ZOJ 1610 Count the Colors (line segment tree lazy + brute force statistics), zojlazy

ZOJ 1610 Count the Colors (line segment tree lazy + brute force statistics), zojlazy

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

The specific method is to perform lazy operations during Interval Update and brute force operations during endpoint statistics ,.

# Include <cstdio> # include <cstring> # include <algorithm> # include <vector> # include <string> # include <iostream> # include <queue> # include <cmath> # include <map> # include <stack> # include <bitset> using namespace std; # define REPF (I, a, B) for (int I = a; I <= B; ++ I) # define REP (I, n) for (int I = 0; I <n; ++ I) # define CLEAR (a, x) memset (a, x, sizeof a) typedef long LL; typedef pair <int, int> p Il; const int maxn = 8000 + 10; int sum [maxn <2]; int cnt, n; int col [maxn], ans [maxn]; void pushdown (int rs) {if (sum [rs]! =-1) {sum [rs <1] = sum [rs <1 | 1] = sum [rs]; sum [rs] =-1 ;}} void update (int x, int y, int c, int l, int r, int rs) {if (l> = x & r <= y) {sum [rs] = c; return;} pushdown (rs); int mid = (l + r)> 1; if (x <= mid) update (x, y, c, l, mid, rs <1); if (y> mid) update (x, y, c, mid + 1, r, rs <1 | 1);} void solve (int l, int r, int rs) {if (l = r) {col [cnt ++] = sum [rs]; return;} pushdown (rs); int mid = (l + r)> 1; solve (l, mid, rs <1); solve (mid + 1, r, rs <1 | 1);} int main () {int l, r, x; while (~ Scanf ("% d", & n) {cnt = 0; CLEAR (sum,-1); CLEAR (ans, 0); int m = 8000; REP (I, n) {scanf ("% d", & l, & r, & x); update (l, R-1, x, 0, m, 1 );} solve (0, m, 1); REP (I, cnt) {int j = I + 1; if (col [I]! =-1) ans [col [I] ++; while (col [j] = col [I] & j <cnt) j ++; I = J-1;} for (int I = 0; I <= maxn; I ++) // for color type if (ans [I]) printf ("% d \ n", I, ans [I]); puts ("");} return 0 ;}