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 separatedby single spaces:

x1 x2 c

x1 and x2 indicate the left endpoint and right endpoint of the segment, c indicatesthe 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 thetop, following the count of the segments of this color, they should be printedaccording 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

題目意思

多組案例，每組一個n，表示操作次數，每次操作三個數字x1,x2,c，表示把x1-x2之間染成顏色c，並且連續的顏色只算一次，求最後各種顏色的出現的次數。

坑點

1.每次的區間[x1 , x2]表示的是平面端點，在一維情況下代表的是[x1+1 , x2]的點值，所以每次讀入x1後都要加1。（建議紙上模擬一下）
2.有顏色0，所以mark初始為-1。
3.不是所有區間的範圍小於n，而是所有區間的範圍小於8000，所以應該是build(1,8000,1)，update，query同理。
4.可能中間有沒有塗顏色（即-1的點），經過這些點要把pre（標記上一次顏色）重置為-1。pushdown也要將mark置為-1。
5.每組案例多輸出一個換行。
6.每次操作的區間左值不一定小於等於右值。（有些題目會卡這種情況，本題沒卡）。

AC程式碼

一個星期前卡了我一個下午，今天重寫突然發現踩了坑點4。。。

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
#define ll long long
#define lson left,mid,k<<1
#define rson mid+1,right,k<<1|1
#define sc scanf
#define pr printf
#define rep(i,a,b) for (int i=a;i<b;i++)
#define per(i,a,b) for (int i=a;i>=b;i--)
#define intmid int mid=(left+right)>>1
using namespace std;
ll a[200000];
int book[20000];
int ql, qr;
ll val, pre = -1;
struct node
{
	int l;
	int r;
	ll mark;
}que[200000 * 4];
void pushdown(int k)
{
	if (que[k].mark != -1)
	{
		que[k << 1].mark = que[k].mark;
		que[k << 1 | 1].mark = que[k].mark;
		que[k].mark = -1;
	}
}
void build(int left, int right, int k)
{
	que[k].l = left;
	que[k].r = right;
	que[k].mark = -1;
	if (left == right)
		return;
	intmid;
	build(lson);
	build(rson);
}
void update(int left, int right, int k)
{
	if (qr < left || right < ql)
		return;
	if (ql <= left && right <= qr)
	{
		que[k].mark = val;
		return;
	}
	pushdown(k);
	intmid;
	update(lson);
	update(rson);
}
void query(int left, int right, int k)
{
	if (que[k].mark != -1)
	{
		if (pre != que[k].mark)
		{
			book[que[k].mark]++;
			pre = que[k].mark;
		}
		return;
	}
	if (left == right)
	{
		pre = -1;
		return;
	}
	intmid;
	query(lson);
	query(rson);
}
int main()
{
	int n;
	while (sc("%d", &n) > 0)
	{
		pre = -1;
		memset(book, 0, sizeof book);
		build(1, 8000, 1);
		ll maxn = 0;
		for (int i = 0; i < n; i++)
		{
			sc("%d%d%lld", &ql, &qr, &val);
			/*
			if (ql > qr)
			{
				int tt = ql;
				ql = qr;
				qr = tt;
			}
			*/
			ql++;
			val++;
			maxn = max(maxn, val);
			update(1, 8000, 1);
		}
		ql = 1;
		qr = n;
		query(1, 8000, 1);
		for (int i = 1; i <= maxn; i++)
			if (book[i])
				pr("%d %d\n", i - 1, book[i]);
		pr("\n");
	}
}