线段树+扫描线 HDU-1542 Atlantis

题目大意：给你n个矩形的左下角坐标和右上角坐标，求矩形相交的面积。

题解

扫描线 + 离散化板子

代码如下

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

double dif_x[500];

struct node
{
    double x1,x2,y;
    int flag;
    void init(double l, double r, double h, int key) {
        x1 = l; x2 = r; y = h; flag = key;
    }
}line[500];

bool cmp(node a, node b)
{
    return a.y < b.y;
}

struct node2
{
    int l, r, cnt;
    double len;
}tree[1000];

void build_tree(int id, int l, int r)
{
    tree[id].l = l;
    tree[id].r = r;
    tree[id].cnt = 0;
    tree[id].len = 0;
    if(l == r)
        return;
    int mid = (r + l) >> 1;
    build_tree(id << 1, l, mid);
    build_tree((id << 1) + 1, mid + 1, r);
}

void getlen(int id)
{
    if(tree[id].cnt >= 1) 
        tree[id].len = dif_x[tree[id].r+1] - dif_x[tree[id].l];
    else
        tree[id].len = tree[id<<1].len + tree[(id<<1)|1].len;
}

void update(int id,int l,int r,int v)
{
    if(tree[id].l==l && tree[id].r==r) {
        tree[id].cnt += v;
        getlen(id);
        return;
    }
    int mid = (tree[id].l + tree[id].r) >> 1;
    if(r <= mid)
        update(id << 1,l,r,v);
    else if(l > mid)
        update((id << 1) + 1, l, r, v);
    else {
        update(id << 1, l, mid, v);
        update((id << 1) + 1, mid + 1, r, v);
    }
    getlen(id);
}

int mySearch(double p, int l, int r)
{
    while(l <= r) {
        int mid = (l + r) >> 1;
        if(dif_x[mid] == p)
            return mid;
        if(dif_x[mid] < p)
            l = mid + 1;
        else
            r = mid - 1;
    }
}

int main(void)
{
    int n, noc = 0;
    while (~scanf("%d", &n)) {
        if (n == 0)
            break;
        noc++;
        double x1, y1, x2, y2;
        int line_num = 0;
        for(int i = 0; i < n; i++) {
            scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2);
            line[line_num].init(x1, x2, y1, 1);
            dif_x[line_num++] = x1;
            line[line_num].init(x1, x2, y2, -1);
            dif_x[line_num++] = x2;
        }
        sort(line, line + line_num, cmp);
        sort(dif_x, dif_x + line_num);
        int dif_x_num = unique(dif_x, dif_x + line_num) - dif_x;
        build_tree(1, 0, dif_x_num - 1);
        double ans = 0;
        for(int i = 0; i < line_num - 1; i++) {
            int line_l = mySearch(line[i].x1, 0, dif_x_num - 1);
            int line_r = mySearch(line[i].x2, 0, dif_x_num-1) - 1;
            update(1, line_l, line_r, line[i].flag);
            ans += tree[1].len * (line[i+1].y - line[i].y);
        }
        printf("Test case #%d\n", noc);
        printf("Total explored area: %.2lf\n\n", ans);
    }

    return 0;
}

题解

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