hdu 1264 Counting Squares(扫描线)

Counting Squares

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1175    Accepted Submission(s): 593

Problem Description

Your input is a series of rectangles, one per line. Each rectangle is specified as two points(X,Y) that specify the opposite corners of a rectangle. All coordinates will be integers in the range 0 to 100. For example, the line
5 8 7 10
specifies the rectangle who's corners are(5,8),(7,8),(7,10),(5,10).
If drawn on graph paper, that rectangle would cover four squares. Your job is to count the number of unit(i.e.,1*1) squares that are covered by any one of the rectangles given as input. Any square covered by more than one rectangle should only be counted once.

 

Input

The input format is a series of lines, each containing 4 integers. Four -1's are used to separate problems, and four -2's are used to end the last problem. Otherwise, the numbers are the x-ycoordinates of two points that are opposite corners of a rectangle.

 

Output

Your output should be the number of squares covered by each set of rectangles. Each number should be printed on a separate line.

 

Sample Input

5 8 7 106 9 7 86 8 8 11-1 -1 -1 -10 0 100 10050 75 12 9039 42 57 73-2 -2 -2 -2

 

Sample Output

810000
#include <iostream>#include <cstdio>#include <vector>#include <cstring>#include <algorithm>using namespace std;const int maxn = 110;struct tree{    int l , r , mark , lz;    tree(int a = 0 , int b = 0, int c = 0 , int d = 0){        l = a , r = b , mark = c , lz = d;    }}a[4*maxn];struct segment{    int y1 , y2 , x , mark;    segment(int a = 0, int b = 0 , int c = 0 , int m = 0){        y1 = a , y2 = b , x = c , mark = m;    }};vector<segment> seg;int x1 , x2 , yy1 , y2;bool cmp(segment s1 , segment s2){    if(s1.x < s2.x) return true;    return false;}void pushdown(int k){    if(a[k].l+1 < a[k].r && a[k].lz != 0){        a[2*k].mark += a[k].lz;        a[2*k].lz += a[k].lz;        a[1+2*k].mark += a[k].lz;        a[1+2*k].lz += a[k].lz;        a[k].lz = 0;    }}void build(int l , int r , int k){    a[k].l = l;    a[k].r = r;    a[k].mark = 0;    a[k].lz = 0;    if(l+1 < r){        int mid = (l+r)/2;        build(l , mid , 2*k);        build(mid , r , 1+2*k);    }}void add(int l , int r , int k , int c){    if(l <= a[k].l && a[k].r <= r){        a[k].mark += c;        a[k].lz += c;    }else{        int mid = (a[k].l+a[k].r)/2;        if(mid >= r){            add(l , r , 2*k , c);        }else if(l >= mid){            add(l , r , 1+2*k , c);        }else{            add(l , mid , 2*k , c);            add(mid , r , 1+2*k , c);        }    }}int querry(int l , int r , int k){    if(a[k].mark > 0){        return r-l;    }    if(a[k].l+1 >= a[k].r){        if(a[k].mark <= 0) return 0;        return 1;    }    pushdown(k);    int mid = (a[k].l+a[k].r)/2;    if(mid>=r) return querry(l , r , 2*k);    else if(mid<=l) return querry(l , r , 1+2*k);    else return querry(l , mid , 2*k)+querry(mid , r , 1+2*k);}void initial(){    build(0 , 100 , 1);    seg.clear();}void readcase(){    if(yy1 > y2) swap(yy1 , y2);    if(x1 > x2) swap(x1 , x2);    seg.push_back(segment(yy1 , y2 , x1 , 1));    seg.push_back(segment(yy1 , y2 , x2 , -1));    while(cin >> x1 >> yy1 >> x2 >> y2){        if(x1 == -1 || x1 == -2) break;        if(yy1 > y2) swap(yy1 , y2);        if(x1 > x2) swap(x1 , x2);        seg.push_back(segment(yy1 , y2 , x1 , 1));        seg.push_back(segment(yy1 , y2 , x2 , -1));    }}void computing(){    sort(seg.begin() , seg.end() , cmp);    int ans = 0;    for(int i = 0; i < seg.size(); i++){        int h = querry(0 , 100 , 1);        add(seg[i].y1 , seg[i].y2 , 1 , seg[i].mark);        if(h > 0) ans += (seg[i].x-seg[i-1].x)*h;    }    cout << ans << endl;}int main(){    while(cin >> x1 >> yy1 >> x2 >> y2){        initial();        readcase();        computing();        if(x1 == -2) break;    }    return 0;}