 0);
        else
            Delete(a[i].y1, a[i].y2, 0);
        sum += ST[0].sequence_line * (a[i+1].x - a[i].x) * 2;
        sum += abs(ST[0].m - t2);
        t2 = ST[0].m;
    }
    printf("%d/n", sum);
}