Hdu 1828(线段树+扫描线)

Hdu 1828
线段树扫描线求周长并,看这篇好文学习:
http://blog.csdn.net/tomorrowtodie/article/details/52048323

#include<cstdio>  
#include<algorithm>  
#include<cstring>  
#include<vector>
#define ms(i,j) memset(i,j, sizeof i);
using namespace std;
const int MAXR = 10000 + 5;
const int MAXN = 5000 + 5;
const int Zinf = 100000;
struct xian
{
    int l, r, h;
    int f;
    bool operator<(const xian &b) const
    {
        return h<b.h;
    }
}a[MAXN*2];
int abss(int x) {return x<0 ? -x : x;}
int n;
#define lc (o*2)
#define rc (o*2+1)
#define M ((l+r)>>1)
struct st
{
    int col[MAXR*4*4];//覆盖次数 
    int num[MAXR*4*4];//当前区间分离线段条数 
    int ls[MAXR*4*4], rs[MAXR*4*4];//左右结点是否被遮挡 
    int sum[MAXR*4*4];//计算时的有效区间长 
    void build(int o, int l ,int r)
    {
        col[o] = 0; num[o] = 0; ls[o] = rs[o] = 0; sum[o] = 0;
        if (l==r) return ;
        build(lc, l, M);
        build(rc, M+1, r);
    }
    void pushup(int o, int l, int r)
    {
        if (col[o])
        {
            num[o] = 1;
            ls[o] = rs[o] = 1;
            sum[o] = r-l+1;
        } else if (l==r)
        {
            num[o] = 0;
            ls[o] = rs[o] = 0;
            sum[o] = 0;
        } else
        {
            ls[o] = ls[lc];
            rs[o] = rs[rc];
            sum[o] = sum[lc] + sum[rc];
            num[o] = num[lc] + num[rc] - (rs[lc]&&ls[rc]);
        }
    }
    void update(int o, int l, int r, int x, int y, int c)
    {
        if (x<=l&&r<=y)
        {
            col[o] += c;
            pushup(o,l,r);
            return ;
        }
        if (x<=M) update(lc, l, M, x, y, c);
        if (M<y) update(rc, M+1, r, x, y, c);
        pushup(o,l,r);
    }
}tree;
int main()  
{  
    while (scanf("%d", &n)==1)
    {
        int minr = Zinf, maxr = -Zinf;
        for (int i=0;i<n;i++)
        {
            int x1, y1, x2, y2;
            scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
            minr = min(minr, min(x1,x2));
            maxr = max(maxr, max(x1,x2));
            a[i*2].l = a[i*2+1].l = x1;
            a[i*2].r = a[i*2+1].r = x2;
            a[i*2].h = y1; a[i*2+1].h = y2;
            a[i*2].f = -1; a[i*2+1].f = 1;
        }
        sort(a, a+2*n);//数小,不用离散化。
        tree.build(1, minr, maxr-1);
        int last = 0;
        int ans = 0;
        for (int i=0;i<2*n;i++)
        {
            tree.update(1,minr,maxr-1,a[i].l, a[i].r-1, a[i].f);
            ans += abss(tree.sum[1]-last);
            ans += (a[i+1].h-a[i].h)*2*tree.num[1];
            last = tree.sum[1];
        }
        printf("%d\n",ans);  
    }
    return 0;  
}
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