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

Sample Output

1 1
2 1
3 1

1 1

0 2
1 1

題意有點坑，是n個染色的片段，不是染點，比如 0,3，是染的1,2,3片段。

然後就是建樹建8000，因為是n個染色的片段，最後查詢，直接單點查詢。

維護的是片段的標號，不是點號，寫的時候，要l+1，我寫的l++，智障了。。。

代碼:

  1 //線段樹(區間染色+統計間斷區間數量)
  2 #include<iostream>
  3 #include<cstdio>
  4 #include<cstring>
  5 #include<algorithm>
  6 #include<bitset>
  7 #include<cassert>
  8 #include<cctype>
  9 #include<cmath>
 10 #include<cstdlib>
 11 #include<ctime>
 12 #include<deque>
 13 #include<iomanip>
 14 #include<list>
 15 #include<map>
 16 #include<queue>
 17 #include<set>
 18 #include<stack>
 19 #include<vector>
 20 using namespace std;
 21 typedef long long ll;
 22 typedef long double ld;
 23 typedef pair<int,int> pii;
 24 
 25 const double PI=acos(-1.0);
 26 const double eps=1e-6;
 27 const ll mod=1e9+7;
 28 const int inf=0x3f3f3f3f;
 29 const int maxn=1e5+10;
 30 const int maxm=100+10;
 31 #define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
 32 #define lson l,m,rt<<1
 33 #define rson m+1,r,rt<<1|1
 34 
 35 int now;
 36 int col[maxn<<2],num[maxn];
 37 
 38 void pushdown(int rt)
 39 {
 40     if(col[rt]!=-1){
 41         col[rt<<1]=col[rt<<1|1]=col[rt];
 42         col[rt]=-1;
 43     }
 44 }
 45 
 46 void build(int l,int r,int rt)
 47 {
 48     col[rt]=-1;
 49     if(l==r){
 50         return ;
 51     }
 52 
 53     int m=(l+r)>>1;
 54     build(lson);
 55     build(rson);
 56 }
 57 
 58 void update(int L,int R,int c,int l,int r,int rt)
 59 {
 60     if(r<L||l>R) return ;
 61     if(L<=l&&r<=R){
 62         col[rt]=c;
 63         return ;
 64     }
 65 
 66     pushdown(rt);
 67     int m=(l+r)>>1;
 68     if(L<=m) update(L,R,c,lson);
 69     if(R> m) update(L,R,c,rson);
 70 }
 71 
 72 void query(int l,int r,int rt)
 73 {
 74     if(l==r){
 75         if(col[rt]!=-1&&col[rt]!=now) num[col[rt]]++;
 76         now=col[rt];
 77         return ;
 78     }
 79 
 80     pushdown(rt);
 81     int m=(l+r)>>1;
 82     query(lson);
 83     query(rson);
 84 }
 85 
 86 int main()
 87 {
 88     int n;
 89     while(~scanf("%d",&n)){
 90         memset(num,0,sizeof(num));
 91         build(1,8000,1);
 92         for(int i=1;i<=n;i++){
 93             int l,r,k;
 94             scanf("%d%d%d",&l,&r,&k);
 95             if(l<r){
 96                 l++;
 97                 update(l,r,k,1,8000,1);
 98             }
 99         }
100         now=-1;
101         query(1,8000,1);
102         for(int i=0;i<=8000;i++){
103             if(num[i]) cout<<i<<" "<<num[i]<<endl;
104         }
105         cout<<endl;
106     }
107 }

ZOJ 1610.Count the Colors-線段樹(區間染色、區間更新、單點查詢)-有點小坑(染色片段)