GCJ 2015r1b (Noisy neighbors-category discussion)

Problem

You be a landlord who owns a building a R x C grid of apartments; Each apartment are a unit square cell with four walls. You want to rent off N of these apartments to tenants, with exactly one tenant per apartment, and leave the other s empty. Unfortunately, all of your potential tenants is noisy, so whenever any of the occupied apartments share a wall (and not just A corner), this would add one point of unhappiness to the building. For example, a 2×2 building in which every apartment is occupied have four walls that was shared by neighboring tenants, an D-the building ' s unhappiness score is 4.

If you place your N tenants optimally, what's 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 th E 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
42 3 1 3 2 0	case #1:7Case #2:0Case #3:8Case #4:0

in the case #1, every the occupied by a tenant and all seven internal walls has tenants on either side.

in case #2, there is various ways to place the and the tenants so and they do not share a wall. One is illustrated below.

in case #3, the optimal strategy are to place the eight tenants in a ring, leaving the middle apartment unoccupied.

Here is illustrations of sample cases 1-3. Each red wall adds a point of unhappiness.

Sub RC for odd, odd , even even discussion

#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)/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&LT;=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&LT;=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&LT;=P4) Ans=k2*4;else if ( K2&LT;=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&LT;=P4) Ans2=k2*4;else if (K2&LT;=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&LT;=P4) Ans=k2*4;else if(K2&LT;=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;}

GCJ 2015r1b (Noisy neighbors-category discussion)