Constructing Roads In JGShining's Kingdom
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16047 Accepted Submission(s): 4580
Half
of these cities are rich in resource (we call them rich cities) while
the others are short of resource (we call them poor cities). Each poor
city is short of exactly one kind of resource and also each rich city is
rich in exactly one kind of resource. You may assume no two poor cities
are short of one same kind of resource and no two rich cities are rich
in one same kind of resource.
With the development of industry,
poor cities wanna import resource from rich ones. The roads existed are
so small that they're unable to ensure the heavy trucks, so new roads
should be built. The poor cities strongly BS each other, so are the rich
ones. Poor cities don't wanna build a road with other poor ones, and
rich ones also can't abide sharing an end of road with other rich ones.
Because of economic benefit, any rich city will be willing to export
resource to any poor one.
Rich citis marked from 1 to n are located in Line I and poor ones marked from 1 to n are located in Line II.
The
location of Rich City 1 is on the left of all other cities, Rich City 2
is on the left of all other cities excluding Rich City 1, Rich City 3
is on the right of Rich City 1 and Rich City 2 but on the left of all
other cities ... And so as the poor ones.
But as you know, two
crossed roads may cause a lot of traffic accident so JGShining has
established a law to forbid constructing crossed roads.
For example, the roads in Figure I are forbidden.
In
order to build as many roads as possible, the young and handsome king
of the kingdom - JGShining needs your help, please help him. ^_^
test case will begin with a line containing an integer n(1 ≤ n ≤
500,000). Then n lines follow. Each line contains two integers p and r
which represents that Poor City p needs to import resources from Rich
City r. Process to the end of file.
You should tell JGShining what's the maximal number of road(s) can be built.
1 2
2 1
3
1 2
2 3
3 1
My king, at most 1 road can be built.
Case 2:
My king, at most 2 roads can be built.
Huge input, scanf is recommended.
//#define LOCAL
#include<cstdio>
#include<cstring>
const int maxn=;
const int inf=0x3f3f3f3f;
int str[maxn],dp[maxn],sac[maxn];
int res; int binary(int v,int n)
{
int ll=,rr=n,mid;
while(ll<=rr)
{
mid=ll+(rr-ll)/;
if(sac[mid]<=v&&sac[mid]!=-)
ll=mid+;
else
rr=mid-;
}
return ll;
}
void LIS(int n)
{
res=;
// for(int i=1;i<=n;i++)
// sac[i]=inf;
memset(sac,-,sizeof(int)*(n+));
for(int i=;i<=n;i++)
{
dp[i]=binary(str[i],res);
if(res<dp[i])
res=dp[i];
if(str[i]<sac[dp[i]]||sac[dp[i]]==-)
sac[dp[i]]=str[i];
}
}
int main()
{
#ifdef LOCAL
freopen("test.in","r",stdin);
#endif int n,i,p,r,ct=;
while(scanf("%d",&n)!=EOF)
{
for( i=;i<=n;i++){
scanf("%d%d",&p,&r);
str[p]=r;
}
LIS(n);
printf("Case %d:\n",ct++);
if(res==)
printf("My king, at most 1 road can be built.\n");
else
printf("My king, at most %d roads can be built.\n",res);
printf("\n");
}
return ;
}