【bzoj1645】[Usaco2007 Open]City Horizon 城市地平线 离散化+线段树

题目描述

Farmer John has taken his cows on a trip to the city! As the sun sets, the cows gaze at the city horizon and observe the beautiful silhouettes formed by the rectangular buildings. The entire horizon is represented by a number line with N (1 <= N <= 40,000) buildings. Building i's silhouette has a base that spans locations A_i through B_i along the horizon (1 <= A_i < B_i <= 1,000,000,000) and has height H_i (1 <= H_i <= 1,000,000,000). Determine the area, in square units, of the aggregate silhouette formed by all N buildings.

N个矩形块,交求面积并.

输入

* Line 1: A single integer: N

* Lines 2..N+1: Input line i+1 describes building i with three space-separated integers: A_i, B_i, and H_i

输出

* Line 1: The total area, in square units, of the silhouettes formed by all N buildings

样例输入

4
2 5 1
9 10 4
6 8 2
4 6 3

样例输出

16


题解

离散化+线段树(话说silver就已经涉及到这么复杂的线段树了?)

由于点的坐标值过大,需要离散化。

然后将矩形按照高度从小到大排序。

使用线段树,依次将区间内所有高度修改为height。

最后区间查询(其实不算区间查询,因为求的就是sum[1])

然而这道题不是一般的储存点的线段树,而是储存线段的线段树。

线段之间是连着的,这与点不同,所以需要在原来的基础上做一些小修改,详见程序。

应该不怎么难理解。

#include <cstdio>
#include <algorithm>
#define lson l , mid , x << 1
#define rson mid , r , x << 1 | 1        //not "mid+1,r,x<<1|1"
#define N 40001
using namespace std;
struct data
{
    long long a , b , h;
    int left , right;
}k[N];
struct pt
{
    long long num;
    int from;
}p[N << 1];
long long sum[N << 3] , tag[N << 3] , pos[N << 1];
int tot;
bool cmp1(pt a , pt b)
{
    return a.num < b.num;
}
bool cmp2(data a , data b)
{
    return a.h < b.h;
}
void pushup(int x)
{
    sum[x] = sum[x << 1] + sum[x << 1 | 1];
}
void pushdown(int l , int r , int x)
{
    if(tag[x])
    {
        int mid = (l + r) >> 1;
        sum[x << 1] = tag[x] * (pos[mid] - pos[l]);
        sum[x << 1 | 1] = tag[x] * (pos[r] - pos[mid]);        //not "pos[mid+1]"
        tag[x << 1] =  tag[x << 1 | 1] = tag[x];
        tag[x] = 0;
    }
}
void update(int b , int e , long long a , int l , int r , int x)
{
    if(b <= l && r <= e)
    {
        sum[x] = a * (pos[r] - pos[l]);
        tag[x] = a;
        return;
    }
    pushdown(l , r , x);
    int mid = (l + r) >> 1;
    if(b < mid) update(b , e , a , lson);        //not "b<=mid"
    if(e > mid) update(b , e , a , rson);
    pushup(x);
}
int main()
{
    int n , i , rnd = 0;
    scanf("%d" , &n);
    for(i = 0 ; i < n ; i ++ )
    {
        scanf("%lld%lld%lld" , &k[i].a , &k[i].b , &k[i].h);
        p[tot].num = k[i].a;
        p[tot ++ ].from = i;
        p[tot].num = k[i].b;
        p[tot ++ ].from = i;
    }
    sort(p , p + tot , cmp1);
    for(i = 0 ; i < tot ; i ++ )
    {
        if(i == 0 || p[i].num != p[i - 1].num)
            rnd ++ ;
        if(k[p[i].from].a == p[i].num) k[p[i].from].left = rnd;
        if(k[p[i].from].b == p[i].num) k[p[i].from].right = rnd;
        pos[rnd] = p[i].num;
    }
    sort(k , k + n , cmp2);
    for(i = 0 ; i < n ; i ++ )
        update(k[i].left , k[i].right , k[i].h , 1 , rnd , 1);
    printf("%lld\n" , sum[1]);
    return 0;
}
posted @ 2017-01-13 10:26  GXZlegend  阅读(328)  评论(0编辑  收藏  举报