HDU 1199 Color the Ball
Color the Ball
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5631 Accepted Submission(s): 1395
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 bSample Output
8 11线段树区间覆盖的问题，问题是数字范围太大，所以要离散化。当然也可以不用离散化作，即动态开点。这道题目的测试数据比较弱，所以我用了动态开点，要是数据强一点，可能会内存超限，#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <stdio.h>
#include <algorithm>
#include <math.h>
using namespace std;
typedef long long int LL;
const int maxn=2*1e3;
LL mmax[maxn*35];
LL lmax[maxn*35];
LL rmax[maxn*35];
int c[maxn*35];
int l[maxn*35];
int r[maxn*35];
LL ll[maxn*35];
LL rr[maxn*35];
int n;
int p;
int newnode()
{
l[p]=r[p]=-1;
lmax[p]=rmax[p]=mmax[p]=0;
ll[p]=rr[p]=-1;
return p++;
}
void pushdown(int node,LL L,LL R)
{
if(!(L==R))
{
if(l[node]==-1) l[node]=newnode();
if(r[node]==-1) r[node]=newnode();
}
if(c[node]!=-1)
{
if(c[node])
{
lmax[node]=rmax[node]=mmax[node]=R-L+1;
ll[node]=L,rr[node]=R;
}
else
{
lmax[node]=rmax[node]=mmax[node]=0;
ll[node]=rr[node]=-1;
}
c[l[node]]=c[r[node]]=c[node];
c[node]=-1;
}
}
void pushup(int node,LL L,LL R)
{
LL mid=(L+R)>>1;
pushdown(l[node],L,mid);
pushdown(r[node],mid+1,R);
if(lmax[l[node]]==(mid-L+1))
lmax[node]=lmax[l[node]]+lmax[r[node]];
else
lmax[node]=lmax[l[node]];
if(rmax[r[node]]==(R-mid))
rmax[node]=rmax[r[node]]+rmax[l[node]];
else
rmax[node]=rmax[r[node]];
mmax[node]=max(lmax[r[node]]+rmax[l[node]],max(mmax[l[node]],mmax[r[node]]));
if(mmax[node]==mmax[l[node]])
ll[node]=ll[l[node]],rr[node]=rr[l[node]];
else if(mmax[node]==lmax[r[node]]+rmax[l[node]])
ll[node]=mid-rmax[l[node]]+1,rr[node]=mid+lmax[r[node]];
else
ll[node]=ll[r[node]],rr[node]=rr[r[node]];
}
void update(int node,LL begin,LL end,LL L,LL R,int tag)
{
if(L<=begin&&end<=R)
{
c[node]=tag;
if(l[node]==-1) l[node]=newnode();
if(r[node]==-1) r[node]=newnode();
pushdown(node,begin,end);
return;
}
if(l[node]==-1) l[node]=newnode();
if(r[node]==-1) r[node]=newnode();
pushdown(node,begin,end);
int mid=(begin+end)>>1;
if(L<=mid) update(l[node],begin,mid,L,R,tag);
if(R>mid) update(r[node],mid+1,end,L,R,tag);
pushup(node,begin,end);
}
int main()
{
while(scanf("%d",&n)!=EOF)
{
p=0;
int root=newnode();
LL a,b;
char cc;
LL len=(1<<31)-1;
int tag;
update(root,1,len,1,len,0);
memset(c,-1,sizeof(c));
for(int i=1;i<=n;i++)
{
scanf("%lld%lld %c",&a,&b,&cc);
if(cc=='w') tag=1;
else tag=0;
update(root,1,len,a,b,tag);
}
if(ll[root]==-1)
printf("Oh, my god\n");
else
printf("%lld %lld\n",ll[root],rr[root]);
}
return 0;
}