hdu 1199 Color the Ball(离散化线段树)

Color the Ball

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3529 Accepted Submission(s): 874

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

Sample Output

    8 11
   

//好事恶习啊，我都快要吐了，第一次做离散化的线段树
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define N 4040
int a[N],r[N],p[N],c[N],f[N];//r[i]排名i的原位置，p[i]原位置i的排名
int cmp(int i,int j){
    return a[i]<a[j];
}
struct node{
    int l,r,lz;
    void init(int _l,int _r){
        l=_l;r=_r;lz=-1;
    }
}t[4*N];
void build(int l,int r,int k){
    if(l>=r)return;
    t[k].init(l,r);
    if(l+1>=r)return;
    int md=(l+r)/2;
    build(l,md,k*2);
    build(md,r,k*2+1);//不是md+1(因为md被覆盖)
}
void down(int k){
	if(t[k].l+1==t[k].r)return;
    if(t[k].lz==-1)return;
    t[k*2].lz=t[k*2+1].lz=t[k].lz;
    t[k].lz=-1;
}
void DownAll(int k){
    if(t[k].lz!=-1){
        for(int i=t[k].l;i<t[k].r;i++)f[i]=t[k].lz;//右端点不标记(只有被覆盖时才标记)
        return ;
    }
    DownAll(k*2);
    DownAll(k*2+1);
}
void update(int l,int r,int cr,int k){
    if(t[k].l>=t[k].r)return;
    if(l==t[k].l&&r==t[k].r){
        t[k].lz=cr;//标记
        return;
    }
    down(k);
    if(t[k].l+1>=t[k].r)return;

    int md=(t[k].l+t[k].r)/2;
    if(md>=r)update(l,r,cr,k*2);
    else if(l>md)update(l,r,cr,k*2+1);
    else{
        update(l,md,cr,k*2);
        update(md,r,cr,k*2+1);
    }
}
int main(){
    int i,j,k,n;
    int x,y;
	char ch;
    while(~scanf("%d",&n)){
        for(i=j=0;i<n;i++){
            scanf("%d%d %c",&x,&y,&ch);
			c[i]=(ch=='w');
            //l-1 或r+1 [)一闭一开离散化
            r[j]=j;a[j]=x;++j;
            r[j]=j;a[j]=y+1;++j;
        }
        
        sort(r,r+j,cmp);
        //去重
        p[r[0]]=k=0;//第0位的a，排第0位
        for(i=1;i<j;i++){
            if(a[r[i]]!=a[r[i-1]])a[r[++k]]=a[r[i]];
            p[r[i]]=k;
        }
        //for(i=0;i<j;i++)cout<<p[r[i]]<<" ";cout<<endl;
        //初始化树
        build(0,k,1);
        memset(f,0,sizeof(f));//0黑色1白色
        //update
        for(i=0,n+=n;i<n;i+=2){
            x=p[i];
            y=p[i+1];
            update(x,y,c[i/2],1);//区间更新
        }
        DownAll(1);
       // for(i=0;i<=k;i++)cout<<f[i]<<" ";cout<<endl;
        int tx,ty;
        for(i=x=y=0;i<=k;i++){
            if(f[i]==0)continue;
            tx=a[r[i]];
            while(f[i]==1)i++;
            ty=a[r[i]];
            if(ty-tx>y-x){x=tx;y=ty;}
        }
        if(x==y) puts("Oh, my god");
        else printf("%d %d
",x, y-1);

    }
return 0;
}