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

Sample Output

1 1
2 1
3 1

1 1

0 2
1 1

Author: Standlove
Source: ZOJ Monthly, May 2003

題意：

給出l,r點的區間，給這些點的區間染上顏色c

問最後每種顏色有幾塊(也就是有幾個不連續的區間塊)

輸出按顏色的標號從小到大輸出

輸出內容為

每行兩個整數

第一個整數為顏色標號

第二個整數位該顏色區間塊的個數

做法：

和

http://poj.org/problem?id=2528

一樣的意思，只不過把區間修改變成點修改，然後去掉離散化，再加上一個統計連續塊就好

代碼

  1 #include<iostream>
  2 using namespace std;
  3 #include<cstdio>
  4 #include<cstring>
  5 #include<map>
  6 #include<algorithm>
  7 const int MAXN=32000;
  8 typedef int LL;
  9 int v[MAXN*3];
 10 int ans[MAXN*3];
 11 int zans[MAXN*3];
 12 inline int read()
 13 {
 14     int k=0;
 15     char f=1;
 16     char c=getchar();
 17     for(;!isdigit(c);c=getchar() )
 18         if(c==‘-‘)
 19             f=-1;
 20     for(;isdigit(c);c=getchar() )
 21         k=k*10+c-‘0‘;
 22     return k*f;
 23 }
 24 template<class T>
 25 class tree{
 26     public:
 27     int ws;
 28     T a[150000];
 29     T gg[150000];
 30     int l[150000],r[150000];
 31     void creat(int k,int ll,int rr){
 32         l[k]=ll;r[k]=rr;
 33         if(ll>=rr){
 34             a[k]=-1;
 35             gg[k]=-1;
 36             return;
 37         }
 38         int mid=(ll+rr)/2; 
 39         creat(k*2,ll,mid);
 40         creat(k*2+1,mid+1,rr);
 41         a[k]=a[k*2]+a[k*2+1];
 42         gg[k]=-1;
 43     }
 44     void change(int k,int ll,int rr,int t){
 45         if(gg[k]!=-1){
 46             a[k]=gg[k]*(r[k]-l[k]+1);
 47             if(l[k]!=r[k]){
 48                 gg[k*2]=gg[k];
 49                 gg[k*2+1]=gg[k];
 50             }
 51             gg[k]=-1;
 52         }
 53         if(ll>r[k]||rr<l[k]) return;
 54         if(ll<=l[k]&&rr>=r[k]){
 55             a[k]=(r[k]-l[k]+1)*t;
 56             if(l[k]!=r[k]){
 57                 gg[k*2]=t;
 58                 gg[k*2+1]=t;
 59             }
 60             return;
 61         }
 62         change(k*2,ll,rr,t);
 63         change(k*2+1,ll,rr,t);
 64         a[k]=a[k*2]+a[k*2+1];
 65     }
 66     LL query(int k,int ll,int rr){
 67         if(gg[k]!=-1){
 68             a[k]=(r[k]-l[k]+1)*gg[k];
 69             if(l[k]!=r[k]){
 70                 gg[k*2]=gg[k];
 71                 gg[k*2+1]=gg[k];
 72             }
 73             gg[k]=-1;
 74         }
 75         if(ll>r[k]||rr<l[k]) return 0; 
 76         if(l[k]==r[k]){
 77             return 0;
 78         }
 79         int ans1=query(k*2,ll,rr);
 80         int ans2=query(k*2+1,ll,rr);
 81         return 0;    
 82     }
 83     void get_ans(int k){
 84         if(l[k]==r[k]){
 85         //    cout<<"data: "<<l[k]<<" "<<a[k]<<endl;
 86             zans[l[k]]=a[k];
 87             if(a[k]<0)
 88                 return;
 89             if(v[a[k]])
 90                 return;
 91             else{
 92                 v[a[k]]=1;
 93                 return;
 94             }
 95         }
 96         get_ans(2*k);
 97         get_ans(2*k+1);
 98     }
 99 };
100 tree<int> line_tree;
101 int a[MAXN*3];
102 int b[MAXN*3];
103 int deal(){
104     int n;
105     if(scanf("%d",&n)==EOF)
106         return 1;
107     memset(zans,0,sizeof(zans));
108     int l,r;
109     int maxn,minn;
110     maxn=-1;
111     minn=0x7fffffff;
112     for(int i=0;i<n;i++){
113         a[i*2]=read();//左區間 
114         a[i*2+1]=read();//右區間 
115         if(maxn<a[i*2])
116             maxn=a[i*2];
117         if(maxn<a[i*2+1])
118             maxn=a[i*2+1];
119         if(minn>a[i*2])
120             minn=a[i*2];
121         if(minn>a[i*2+1])
122             minn=a[i*2+1];
123         b[i]=read();//染色顏色 
124     }
125     memset(v,0,sizeof(v));
126     line_tree.creat(1,minn,maxn);
127     for(int i=0;i<n;i++){
128         if(a[i*2]>a[i*2+1])
129             swap(a[i*2],a[i*2+1]);
130         line_tree.change(1,a[i*2],a[i*2+1]-1,b[i]);
131     }
132     line_tree.query(1,minn,maxn);
133     line_tree.get_ans(1);
134     memset(ans,0,sizeof(ans));
135     for(int i=minn;i<=maxn;i++){
136         if((i==minn||zans[i]!=zans[i-1])&&v[zans[i]])
137             ans[zans[i]]++;
138     }
139     for(int i=0;i<=8010;i++){
140         if(v[i]==0)
141             continue;
142         else
143         printf("%d %d\n",i,ans[i]);
144     }
145     return 0;
146 }
147 int main(){
148     while(1){
149         if (deal())
150             break;
151         putchar(‘\n‘);
152     }
153     return 0;
154 }

ZOJ - 1610 Count the Colors 線段樹區間修改