标签：style blog http java color os
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
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 #include <climits> 7 #include <vector> 8 #include <queue> 9 #include <cstdlib> 10 #include <string> 11 #include <set> 12 #define LL long long 13 #define INF 0x3f3f3f 14 using namespace std; 15 const int maxn = 8010; 16 struct node { 17 int lt,rt,color; 18 } tree[maxn<<2]; 19 int col[maxn],temp; 20 void build(int lt,int rt,int v) { 21 int mid = (lt+rt)>>1; 22 tree[v].lt = lt; 23 tree[v].rt = rt; 24 tree[v].color = -1; 25 if(lt+1 == rt) return; 26 build(lt,mid,v<<1); 27 build(mid,rt,v<<1|1); 28 } 29 void update(int lt,int rt,int c,int v) { 30 if(lt == rt || tree[v].color == c) return; 31 if(tree[v].lt >= lt && tree[v].rt <= rt) { 32 tree[v].color = c; 33 return; 34 } 35 if(tree[v].color >= 0) { 36 tree[v<<1].color = tree[v<<1|1].color = tree[v].color; 37 tree[v].color = -2; 38 } 39 int mid = (tree[v].lt+tree[v].rt)>>1; 40 if(rt <= mid) { 41 update(lt,rt,c,v<<1); 42 } else if(lt >= mid) { 43 update(lt,rt,c,v<<1|1); 44 } else { 45 update(lt,mid,c,v<<1); 46 update(mid,rt,c,v<<1|1); 47 } 48 tree[v].color = -2; 49 } 50 void query(int v) { 51 if(tree[v].color == -1) { 52 temp = -1; 53 return; 54 } 55 if(tree[v].color != -2) { 56 if(tree[v].color != temp) { 57 temp = tree[v].color; 58 col[temp]++; 59 } 60 return; 61 } 62 if(tree[v].lt+1 != tree[v].rt) { 63 query(v<<1); 64 query(v<<1|1); 65 } 66 } 67 int main() { 68 int n,i,j,x,y,c,mx; 69 while(~scanf("%d",&n)) { 70 build(0,8002,1); 71 memset(col,0,sizeof(col)); 72 mx = 0; 73 for(i = 0; i < n; i++) { 74 scanf("%d%d%d",&x,&y,&c); 75 update(x,y,c,1); 76 if(c > mx) mx = c; 77 } 78 temp = -1; 79 query(1); 80 for(i = 0; i <= mx; i++) 81 if(col[i]) printf("%d %d\n",i,col[i]); 82 printf("\n"); 83 } 84 return 0; 85 }
xtu数据结构 G. Count the Colors,布布扣,bubuko.com
标签：style blog http java color os
原文地址：http://www.cnblogs.com/crackpotisback/p/3858359.html
