Count The colors_zoj1610_线段树

Painting some colored segments on a line, some previously painted segments may be covered by some the subsequent ones.

Your task is counting the segments of different colors you can see at last.

 

Input

 

The first line of each data set contains exactly one integer n, 1 <= n <= 8000, equal to the number of colored segments. 

Each of the following n lines consists of exactly 3 nonnegative integers separated by single spaces: 

x1 x2 c 

x1 and x2 indicate the left endpoint and right endpoint of the segment, c indicates the color of the segment. 

All the numbers are in the range [0, 8000], and they are all integers. 

Input may contain several data set, process to the end of file. 

 

Output 

Each line of the output should contain a color index that can be seen from the top, following the count of the segments of this color, they should be printed according to the color index. 

If some color can't be seen, you shouldn't print it. 

Print a blank line after every dataset. 

 

Sample Input 

5

0 4 4

0 3 1

3 4 2

0 2 2

0 2 3

4

0 1 1

3 4 1

1 3 2

1 3 1

6

0 1 0

1 2 1

2 3 1

1 2 0

2 3 0

1 2 1

  

Sample Output 

1 1

2 1

3 1

 

1 1

 

0 2

1 1

 

翻完一堆英语，讲讲题目大意 

已知有n条不同颜色的线段且新线段可以覆盖在旧线段上。求各颜色的线段各出现多少次

思路：

颜色覆盖！线段树！color记录当前节点的颜色，最后统计输出。比较简单而且和例3类似就不细说了。

源代码/pas:

typetree=record  l,r,c:longint;end;var  tmp,n:longint;  t,g:array[0..40000]of tree;  ans:array[0..10000]of longint;procedure insert(f,x,y,color:longint);var  mid:longint;begin  mid:=(t[f].l+t[f].r)div 2;  if (t[f].c<>-1)and(t[f].l+1<>t[f].r) then  begin    t[f*2].c:=t[f].c;    t[f*2+1].c:=t[f].c;  end;  if t[f].l+1<>t[f].r then t[f].c:=-1;  if (x=t[f].l)and(y=t[f].r) then  t[f].c:=color  else  begin    if (x>=mid) then    insert(f*2+1,x,y,color)    else    if (y<=mid) then    insert(f*2,x,y,color)    else    begin      insert(f*2,x,mid,color);      insert(f*2+1,mid,y,color);    end;  end;end;procedure count(f:longint);var  mid:Longint;begin  if t[f].c<>-1 then  begin    if tmp<>t[f].c then    begin      inc(ans[t[f].c]);      tmp:=t[f].c;    end;    exit;  end;  if t[f].l+1<>t[f].r then  begin    count(f*2);    count(f*2+1);  end  else  tmp:=-1;end;procedure build(f:longint);var  mid:longint;begin  if t[f].r-t[f].l=1 then exit;  mid:=(t[f].l+t[f].r)div 2;  t[f*2].l:=t[f].l;  t[f*2].c:=t[f].c;  t[f*2].r:=mid;  t[f*2+1].l:=mid;  t[f*2+1].r:=t[f].r;  t[f*2+1].c:=t[f].c;  build(f*2);  build(f*2+1);end;procedure init;var  n,m,i,j,x,y,z:longint;begin  t[1].l:=0;  t[1].r:=8000;  t[1].c:=-1;  build(1);  g:=t;  while not eof do  begin    readln(m);    for i:=1 to m do    begin      readln(x,y,z);      insert(1,x,y,z);    end;    tmp:=-1;    count(1);    for i:=0 to 8001 do    if ans[i]<>0 then    writeln(i,' ',ans[i]);    writeln;    t:=g;    fillchar(ans,sizeof(ans),0);  end;end;begin  init;end.