hdu 1542 线段树

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Atlantis

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12402    Accepted Submission(s): 5203

Problem Description

There are several ancient Greek texts that contain descriptions of the fabled island Atlantis. Some of these texts even include maps of parts of the island. But unfortunately, these maps describe different regions of Atlantis. Your friend Bill has to know the total area for which maps exist. You (unwisely) volunteered to write a program that calculates this quantity.

 

Input

The input file consists of several test cases. Each test case starts with a line containing a single integer n (1<=n<=100) of available maps. The n following lines describe one map each. Each of these lines contains four numbers x1;y1;x2;y2 (0<=x1
The input file is terminated by a line containing a single 0. Don’t process it.

 

Output

For each test case, your program should output one section. The first line of each section must be “Test case #k”, where k is the number of the test case (starting with 1). The second one must be “Total explored area: a”, where a is the total explored area (i.e. the area of the union of all rectangles in this test case), printed exact to two digits to the right of the decimal point.

Output a blank line after each test case.

 

Sample Input

2 10 10 20 20 15 15 25 25.5 0

 

Sample Output

Test case #1 Total explored area: 180.00

 

Source

Mid-Central European Regional Contest 2000

 

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模拟：

#include 
#include 
using namespace std;
const int maxn=110;


struct LINE
{
    double  x, y_down, y_up;
    int  flag;
    bool operator<(const LINE &a)const
    {
        return  x>1;
       build(2*i, l, mid);
       build(2*i+1, mid, r);
}


double insert(int i, double x, double l, double r, int flag) //flag表示为左边还是右边
{
    if (r<=tree[i].y_down || l>=tree[i].y_up) 
        return 0;
    if (tree[i].flag) 
    {
        if (tree[i].cover > 0) //递归到了叶子节点
        {
             double temp_x = tree[i].x;
             double ans=(x-temp_x)*(tree[i].y_up - tree[i].y_down);
             tree[i].x = x;   //定位上一次的x
             tree[i].cover += flag;
             return ans;
        }
        else 
        {
            tree[i].cover += flag;
            tree[i].x = x;
            return 0;
        }
    }
    double ans1, ans2;
    ans1 = insert(2*i, x, l, r, flag);
    ans2 = insert(2*i+1, x, l, r, flag);
    return ans1+ans2;
}


int main( )
{
   // freopen("d:\\in.txt","r",stdin);
    int  count=0;
    while (scanf("%d", &n)!=EOF&&n)
    {
        index = 1;
        for (int i=1; i<=n; i++)
        {
            scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2);
            y[index] = y1;
            line[index].x = x1;
            line[index].y_down = y1;
            line[index].y_up = y2;
            line[index].flag = 1; //1表示左边


            index++;
            y[index] = y2;
            line[index].x = x2;
            line[index].y_down = y1;
            line[index].y_up = y2;
            line[index].flag = -1; //-1表示右边
            index++;
        }
        sort(&y[1], &y[index]); //把所有的纵坐标按从小到大排序,把1写成了0，WA一次
        sort(&line[1], &line[index]);
        build(1, 1, index-1);
        double ans=0;
        for (i=1;i

树形DP

#include
#include
#include
#include
using namespace std;
struct node
{
    int l;
    int r;
    int cover;
    double len;
};
node tree[2000];
double yy[250];
int n,len;
struct Line
{
    double y_down;
    double y_up;
    double x;
    int cover;
};
Line line[250];
int cmp(Line a,Line b)//根据线段的x左边排序
{
    return a.xr || tree[i].r=l && tree[i].r<=r)
    {
        tree[i].cover+=cover;
        fun(i);
        return;
    }
    updata(2*i,l,r,cover);
    updata(2*i+1,l,r,cover);
    fun(i);
}


int main()
{
    double x1,y1,x2,y2;
    int i,m,a,b,cas=1;
    while(scanf("%d",&n)==1 && n)
    {
        m=0;
        for(i=0;i