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HDU 1264 Counting Squares（线段树求面积的并）
Counting Squares
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1885 Accepted Submission(s): 946
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.
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
题目链接：HDU 1264
连离散化都不用的水题，有一个坑点就是题目给的两个对角线坐标不一定是左下、右上这样一个顺序，或者也可能是副对角线上的点，需要判断一下
代码：
#include <stdio.h>#include <bits/stdc++.h>using namespace std;#define INF 0x3f3f3f3f#define CLR(arr,val) memset(arr,val,sizeof(arr))#define LC(x) (x<<1)#define RC(x) ((x<<1)+1)#define MID(x,y) ((x+y)>>1)typedef pair<int,int> pii;typedef long long LL;const double PI=acos(-1.0);const int N=1e3+7;struct seg{ int l,mid,r; int cnt,len;};struct Line{ int l,r,h,flag; bool operator<(const Line &t)const { return h<t.h; }};seg T[N<<3];Line xline[N<<1];inline void pushup(int k){ if(T[k].cnt>0) T[k].len=T[k].r-T[k].l+1; else { if(T[k].l==T[k].r) T[k].len=0; else T[k].len=T[LC(k)].len+T[RC(k)].len; }}void build(int k,int l,int r){ T[k].l=l; T[k].r=r; T[k].mid=MID(l,r); T[k].len=T[k].cnt=0; if(l==r) return ; build(LC(k),l,T[k].mid); build(RC(k),T[k].mid+1,r); pushup(k);}void update(int k,int l,int r,int flag){ if(l<=T[k].l&&T[k].r<=r) { T[k].cnt+=flag; pushup(k); } else { if(r<=T[k].mid) update(LC(k),l,r,flag); else if(l>T[k].mid) update(RC(k),l,r,flag); else update(LC(k),l,T[k].mid,flag),update(RC(k),T[k].mid+1,r,flag); pushup(k); }}int main(void){ int n,i; int xa,ya,xb,yb; int cnt=0; while (scanf("%d%d%d%d",&xa,&ya,&xb,&yb)) { if(xa==-1&&xb==-1&&ya==-1&&yb==-1) { int ans=0; build(1,0,N); sort(xline,xline+cnt); for (i=0; i<cnt-1; ++i) { update(1,xline[i].l,xline[i].r-1,xline[i].flag); ans=ans+(xline[i+1].h-xline[i].h)*T[1].len; } printf("%d\n",ans); cnt=0; } else if(xa==-2&&xb==-2&&ya==-2&&yb==-2) { int ans=0; build(1,0,N); sort(xline,xline+cnt); for (i=0; i<cnt-1; ++i) { update(1,xline[i].l,xline[i].r-1,xline[i].flag); int dh=(xline[i+1].h-xline[i].h); ans=ans+dh*T[1].len; } printf("%d\n",ans); cnt=0; break; } else { if(xa>xb) swap(xa,xb); if(ya>yb) swap(ya,yb); xline[cnt++]=(Line){xa,xb,ya,1}; xline[cnt++]=(Line){xa,xb,yb,-1}; } } return 0;}HDU 1264 Counting Squares（线段树求面积的并）
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