Color the Ball

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3457    Accepted Submission(s): 849

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

31 4 w8 11 w3 5 b

 

 

Sample Output

8 11

 

 

 

#include <cmath>#include <cstdio>#include <cstdlib>#include <iostream>#include <cstring>#include <string>#include <algorithm>#include <queue>#include <stack>#include <climits>#define MID(x,y) ((x+y)>>1)#define L(x) ( x << 1 )#define R(x) ( x << 1 | 1 )using namespace std;const int MAX = 4010;int x[MAX];struct Tnode{int l,r;int col;};struct NODE{int x,y;bool col;};NODE nn[MAX];Tnode node[MAX*4];int col[MAX];void init(){    memset(col,0,sizeof(col));    memset(nn,0,sizeof(nn));    memset(node,0,sizeof(node));}void Build(int t,int l,int r){if( l >= r ) return ;    node[t].l = l; node[t].r = r;    node[t].col = 0;    if( l == r - 1 ) return ;    int mid = MID(l,r);    Build(L(t),l,mid);    Build(R(t),mid,r);}void Updata(int t,int l,int r,int col){if( l >= r ) return ;    if( node[t].l == l && node[t].r == r )    {        node[t].col = col;        return ;    }    if( node[t].col !=-1 )    {        node[R(t)].col = node[L(t)].col = node[t].col;        node[t].col = -1;    }    int mid = MID(node[t].l,node[t].r);    if( l >= mid )        Updata(R(t),l,r,col);    else        if( r <= mid )            Updata(L(t),l,r,col);        else        {            Updata(L(t),l,mid,col);            Updata(R(t),mid,r,col);        }}void Query(int t,int l,int r){if( l >= r ) return ;    if( node[t].col != -1 )    {        for(int i=node[t].l; i<node[t].r; i++)            col[i] = node[t].col;        return ;    }    int mid = MID(node[t].l,node[t].r);    if( l >= mid )        Query(R(t),l,r);    else        if( r <= mid )            Query(L(t),l,r);        else        {            Query(L(t),l,mid);            Query(R(t),mid,r);        }}void solve(int *x,int cnt,int n){    Build(1,0,cnt);    for(int i=0; i<n; i++)    {        int xx = lower_bound(x,x+cnt,nn[i].x) - x;        int yy = lower_bound(x,x+cnt,nn[i].y) - x;        Updata(1,xx,yy,nn[i].col);        }    Query(1,0,cnt);    int s = 0,e = 0,te,ts;    int mmax = 0;    x[cnt] = 0; x[cnt] = x[cnt-1];    for( i=0; i<cnt; i++)    {        if( col[i] != 1 ) continue;        ts = x[i];        while( col[i] == 1 )            i++;        if( i > cnt ) break;        te = x[i];        if( te - ts > e - s )        {            e = te;            s = ts;        }    }    if( s == e )        printf("Oh, my god\n");    else        printf("%d %d\n",s,e-1);}int main(){    int n;    char s[5];    while( ~scanf("%d",&n) )    {        init();        int cnt = 0;        for(int i=0; i<n; i++)        {            scanf("%d%d%s",&nn[i].x,&nn[i].y,s);            nn[i].y++;             x[cnt++] = nn[i].x;            x[cnt++] = nn[i].y;            if( s[0] == 'w' )                nn[i].col = 1;        }        sort(x,x+cnt);        cnt = unique(x,x+cnt) - x;         solve(x,cnt,n);    }return 0;}