ural 1019

題意：給定一個數N 下面有N 行操作，每行有兩個數a，b和一個字符c，把區間a-b染成c 的顏色，求最後全部染完色後最長白色區間的左右端點座標。

線段樹成段更新+染色，由於數據量很大數據要進行離散。離散後映射到數組上，按條件查詢最長的白色區間就可以了。

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
#define Max 10010
#pragma warning(disable:4996)

int Seg[Max << 2], Lazy[Max << 2], cnt;
bool Vis[Max << 2];
struct node
{
	int x, y;
	char c;
}Edge[Max];

void Pushdown(int id)
{
	if (Lazy[id] != -1)
	{
		Lazy[id << 1] = Lazy[id << 1 | 1] = Lazy[id];
		Lazy[id] = -1;
	}
}

void Update(int op, int L, int R, int l, int r, int id)
{
	if (L <= l&&r <= R)
	{
		Lazy[id] = op;
		return;
	}
	Pushdown(id);
	int mid = l + r >> 1;
	if (L <= mid)Update(op, L, R, l, mid, id << 1);
	if (R > mid)Update(op, L, R, mid + 1, r, id << 1 | 1);
}

void Query(int l, int r, int id)
{
	if (Lazy[id] != -1)
	{
		for (int i = l; i <= r; i++)
		{
			Vis[i] = Lazy[id];
		}
		return;
	}
	if (l == r)return;
	Pushdown(id);
	int mid = l + r >> 1;
	Query(l, mid, id << 1);
	Query(mid + 1, r, id << 1 | 1);
}

int Solve(int x)
{
	int l = 0, r = cnt - 1, mid;
	while (l <= r)
	{
		mid = l + r >> 1;
		if (Seg[mid] == x)return mid;
		else if (Seg[mid] > x)r = mid - 1;
		else l = mid + 1;
	}
	return -1;
}

int main()
{
	memset(Lazy, -1, sizeof(Lazy));
	int n;
	scanf("%d", &n);
	Seg[cnt++] = 0;
	Seg[cnt++] = 1000000000;
	for (int i = 0; i < n; i++)
	{
		scanf("%d %d %c", &Edge[i].x, &Edge[i].y, &Edge[i].c);
		getchar();
		Seg[cnt++] = Edge[i].x;
		Seg[cnt++] = Edge[i].y;
	}
	cnt = unique(Seg, Seg + cnt) - Seg;
	sort(Seg, Seg + cnt);
	int t = cnt;
	for (int i = 1; i<t; i++)
		if (Seg[i] - Seg[i - 1]>1)
			Seg[cnt++] = Seg[i - 1] + 1;
	sort(Seg, Seg + cnt);
	for (int i = 0; i < n; i++)
	{
		Edge[i].x = Solve(Edge[i].x);
		Edge[i].y = Solve(Edge[i].y);
		Update(Edge[i].c == 'b', Edge[i].x, Edge[i].y - 1, 0, cnt - 1, 1);
	}
	Query(0, cnt - 1, 1);
	int len = 0, l, r, k;
	for (int i = 0; i < cnt;)
	{
		int j = i;
		while (j < cnt&&Vis[j] == Vis[i])j++;
		if (j == cnt)k = 1000000000;
		else k = Seg[j];
		if (0 == Vis[i] && len < k - Seg[i])
		{
			len = k - Seg[i];
			l = Seg[i], r = k;
		}
		i = j;
	}
	printf("%d %d\n", l, r);
	return 0;
}