Problem
You are a landlord who owns a building that is an R x C grid of apartments; each apartment is a unit square cell with four walls. You want to rent out N of
these apartments to tenants, with exactly one tenant per apartment, and leave the others empty. Unfortunately, all of your potential tenants are noisy, so whenever any two occupied apartments share a wall (and not just a corner), this will add one point of unhappiness to
the building. For example, a 2x2 building in which every apartment is occupied has four walls that are shared by neighboring tenants, and so the building‘s unhappiness score is 4.
If you place your N tenants optimally, what is the minimum unhappiness value for your building?
Input
The first line of the input gives the number of test cases, T. T lines follow; each contains three space-separated integers: R, C, and N.
Output
For each test case, output one line containing "Case #x: y", where x is the test case number (starting from 1) and y is the minimum possible unhappiness for the building.
Limits
1 ≤ T ≤ 1000.
0 ≤ N ≤ R*C.
Small dataset
1 ≤ R*C ≤ 16.
Large dataset
1 ≤ R*C ≤ 10000.
Sample
|
Input |
Output |
4 2 3 6 4 1 2 3 3 8 5 2 0 |
Case #1: 7 Case #2: 0 Case #3: 8 Case #4: 0 |
In Case #1, every room is occupied by a tenant and all seven internal walls have tenants on either side.
In Case #2, there are various ways to place the two tenants so that they do not share a wall. One is illustrated below.
In Case #3, the optimal strategy is to place the eight tenants in a ring, leaving the middle apartment unoccupied.
Here are illustrations of sample cases 1-3. Each red wall adds a point of unhappiness.
分RC为奇偶，奇奇，偶偶讨论
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXR (10000+10)
typedef long long ll;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int main()
{
// freopen("B-large.in","r",stdin);
// freopen("B-large.out","w",stdout);
int T;
cin>>T;
For(kcase,T)
{
ll r,c,k;
cin>>r>>c>>k;
if (r<c) swap(r,c);
if (r%2==1&&c%2==0) swap(r,c);
ll t1=r*c/2,S=r*c;
ll t2=r*c-t1;
if (t1>t2) swap(t1,t2);
ll ans=0;
ll k2=S-k;
ll allsid=(r-1)*c+(c-1)*r;
if (k<=t2) ans=0;
else if (c==1)
{
ans=2*k2;
ans=allsid-ans;
}
else
{
if (r%2==0&&c%2==0)
{
ll p4=(r-2)*(c-2)/2,p3=r+c-4,p2=2;
if (k2<=p4)
ans=k2*4;
else if (k2<=p4+p3)
ans=p4*4+(k2-p4)*3;
else
ans=p4*4+p3*3+(k2-p4-p3)*2;
ans=allsid-ans;
}
else if (r%2==1&&c%2==1)
{
ll p4=(r-2)*(c-2)/2+1,p3=(r-2)/2*2+(c-2)/2*2,p2=4;
if (k2<=p4)
ans=k2*4;
else if (k2<=p4+p3)
ans=p4*4+(k2-p4)*3;
else
ans=p4*4+p3*3+(k2-p4-p3)*2;
ans=allsid-ans;
{
ll p4=(r-2)*(c-2)/2,p3=(r-2+1)/2*2+(c-2+1)/2*2,p2=0;
ll ans2=0;
if (k2<=p4)
ans2=k2*4;
else if (k2<=p4+p3)
ans2=p4*4+(k2-p4)*3;
else
ans2=p4*4+p3*3+(k2-p4-p3)*2;
ans2=allsid-ans2;
ans=min(ans,ans2);
}
}
else if (r%2==0&&c%2==1)
{
ll p4=(r-2)*(c-2)/2,p3=(r-2)+(c-2),p2=2;
if (k2<=p4)
ans=k2*4;
else if (k2<=p4+p3)
ans=p4*4+(k2-p4)*3;
else
ans=p4*4+p3*3+(k2-p4-p3)*2;
ans=allsid-ans;
}
}
printf("Case #%d: %lld\n",kcase,ans);
}
return 0;
}