题目来源：URAL 1019. Line Painting

题意：求最长的一段全部为白色的区间

思路：线段树成段更新 区间合并 离散化 这里对应的是一段区间 所以每次不是m+1 而是 l m 和 m r 了 另外我加上了0 和 10^9 这两个点

每一段区间(l, r)我记录的是l和r之间有多少条线段

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 50010;
int sum[maxn<<2], lsum[maxn<<2], rsum[maxn<<2], co[maxn<<2];

struct node
{
	int x, v, id;
}c[maxn*2];
int num[maxn*2];
char s[maxn][5];

void pushup(int l, int r, int rt)
{
	int k = r-l+1;
	int m = (l+r) >> 1;
	lsum[rt] = lsum[rt<<1];
	rsum[rt] = rsum[rt<<1|1];
	
	if(lsum[rt] == num[m]-num[l])
		lsum[rt] += lsum[rt<<1|1];
	if(rsum[rt] == num[r]-num[m])
		rsum[rt] += rsum[rt<<1];
	
	sum[rt] = max(sum[rt<<1], sum[rt<<1|1]);
	sum[rt] = max(sum[rt], rsum[rt<<1]+lsum[rt<<1|1]);
}
void build(int l, int r, int rt)
{
	sum[rt] = lsum[rt] = rsum[rt] = num[r]-num[l];
	co[rt] = -1;
	if(l + 1 == r)
		return;
	int m = (l+r) >> 1;
	build(l, m, rt<<1);
	build(m, r, rt<<1|1);
	
	//pushup(l, r, rt);
}

void pushdown(int l, int r, int rt)
{
	if(co[rt] != -1)
	{
		int k = r-l+1;
		int m = (l + r) >> 1;
		sum[rt<<1] = co[rt]*(num[m]-num[l]);
		lsum[rt<<1] = co[rt]*(num[m]-num[l]);
		rsum[rt<<1] = co[rt]*(num[m]-num[l]);
		
		sum[rt<<1|1] = co[rt]*(num[r]-num[m]);
		lsum[rt<<1|1] = co[rt]*(num[r]-num[m]);
		rsum[rt<<1|1] = co[rt]*(num[r]-num[m]);
		
		co[rt<<1] = co[rt<<1|1] = co[rt];
		co[rt] = -1;
	}
}
int query(int l, int r, int rt, int a)
{
	if(l + 1 == r)
		return r;
	
	pushdown(l, r, rt);
	
	int m = (l+r) >> 1;
	//printf("*****%d %d %d %d %d\n", l, r, sum[rt<<1], sum[rt<<1|1], a);
	if(sum[rt<<1] >= a)
		return query(l, m, rt<<1, a);
	else if(rsum[rt<<1] + lsum[rt<<1|1] >= a)
		return query(m, r, rt<<1|1, a-rsum[rt<<1]);
	else
		return query(m, r, rt<<1|1, a);
}

void update(int x, int y, int l, int r, int rt, int c)
{
	if(x == l && y == r)
	{
		sum[rt] = lsum[rt] = rsum[rt] = c ? (num[r]-num[l]) : 0;
		co[rt] = c;
		return;
	}
	
	pushdown(l, r, rt);
	
	int m = (l+r) >> 1;
	if(y <= m)
		update(x, y, l, m, rt<<1, c);
	else if(x >= m)
		update(x, y, m, r, rt<<1|1, c);
	else
	{
		update(x, m, l, m, rt<<1, c);
		update(m, y, m, r, rt<<1|1, c);
	}
	pushup(l, r, rt);
}

bool cmp1(node a, node b)
{
	return a.x < b.x;
}

bool cmp2(node a, node b)
{
	return a.id < b.id;
}

int main()
{
	int q;
	while(scanf("%d", &q) != EOF)
	{
		for(int i = 0; i < q*2; i++)
		{
			scanf("%d", &c[i].x);
			c[i].id = i;
			if(i&1)
				scanf("%s", s[(i-1)/2]);
		}
		sort(c, c+2*q, cmp1);
		int now = c[0].x, n = 2;
		num[1] = 0;
		for(int i = 0; i < q*2; i++)
		{
			if(now != c[i].x)
			{
				n++;
				now = c[i].x;
			}
			c[i].v = n;
			num[n] = now;
		}
		num[++n] = 1000000000;
		//for(int i = 1; i <= n; i++)
			//printf("***%d\n", num[i]);
		//printf("%d\n", n);
		sort(c, c+2*q, cmp2);
		build(1, n, 1);
		for(int i = 0; i < q; i++)
		{
			int a = c[i<<1].v;
			int b = c[i<<1|1].v;
			int color = 1;
			if(s[i][0] == 'b')
				color = 0;	
			update(a, b, 1, n, 1, color);
			//printf("***%d %d %d\n", a, b, sum[1]);	
		}
		int ans = query(1, n, 1, sum[1]);
		printf("%d %d\n", num[ans]-sum[1], num[ans]);
	}
	return 0;
}

URAL 1019. Line Painting 线段树 区间合并 离散化