데이터 구조 - 선분 트 리 - 구간 색칠 문제

Count the Colors
Time Limit: 2 Seconds      
Memory Limit: 65536 KB
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
선분 나무의 전형 적 인 응용
코드:

  1 #include "cstdio"  //zju 1610         

  2 #include "cstring"

  3 #include "iostream"

  4 using namespace std;

  5 

  6 #define N 8005

  7 const int NoColor = -1; //     

  8 const int MulColor = -2; //     

  9 

 10 struct node

 11 {

 12     int l,r;

 13     int col;

 14 }Tree[4*N];

 15 

 16 int a[N],many[N]; //a[]      ，many[]        

 17 

 18 void Build(int t,int l,int r)  //  

 19 {

 20     Tree[t].l = l;

 21     Tree[t].r = r;

 22     Tree[t].col = NoColor;  //         

 23     if(Tree[t].l+1==Tree[t].r)  

 24         return ;

 25     int mid = (Tree[t].l + Tree[t].r)/2;

 26     Build(t<<1,l,mid);

 27     Build(t<<1|1,mid,r);

 28 }

 29 

 30 void Update(int t,int l,int r,int k)

 31 {

 32     if(Tree[t].col == k)    //    ，      

 33         return ; 

 34     if(Tree[t].l==l && Tree[t].r==r)    //                  ，          ，return ;

 35         {  Tree[t].col = k;    return ; }  

 36     if(Tree[t].l+1==Tree[t].r)  //     ，       ，return ;

 37     {    Tree[t].col = k; return ; }

 38     if(Tree[t].col != MulColor)  //          ，         ;

 39     {

 40         Tree[t<<1].col   = Tree[t].col;

 41         Tree[t<<1|1].col = Tree[t].col;

 42     }

 43     Tree[t].col = MulColor;  //      

 44     int mid = (Tree[t].l+Tree[t].r)/2;

 45     if(r<=mid)

 46         Update(t<<1,l,r,k);

 47     else if(l>=mid)

 48         Update(t<<1|1,l,r,k);

 49     else

 50     {

 51         Update(t<<1,l,mid,k);

 52         Update(t<<1|1,mid,r,k);

 53     }

 54 }

 55 

 56 void Stats(int t)   //           

 57 {

 58     int i,j;

 59     if(Tree[t].l+1==Tree[t].r)

 60     {

 61         a[Tree[t].l] = Tree[t].col;

 62         return ;

 63     }

 64     if(Tree[t].col>=0)

 65     {

 66         for(i=Tree[t].l; i<Tree[t].r; ++i)

 67             a[i] = Tree[t].col;

 68         return ;

 69     }

 70     Stats(t<<1);

 71     Stats(t<<1|1);

 72 }

 73 

 74 int main()

 75 {

 76     int n;

 77     int i,j;

 78     int x,y,k;

 79     while(scanf("%d",&n)!=-1)

 80     {

 81         Build(1,1,N); //  

 82         while(n--)

 83         {

 84             scanf("%d %d %d",&x,&y,&k);

 85             x++; y++;

 86             Update(1,x,y,k);  //        ，       

 87         }

 88         memset(a,0,sizeof(a));

 89         memset(many,0,sizeof(many));

 90         Stats(1);

 91         int t = -1;

 92 

 93         for(i=1; i<=8000; ++i)

 94         {

 95             if(a[i]==t) continue;

 96             t = a[i];

 97             if(t!=-1)

 98                 many[t]++;

 99         }

100         for(i=0; i<=8000; ++i)

101         {

102             if(many[i])

103                 printf("%d %d
",i,many[i]);

104         }

105         printf("
");

106     }

107     return 0;

108 }