Zoj 1610 count the colors [brute force]

Link: http://acm.zju.edu.cn/onlinejudge/showProblem.do? Problemcode = 1610 http://acm.hust.edu.cn/vjudge/contest/view.action? Cid = 28619 # Problem/F

Count the colors

Time Limit: 2 seconds memory limit: 65536 KB

Painting some colored segments on a line, some previusly painted segments may be covered by some of 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 shoshould contain a color index that can be seen from the top, following the count of the segments of this color, they shocould 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

Author: standlove
Source: zoj monthly, May 2003 code:

// I thought we wanted to discretization the line tree. In this way, the AC would be too unreasonable. orz # include <stdio. h> # include <string. h >#include <algorithm> # include <iostream> using namespace STD; const int maxn = 8000 + 10; // int left, right; int A [maxn]; int node [maxn]; int main () {int N; while (scanf ("% d", & N )! = EOF) {int left = maxn; int right =-maxn; memset (node, 0, sizeof (node); For (INT I = 0; I <maxn; I ++) A [I] =-1; int St, en, color; while (n --) {scanf ("% d", & St, & en, & color); For (INT I = sT; I <en; I ++) A [I] = color; left = min (left, St ); right = max (right, en) ;}// for (INT I = left; I <right; I ++) printf ("% d", a [I]); printf ("\ n"); node [A [left] = 1; for (INT I = left + 1; I <right; I ++) {If (A [I] = A [I-1]) {continue;} else {if (a [I]! =-1) node [A [I] ++ ;}} for (INT I = 0; I <maxn; I ++) {If (node [I]> 0) printf ("% d \ n", I, node [I]) ;}printf ("\ n ");} return 0 ;}