URAL1019. Line Painting

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.

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	output
4 1 999999997 b 40 300 w 300 634 w 43 47 b	47 634

涂色问题：

将[0，1E9]区间涂色，只能涂白色和黑色，可覆盖，求最长的白色区间。

思路：

离散化，可以将点的数量下降到2*N  接下来看可以用线段树(NlogN)或者暴力(N*N)查找答案，这题是被自己坑了，在test2弹了好几次。

只要注意这组样例：

1

10 11 b

answer is 11 1000000000

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <map>
#include <algorithm>
using namespace std;
struct paint{
	int  l,r;
	char op;
};
int len;
vector<bool>vis;
vector<paint> vp,tp;
vector<int> tn;
map<int,int> yp,kp;
void init(){
	
}
void disret(){
	vis.resize(tn.size());
	for(int i = 0; i < vis.size(); i++)
		vis[i] = 0;
	sort(tn.begin(),tn.end());
	int k = 0;
	for(int i = 0; i < tn.size(); i++){
		while(tn[i]==tn[i+1]&&i<tn.size()-1){
			i++;
		}
		kp[k] = tn[i];
		yp[tn[i]] = k++;
	}
	//for(int i = 0; i < k; i++){
		//cout<<kp[i]<<endl;
	//}
	for(int i = 0; i < vp.size(); i++){
		int l = yp[vp[i].l];
		int r = yp[vp[i].r];
		//cout<<l<<" "<<r<<endl;
		for(int j = l; j <= r-1; j++){
			if(vp[i].op=='b'){
				vis[j] = 1;
			}else{
				vis[j] = 0;
			}
		}
	}
	//for(int i =0 ;i < k-1; i++)
		//cout<<vis[i]<<endl;
	len = k; 
}
void solve(){
	disret();
	int ans = 0,l=0,r=0;
	for(int i = 0; i < len-1; i++){
		if(vis[i]==0){
			int j = i;
			while(vis[j]==0&&j<len-1)
				j++;
			if(ans < kp[j]-kp[i]){
				l = kp[i];
				r = kp[j];
				ans = kp[j]-kp[i];
			}
			i = j;	
		}
	}
	cout<<l<<" "<<r<<endl;
}
int main(){
	int n;
	scanf("%d",&n);
	tn.push_back(0);
	tn.push_back(1000000000);
	while(n--){
		paint tmp;
		cin >> tmp.l >> tmp.r >> tmp.op;
		tn.push_back(tmp.r);
		tn.push_back(tmp.l);
		vp.push_back(tmp);	
	}
	solve();
	return 0;
}