题目大意;说是可以吧一段区间变成白色或者黑色， 区间（0-10^9）初始都是白色，问经过n次操作以后最大的连续白色区间
Problem Description
The segment of numerical axis from 0 to 109 is painted into white color. After that some parts of this segment are painted into black, then some into white again and so on. In total there have been made N re-paintings (1 ≤ N ≤ 5000). You are to write a program that finds the longest white open interval after this sequence of re-paintings.
Input
The first line of input contains the only number N. Next N lines contain information about re-paintings. Each of these lines has a form:
ai bi ci
where ai and bi are integers, ci is symbol 'b' or 'w', ai, bi, ci are separated by spaces.
This triple of parameters represents repainting of segment from ai to bi into color ci ('w' white, 'b' black). You may assume that 0 < ai < bi < 109.
ai bi ci
where ai and bi are integers, ci is symbol 'b' or 'w', ai, bi, ci are separated by spaces.
This triple of parameters represents repainting of segment from ai to bi into color ci ('w' white, 'b' black). You may assume that 0 < ai < bi < 109.
Output
Output should contain two numbers x and y (x < y) divided by space(s). These numbers should define the longest white open interval. If there are more than one such an interval output should contain the one with the smallest x.
Sample Input
input output
#include<iostream>
#include<algorithm>
#include<cstdio>
using namespace std;
#define maxn 5000
struct node
{
int L, R, color;
int Mid(){return (L+R)/2;}
};
node a[maxn*4*2];
int point[maxn*2+10], npoint;
int maxv, minv, First, Last, color;
void BuildTree(int r, int L, int R);
void Insert(int r, int L, int R, int color);
void Query(int r);
int main()
{
int N;
while(scanf("%d", &N) != EOF)
{
int i, L[maxn+10], R[maxn+10], c[maxn+10];
char ch;
for(npoint=i=0; i<N; i++)
{
cin >> L[i] >> R[i] >> ch;
c[i] = (ch == 'w' ? 0 : 1);
point[npoint++] = L[i];
point[npoint++] = R[i];
}
point[npoint++] = 0, point[npoint++] = 1000000000;
sort(point, point+npoint);
npoint = unique(point, point+npoint) - point;
BuildTree(1, 0, npoint-1);
for(i=0; i<N; i++)
{
L[i] = lower_bound(point, point+npoint, L[i]) - point;
R[i] = lower_bound(point, point+npoint, R[i]) - point;
Insert(1, L[i], R[i], c[i]);
}
maxv = minv = First = Last = 0;
color = 0;
Query(1);
printf("%d %d\n", minv, maxv);
}
return 0;
}
void BuildTree(int r, int L, int R)
{
a[r].L = L, a[r].R = R, a[r].color = 0;
if(R - L == 1)return ;
BuildTree(r*2, L, a[r].Mid());
BuildTree(r*2+1, a[r].Mid(), R);
}
void Insert(int r, int L, int R, int C)
{
if(a[r].color == C)return ;
if(a[r].L == L && a[r].R == R)
{
a[r].color = C;
return ;
}
if(a[r].color >= 0)
a[r*2].color = a[r*2+1].color = a[r].color;
a[r].color = -1;
if(R <= a[r].Mid())
Insert(r*2, L, R, C);
else if(L >= a[r].Mid())
Insert(r*2+1, L, R, C);
else
{
Insert(r*2, L, a[r].Mid(), C);
Insert(r*2+1, a[r].Mid(), R, C);
}
}
void Query(int r)
{
if(a[r].color == 0)
{
Last = a[r].R;
if(color == 1)
{
color = a[r].color;
First = a[r].L;
}
if(maxv-minv < point[Last] - point[First])
maxv = point[Last], minv = point[First];
return ;
}
if(a[r].color == 1)
{
color = 1;
return ;
}
Query(r*2);
Query(r*2+1);
}