POJ 1151HDU 1542 矩阵面积并

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POJ 1151|HDU 1542 矩阵面积并

题意

  求n个矩形的覆盖面积。

解题思路

  扫描线例题。

代码

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1
const int maxn = 205;

int n;
struct Seg
{
    double l, r, h;
    int d;
    Seg() {}
    Seg(double l, double r, double h, int d): l(l), r(r), h(h), d(d) {}
    bool operator< (const Seg& rhs) const
    {
        return h < rhs.h;
    }
} arr[maxn];

int cnt[maxn << 2]; //根节点维护的是[l, r+1]的区间
double sum[maxn << 2], all[maxn];

void pushup(int l, int r, int rt)
{
    if(cnt[rt]) sum[rt] = all[r + 1] - all[l];
    else if(l == r) sum[rt] = 0; //leaves have no sons
    else sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
}

void update(int L, int R, int v, int l, int r, int rt)
{
    if(L <= l && R >= r)
    {
        cnt[rt] += v;
        pushup(l, r, rt);
        return;
    }
    int m = l + r >> 1;
    if(L <= m) update(L, R, v, lson);
    if(R > m) update(L, R, v, rson);
    pushup(l, r, rt);
}

int main()
{
#ifdef LOCAL
    freopen("in.txt", "r", stdin);
//  freopen("out.txt","w",stdout);
#endif
    int kase = 0;
    while(scanf("%d", &n) == 1 && n)
    {
        for(int i = 1; i <= n; ++i)
        {
            double x1, y1, x2, y2;
            scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2);
            arr[i] = Seg(x1, x2, y1, 1);
            arr[i + n] = Seg(x1, x2, y2, -1);
            all[i] = x1;
            all[i + n] = x2;
        }
        n <<= 1;
        sort(arr + 1, arr + 1 + n);
        sort(all + 1, all + 1 + n);
        int m = unique(all + 1, all + 1 + n) - all - 1;

        memset(cnt, 0, sizeof cnt);
        memset(sum, 0, sizeof sum);

        double ans = 0;
        for(int i = 1; i < n; ++i)
        {
            int l = lower_bound(all + 1, all + 1 + m, arr[i].l) - all;
            int r = lower_bound(all + 1, all + 1 + m, arr[i].r) - all;
            update(l, r - 1, arr[i].d, 1, m-1, 1);
            ans += sum[1] * (arr[i + 1].h - arr[i].h);
        }
        printf("Test case #%d\nTotal explored area: %.2f\n\n", ++kase, ans);
    }
    return 0;
}