Counting Squares
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1072 Accepted Submission(s): 529
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 10
6 9 7 8
6 8 8 11
-1 -1 -1 -1
0 0 100 100
50 75 12 90
39 42 57 73
-2 -2 -2 -2
Sample Output
8
10000
Source
浙江工业大学第四届大学生程序设计竞赛
Recommend
JGShining
漂浮法的简单应用
#include<stdio.h>
int l[5000],r[5000],u[5000],d[5000],N,ans;
void dfs(int ll,int rr,int uu,int dd,int dep)
{
if (dep==N)
{
ans+=(rr-ll)*(uu-dd);
return;
}
if (ll==rr || uu==dd) return;
if (rr<=l[dep+1] || r[dep+1]<=ll || u[dep+1]<=dd || uu<=d[dep+1]) dfs(ll,rr,uu,dd,dep+1);
else
{
if (ll<l[dep+1] && l[dep+1]<rr)
{
dfs(ll,l[dep+1],uu,dd,dep+1);
ll=l[dep+1];
}
if (ll<r[dep+1] && r[dep+1]<rr)
{
dfs(r[dep+1],rr,uu,dd,dep+1);
rr=r[dep+1];
}
if (dd<u[dep+1] && u[dep+1]<uu)
{
dfs(ll,rr,uu,u[dep+1],dep+1);
uu=u[dep+1];
}
if (dd<d[dep+1] && d[dep+1]<uu)
{
dfs(ll,rr,d[dep+1],dd,dep+1);
dd=d[dep+1];
}
}
}
int main()
{
while (true)
{
N=0;
int xx1,xx2,yy1,yy2,i;
while (scanf("%d%d%d%d",&xx1,&yy1,&xx2,&yy2)!=EOF)
{
N++;
if (xx1<xx2)
{
l[N]=xx1;
r[N]=xx2;
}
else
{
l[N]=xx2;
r[N]=xx1;
}
if (yy1<yy2)
{
u[N]=yy2;
d[N]=yy1;
}
else
{
u[N]=yy1;
d[N]=yy2;
}
if (xx1<0)
{
N--;
break;
}
}
ans=0;
for (i=1;i<=N;i++) dfs(l[i],r[i],u[i],d[i],i);
printf("%d\n",ans);
if (xx1==-2) return 0;
}
return 0;
}
