线段树 1828（扫描+离散化）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1828

具体看 http://www.cnblogs.com/shuaiwhu/archive/2012/04/22/2464876.html 的讲解

我一开始连讲解都听不懂  直接看代码才慢慢明白的

看了一遍他的代码 自己手打了一遍  所以几乎一样~

#include <stdio.h>
#include <string.h>
#include <algorithm>
#define MAX 5010
#define ABS(X) ((X)>=0?(X):(-(X)))
using namespace std;
struct node
{
    int x, y1, y2, flag;
}scan[MAX<<1];

int y[MAX<<1];
struct Tree
{
    int l, r, cnt, line, lbd, rbd, sum;
}tree[MAX<<4];
int cmp1( const void *a, const void *b )
{
    node *c = (node *)a ;
    node *d = (node *)b ;
    if( c->x == d->x )
        return c->flag <= d->flag ?1:-1;   //先处理输入再处理输出，两条竖线重叠时才不计算ans内
    else
        return c->x >= d->x ?1:-1;
}
int cmp2( const void *a, const void *b )
{
    return *(int *)a >= *(int *)b ?1: -1;
}
void build( int l, int r, int pos )
{
    tree[pos].l = l, tree[pos].r = r;
    tree[pos].cnt = tree[pos].line = tree[pos].lbd = tree[pos].rbd =tree[pos].sum = 0;
    if( l + 1 == r )  //叶节点不需要再分 再中分就再跳不出去
        return ;
    else
    {
        int mid = (tree[pos].l + tree[pos].r)/2 ;
        build( l, mid, pos*2 );
        build(mid, r, pos*2+1 );
    }
}
void update_sum( int pos )
{
    if(tree[pos].cnt > 0)
        tree[pos].sum = y[tree[pos].r] - y[tree[pos].l];
//    else if( tree[pos].l == tree[pos].r )
//        return ;
    else
    {
        tree[pos].sum = tree[pos*2].sum + tree[pos*2+1].sum ;
    }
}
void update_line( int pos )
{
    if( tree[pos].cnt > 0 )
    {
        tree[pos].lbd = tree[pos].rbd = 1;
        tree[pos].line = 1;    
    }
    else
    {
        tree[pos].lbd = tree[pos*2].lbd;
        tree[pos].rbd = tree[pos*2+1].rbd;
        //注意两个子节点连续的情况，即tree[pos*2].rbd=tree[pos*2+1].lbd=1
        tree[pos].line = tree[pos*2].line + tree[pos*2+1].line - rbd*tree[pos*2+1].lbd ;
       
    }
}
void update( int l, int r, int pos, int mark )
{
    int mid = (tree[pos].l + tree[pos].r)/2 ;
    if( y[tree[pos].l] == l && y[tree[pos].r] == r )  //y储存的是真正的纵坐标， tree[pos].l,tree[pos].r储存的只是映射的 
        mark == 1 ?tree[pos].cnt ++ : tree[pos].cnt --;
    else if( tree[pos].l + 1 == tree[pos].r )  //是叶节点且不是匹配的l r直接返回 不update_sum,update_line。 注意这里else if 的意思 是前面两个条件缺一不可
        return ;    
    else if( r <= y[mid] )
        update(l, r, pos*2, mark);
    else if( l >= y[mid] )
        update(l, r, pos*2+1,mark);
    else
    {
        update(l, y[mid], pos*2, mark);
        update(y[mid], r, pos*2+1, mark);
    }
    update_sum( pos );
    update_line( pos );    
}
int main()
{
    int i, x1, y1, x2, y2, n, ans;
    while( scanf("%d", &n) !=EOF )
    {
        ans = 0, i = -1;
        while( n-- )
        {
            scanf("%d %d %d %d", &x1, &y1, &x2, &y2);
            scan[++i].x = x1;
            scan[i].y1 = y1;
            scan[i].y2 = y2;
            scan[i].flag = 1;
            y[i+1] = y1;
            scan[++i].x = x2;
            scan[i].y1 = y1;
            scan[i].y2 = y2;
            scan[i].flag = 0;
            y[i+1] = y2;
            
        }
        int count = i+1;
        qsort( scan, count, sizeof(scan[0]), cmp1 );
        qsort( y+1, count, sizeof(y[1]), cmp2 );
        int unique_cnt = unique( y+1, y+1+count ) - y - 1;  //unique 所有的纵坐标
        build( 1, unique_cnt, 1 );
        int pre_sum, pre_line ;
        for( int j=0; j<count; j++ )
        {
            update( scan[j].y1, scan[j].y2, 1, scan[j].flag );
            if(j >= 1)
            {
                ans += 2*pre_line*( scan[j].x - scan[j-1].x );
                ans += ABS(tree[1].sum - pre_sum);  //跟之前的对比！！很巧妙！！
            }
            else
                ans += tree[1].sum ;
            pre_sum = tree[1].sum ;
            pre_line = tree[1].line ;
        }
        printf("%d\n", ans);
    }
    return 0;
}