Description Mom and dad have a problem: their child, Reza, never puts his toys away when he is finished playing with them. They gave Reza a rectangular box to put his toys in. Unfortunately, Reza is rebellious and obeys his parents by simply throwing his toys into the box. All the toys get mixed up, and it is impossible for Reza to find his favorite toys anymore.
Reza's parents came up with the following idea. They put cardboard partitions into the box. Even if Reza keeps throwing his toys into the box, at least toys that get thrown into different partitions stay separate. The box looks like this from the top:  We want for each positive integer t, such that there exists a partition with t toys, determine how many partitions have t, toys. Input The input consists of a number of cases. The first
line consists of six integers n, m, x1, y1, x2, y2. The number of cardboards to
form the partitions is n (0 < n <= 1000) and the number of toys is given
in m (0 < m <= 1000). The coordinates of the upper-left corner and the
lower-right corner of the box are (x1, y1) and (x2, y2), respectively. The
following n lines each consists of two integers Ui Li, indicating that the ends
of the ith cardboard is at the coordinates (Ui, y1) and (Li, y2). You may assume
that the cardboards do not intersect with each other. The next m lines each
consists of two integers Xi Yi specifying where the ith toy has landed in the
box. You may assume that no toy will land on a cardboard.
A line consisting of a single 0 terminates the input. Output For each box, first provide a header stating "Box" on
a line of its own. After that, there will be one line of output per count (t
> 0) of toys in a partition. The value t will be followed by a colon and a
space, followed the number of partitions containing t toys. Output will be
sorted in ascending order of t for each box.
Sample Input 4 10 0 10 100 0 20 20 80 80 60 60 40 40 5 10 15 10 95 10 25 10 65 10 75 10 35 10 45 10 55 10 85 10 5 6 0 10 60 0 4 3 15 30 3 1 6 8 10 10 2 1 2 8 1 5 5 5 40 10 7 9 0 Sample Output Box 2: 5 Box 1: 4 2: 1 Source |
[Submit] [Go Back] [Status] [Discuss]
这题和上上道题的思路和算法都是一样的，
只不过需要改一下输出格式，
而且这道题不保证读入数据是有序的，
所以需要排一下序
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 using namespace std; 6 const int MAXN=300001; 7 inline void read(int &n) 8 { 9 char c=getchar();bool flag=0;n=0; 10 while(c<'0'||c>'9') c=='-'?flag=1,c=getchar():c=getchar(); 11 while(c>='0'&&c<='9') n=n*10+(c-48),c=getchar();if(flag==1)n=-n; 12 } 13 int n,m,x1,yy1,x2,y2; 14 struct node 15 { 16 int x,y; 17 }point[MAXN]; 18 struct Line 19 { 20 node a,b; 21 }line[MAXN]; 22 int cross(node a,node b,node c) 23 {return (c.x-a.x)*(c.y-b.y)-(c.x-b.x)*(c.y-a.y);} 24 int ans[MAXN]; 25 int happen[MAXN]; 26 void calc(int num) 27 { 28 int l=0,r=n-1; 29 while(l<r) 30 { 31 int mid=(l+r)>>1; 32 if(cross(point[num],line[mid].a,line[mid].b)>0) l=mid+1; 33 else r=mid; 34 } 35 if(cross(point[num],line[l].a,line[l].b)<0) ans[l]++; 36 else ans[l+1]++; 37 } 38 bool cmp(const Line &p,const Line &q) 39 { return min(p.a.x,p.b.x)<min(q.a.x,q.b.x)||((min(p.a.x,p.b.x)==min(q.a.x,q.b.x))&&(max(p.a.x,p.b.x)<max(q.a.x,q.b.x))); } 40 int main() 41 { 42 while(scanf("%d",&n)&&n) 43 { 44 scanf("%d%d%d%d%d",&m,&x1,&yy1,&x2,&y2); 45 for(int i=0;i<n;i++) 46 { 47 int p,q;read(p);read(q); 48 line[i].a.x=p; line[i].a.y=yy1; 49 line[i].b.x=q; line[i].b.y=y2; 50 } 51 sort(line,line+n,cmp); 52 memset(ans,0,sizeof(ans)); memset(happen,0,sizeof(happen)); 53 for(int i=0;i<m;i++) 54 { 55 int p,q;read(p);read(q); 56 point[i].x=p; point[i].y=q; 57 } 58 for(int i=0;i<m;i++) calc(i); 59 for(int i=0;i<=n;i++) happen[ans[i]]++; 60 printf("Box "); 61 for(int i=1;i<=n;i++) 62 if(happen[i]>0) printf("%d: %d ",i,happen[i]); 63 64 } 65 return 0; 66 }