Count the Colors
Time Limit: 1 Second
Memory Limit: 32768 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
题目大意：给出一系列线段和线段所要覆盖的颜色，最后求每种颜色有多少条线段。
总结：
线段树的入门好题。
代码：
1 /*
2 * Problem: ZJU 1610
3 * Author: Xu Fei
4 * Time: 2010.8.13 11:43
5 * Method: 线段树-指针实现
6 */
7 #include<iostream>
8 #include<cstdio>
9 #include<cstring>
10 using namespace std;
11
12 const int N=8000,M=10000,Noc=-1,Muc=-2;
13
14 struct Node
15 {
16 int left,right,cover;
17 Node *lc;
18 Node *rc;
19 }*root;
20 int color[M];
21 Node *build(Node *Tree,int left,int right)
22 {
23 Node *P=new Node;
24 P->left=left;
25 P->right=right;
26 P->cover=Noc;
27 P->lc=NULL;
28 P->rc=NULL;
29 if(left<right)
30 {
31 int mid=(left+right)>>1;
32 P->lc=build(P->lc,left,mid);
33 P->rc=build(P->rc,mid+1,right);
34 }
35 return P;
36 }
37 void insert(Node *Tree,int left,int right,int col)
38 {
39 if(Tree->left == left && Tree->right == right)
40 {
41 Tree->cover=col;
42 return;
43 }
44 if(Tree->cover >= 0)
45 Tree->lc->cover=Tree->rc->cover=Tree->cover;
46 Tree->cover=Muc;
47 int mid=(Tree->left+Tree->right)>>1;
48 if(right <= mid)
49 insert(Tree->lc,left,right,col);
50 else if(left > mid)
51 insert(Tree->rc,left,right,col);
52 else
53 {
54 insert(Tree->lc,left,mid,col);
55 insert(Tree->rc,mid+1,right,col);
56 }
57 }
58 void getcount(Node *Tree,int &col)
59 {
60 if(Tree->cover >= 0 && Tree->cover != col)
61 {
62 col=Tree->cover;
63 color[col]++;
64 }
65 else if(Tree->cover == Muc)
66 {
67 getcount(Tree->lc,col);
68 getcount(Tree->rc,col);
69 }
70 else
71 col=Tree->cover;
72 }
73 int main()
74 {
75 int n,i,a,b,c;
76 while(scanf("%d",&n)!=EOF)
77 {
78 root=build(root,1,N);
79 for(i=1;i<=n;++i)
80 {
81 scanf("%d%d%d",&a,&b,&c);
82 insert(root,a+1,b,c);
83 }
84 memset(color,0,sizeof(color));
85 int col=Noc;
86 getcount(root,col);
87 for(i=0;i<M;++i)
88 if(color[i])
89 printf("%d %d\n",i,color[i]);
90 printf("\n");
91 }
92 return 0;
93 }
1 /*
2 * Problem: ZJU 1610
3 * Author: Xu Fei
4 * Time: 2010.8.13 15:33
5 * Method: 线段树-静态实现
6 */
7 #include<iostream>
8 #include<cstdio>
9 #include<cstring>
10 using namespace std;
11
12 const int N=8000,M=10000,Noc=-1,Muc=-2;
13
14 struct Node
15 {
16 int left,right,cover;
17 }tree[N*4];
18 int color[M];
19
20 void build(int len,int left,int right)
21 {
22 tree[len].left=left;
23 tree[len].right=right;
24 tree[len].cover=Noc;
25 if(left<right)
26 {
27 int mid=(left+right)>>1;
28 build(len<<1,left,mid);
29 build((len<<1)+1,mid+1,right);
30 }
31 }
32 void insert(int len,int left,int right,int col)
33 {
34 if(tree[len].left == left && tree[len].right == right)
35 {
36 tree[len].cover=col;
37 return;
38 }
39 if(tree[len].cover >= 0)
40 tree[len<<1].cover=tree[(len<<1)+1].cover=tree[len].cover;
41 tree[len].cover=Muc;
42 int mid=(tree[len].left+tree[len].right)>>1;
43 if(right <= mid)
44 insert(len<<1,left,right,col);
45 else if(left > mid)
46 insert((len<<1)+1,left,right,col);
47 else
48 {
49 insert(len<<1,left,mid,col);
50 insert((len<<1)+1,mid+1,right,col);
51 }
52 }
53 void getcount(int len,int &col)
54 {
55 if(tree[len].cover >= 0 && tree[len].cover != col)
56 {
57 col=tree[len].cover;
58 color[col]++;
59 }
60 else if(tree[len].cover == Muc)
61 {
62 getcount(len<<1,col);
63 getcount((len<<1)+1,col);
64 }
65 else
66 col=tree[len].cover;
67 }
68 int main()
69 {
70 int n,i,a,b,c;
71 while(scanf("%d",&n)!=EOF)
72 {
73 build(1,1,N);
74 for(i=0;i<n;++i)
75 {
76 scanf("%d%d%d",&a,&b,&c);
77 insert(1,a+1,b,c);
78 }
79 memset(color,0,sizeof(color));
80 int col=Noc;
81 getcount(1,col);
82 for(i=0;i<M;++i)
83 if(color[i])
84 printf("%d %d\n",i,color[i]);
85 printf("\n");
86 }
87 return 0;
88 }
posted on 2010-08-13 11:45
xfstart07 阅读(584)
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