Color the Ball

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5481    Accepted Submission(s): 1346


Problem Description There are infinite balls in a line (numbered 1 2 3 ....), and initially all of them are paint black. Now Jim use a brush paint the balls, every time give two integers a b and follow by a char 'w' or 'b', 'w' denotes the ball from a to b are painted white, 'b' denotes that be painted black. You are ask to find the longest white ball sequence.

Input First line is an integer N (<=2000), the times Jim paint, next N line contain a b c, c can be 'w' and 'b'.

There are multiple cases, process to the end of file.

Output Two integers the left end of the longest white ball sequence and the right end of longest white ball sequence (If more than one output the small number one). All the input are less than 2^31-1. If no such sequence exists, output "Oh, my god".

Sample Input 3 1 4 w 8 11 w 3 5 b
Sample Output 8 11 只儲存白色的求，然後用黑色區間更新白色球。（我用的結構體暴力求解）

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAX = 2005;
struct ss
{
    int l, r;
    bool operator< (const ss a) const
    {
        return l < a.l;
    }
} L[MAX];
int N, l, r, k;
char wb;
int main()
{
    while(scanf("%d", &N) != EOF)
    {
        k = 0;
        while(N--)
        {
            scanf("%d %d %c", &l, &r, &wb);
            if(l > r) swap(l, r);
            if(wb == 'w')
            {
                L[k].l = l;
                L[k++].r = r;
            }
            else
            {
                int k2 = k;
                for(int i = 0; i < k2; ++i)
                {
                    if(l <= L[i].l && r >= L[i].l && r < L[i].r) L[i].l = r+1;
                    else if(l > L[i].l && l <= L[i].r && r >= L[i].r) L[i].r = l-1;
                    else if(l > L[i].l && r < L[i].r)
                    {
                        L[k].l = r+1;
                        L[k++].r = L[i].r;
                        L[i].r = l-1;
                    }
                    else if(l <= L[i].l && r >= L[i].r)
                    {
                        L[i].l = -1;
                        L[i].r = -1;
                    }
                }
            }
        }
        sort(L, L+k);
        int ans = 0, ansl, ansr;
        l = -1;
        r = -2;
        for(int i = 0; i < k; ++i)
        {
            if(!L[i].l == -1) continue;
            if(L[i].l > r+1)
            {
                if(r-l+1 > ans)
                {
                    ansl = l;
                    ansr = r;
                    ans = r-l+1;
                }
                l = L[i].l;
                r = L[i].r;
            }
            else r = max(r, L[i].r);
        }
        if(r-l+1 > ans)
        {
            ansl = l;
            ansr = r;
            ans = r-l+1;
        }
        if(ans == 0) cout <<"Oh, my god"<< endl;
        else cout << ansl <<" "<< ansr << endl;
    }
    return 0;
}