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
Author: Standlove
Source: ZOJ Monthly, May 2003
乱的代码
1 #include <bits/stdc++.h> 2 const int MAXN=16000; 3 using namespace std; 4 struct node{ 5 int l,r; 6 int c;//color 7 }tree[MAXN]; 8 int n; 9 int color[MAXN]; 10 //每个节点左右端点的颜色 11 int leftC[MAXN]; 12 int rightC[MAXN]; 13 14 void print(){ 15 for(int i=1;i<=7;i++) cout<<tree[i].l<<" "<<tree[i].r<<" "<<tree[i].c<<endl; 16 } 17 18 void build(int p,int l,int r){ 19 tree[p].l=l; 20 tree[p].r=r; 21 tree[p].c=0; 22 //拓展节点 23 if(l+1<r){ 24 int mid=(l+r)/2; 25 build(2*p,l,mid); 26 build(2*p+1,mid,r); 27 } 28 } 29 30 void insert(int p,int l,int r,int c){//这里的l和r代表线段的左端点和右端点 31 //颜色不同才有涂色的必要 32 if(tree[p].c!=c){ 33 cout<<c<<endl; 34 int mid=(tree[p].l+tree[p].r)/2;//日常拆区间 35 if(l==tree[p].l&&r==tree[p].r){ 36 37 tree[p].c=c; 38 } 39 else if(tree[p].l+1==tree[p].r) return;//树的叶子节点 40 //区间不合适，我这是要拆区间的节奏，肯定会给加一种颜色造成混色 41 //可能没有交集么？？？不可能没有交集，不在左，毕在右，第一轮就给你分好了，所以肯定要进行母树颜色往下分的操作 42 else if(tree[p].c>=0) { 43 //母树的颜色往下分 44 tree[2*p].c=tree[p].c; 45 tree[2*p+1].c=tree[p].c; 46 tree[p].c=-1; 47 if(r<mid) insert(2*p,l,r,c); 48 else if(l>mid) insert(2*p+1,l,r,c); 49 else{//把线段拆了 50 insert(2*p,l,mid,c); 51 insert(2*p+1,mid,r,c); 52 } 53 } 54 55 56 } 57 } 58 59 void count1(int p,int lc,int rc){ 60 cout<<"p:"<<p<<" lc:"<<lc<<" rc"<<rc<<endl; 61 int tl=0,tr=0; 62 //单一颜色才计数 63 if(tree[p].c>=0){ 64 cout<<"1"<<endl; 65 cout<<"tree[p].c:"<<tree[p].c<<endl; 66 lc=tree[p].c; 67 rc=tree[p].c; 68 color[tree[p].c]++;//这种颜色的线段数加1 69 cout<<"p:"<<p<<" lc:"<<lc<<" rc"<<rc<<endl; 70 } 71 else if(tree[p].r-tree[p].l>1){ 72 count1(2*p,lc,tl); 73 count1(2*p+1,tr,rc); 74 } 75 //每一轮做完就看p的左右孩子是否同色或者部分同色 76 if(tree[2*p].r==tree[2*p+1].l){ 77 color[tree[p].c]--; 78 } 79 } 80 81 void count2(int p){ 82 //单一颜色才计数 83 if(tree[p].c>=0){ 84 leftC[p]=tree[p].c; 85 rightC[p]=tree[p].c; 86 color[tree[p].c]++;//这种颜色的线段数加1 87 return; 88 } 89 else if(tree[p].r-tree[p].l>1){ 90 count2(2*p); 91 leftC[p]=leftC[2*p]; 92 count2(2*p+1); 93 rightC[p]=rightC[2*p+1]; 94 } 95 //每一轮做完就看p的左右孩子是否同色或者部分同色 96 if(rightC[2*p]==leftC[2*p+1]){ 97 color[rightC[2*p]]--; 98 } 99 } 100 101 102 void printColor(){ 103 for(int i=1;i<=5;i++) cout<<color[i]<<" "; cout<<endl; 104 } 105 106 int main(){ 107 build(1,1,5); 108 insert(1,1,4,4); 109 insert(1,4,5,5); 110 count2(1); 111 print(); 112 printColor(); 113 return 0; 114 }
题解/solution：
这题大体和我写的解题报告（PPT1 例2）相同，只是在统计算法上要改一下。Look down！
图，come out. 
用ls表示上一个颜色，如果当前颜色与ls不同，那给这个颜色加一。例：
ls颜色为空，而一区间为红，红加一，ls=红。
ls颜色为红，而三区间为蓝，蓝加一，ls=蓝。
以此类推......
1 type 2 arr=record 3 l,r:longint; 4 color:longint; 5 end; 6 var 7 tree:array [0..32001] of arr; 8 ans:array [0..8001] of longint; 9 n,m,ls,max_co:longint; 10 procedure cre(p,b,e:longint); 11 var 12 m:longint; 13 begin 14 with tree[p] do 15 begin 16 l:=b; r:=e; color:=-1; 17 if e-b=1 then exit; 18 m:=(b+e) div 2; 19 cre(p*2,b,m); 20 cre(p*2+1,m,e); 21 end; 22 end; 23 24 procedure ins(p,a,b,c:longint); 25 var 26 m:longint; 27 begin 28 with tree[p] do 29 begin 30 if color<>c then 31 begin 32 m:=(l+r) div 2; 33 if (a=l) and (b=r) then color:=c else 34 begin 35 if color>=0 then 36 begin 37 tree[p*2].color:=color; 38 tree[p*2+1].color:=color; 39 end; 40 color:=-2; 41 if b<=m then ins(p*2,a,b,c) else 42 if a>=m then ins(p*2+1,a,b,c) else 43 begin 44 ins(p*2,a,m,c); 45 ins(p*2+1,m,b,c); 46 end; 47 end; 48 end; 49 end; 50 end; 51 52 procedure count(p:longint); 53 begin 54 with tree[p] do 55 begin 56 if color>=0 then 57 begin 58 if color<>ls then 59 begin 60 inc(ans[color]); 61 ls:=color; 62 end; 63 exit; 64 end; 65 if color=-1 then 66 begin 67 ls:=color; 68 exit; 69 end; 70 count(p*2); 71 count(p*2+1); 72 end; 73 end; 74 75 procedure main; 76 var 77 i,x,y,z:longint; 78 begin 79 m:=8000; 80 while not eof do 81 begin 82 fillchar(ans,sizeof(ans),0); 83 readln(n); 84 ls:=-1; max_co:=-(maxlongint div 3); 85 cre(1,0,m); 86 for i:=1 to n do 87 begin 88 readln(x,y,z); 89 ins(1,x,y,z); 90 if z>max_co then max_co:=z; 91 end; 92 count(1); 93 for i:=0 to max_co do 94 if ans[i]>0 then 95 writeln(i,' ',ans[i]); 96 writeln; 97 end; 98 end; 99 100 begin 101 main; 102 end.