GCJ Practice Contest

https://codejam.withgoogle.com/codejam/contest/32004/dashboard

A. Old Magician

方法：

打表找规律，发现结果只和black个数的奇偶性有关。

code:

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
#include <stack>
#include <bitset>
#include <cstdlib>
#include <cmath>
#include <set>
#include <list>
#include <deque>
#include <map>
#include <queue>
#include <fstream>
#include <cassert>
#include <unordered_map>
#include <unordered_set>
#include <cmath>
#include <sstream>
#include <time.h>
#include <complex>
#include <iomanip>
#include <tuple>

#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define FOR(a,b,c) for (ll (a)=(b);(a)<(c);++(a))
#define FORN(a,b,c) for (ll (a)=(b);(a)<=(c);++(a))
#define DFOR(a,b,c) for (ll (a)=(b);(a)>=(c);--(a))
#define FORSQ(a,b,c) for (ll (a)=(b);(a)*(a)<=(c);++(a))
#define FORC(a,b,c) for (char (a)=(b);(a)<=(c);++(a))
#define FOREACH(a,b) for (auto &(a) : (b))
#define rep(i,n) FOR(i,0,n)
#define repn(i,n) FORN(i,1,n)
#define drep(i,n) DFOR(i,n-1,0)
#define drepn(i,n) DFOR(i,n,1)
#define MAX(a,b) a = Max(a,b)
#define MIN(a,b) a = Min(a,b)
#define SQR(x) ((LL)(x) * (x))
#define Reset(a,b) memset(a,b,sizeof(a))
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define all(v) v.begin(),v.end()
#define ALLA(arr,sz) arr,arr+sz
#define SIZE(v) (int)v.size()
#define SORT(v) sort(all(v))
#define REVERSE(v) reverse(ALL(v))
#define SORTA(arr,sz) sort(ALLA(arr,sz))
#define REVERSEA(arr,sz) reverse(ALLA(arr,sz))
#define PERMUTE next_permutation
#define TC(t) while(t--)
#define forever for(;;)
#define PINF 1000000000000
#define newline '
'

#define test if(1)if(0)cerr
using namespace std;
using namespace std;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int,int> ii;
typedef pair<double,double> dd;
typedef pair<char,char> cc;
typedef vector<ii> vii;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> l4;





int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n;
    cin >> n;
    repn(i, n)
    {
        int x, y;   cin >> x >> y;
        cout << "Case #" << i << ": ";
        if (y%2) cout << "BLACK
";
        else cout << "WHITE
";
    }
}

B. Square Fields

方法：

最优解一定可以把所有的正方形移动，使得每个正方形至少有两条边上有点。首先二分长度，根据所有可能的横纵坐标建立n^2个正方形，对于每个正方形，可以用一个不超过2^15的整数表示哪些点可以被其覆盖，然后只需对这n^2个正方形进行一次背包，即可求出cover所有点所需的正方形的最小数目m。如果m <＝ k， 则可行。

code：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
#include <stack>
#include <bitset>
#include <cstdlib>
#include <cmath>
#include <set>
#include <list>
#include <deque>
#include <map>
#include <queue>
#include <fstream>
#include <cassert>
#include <unordered_map>
#include <unordered_set>
#include <cmath>
#include <sstream>
#include <time.h>
#include <complex>
#include <iomanip>
#include <tuple>

#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define FOR(a,b,c) for (ll (a)=(b);(a)<(c);++(a))
#define FORN(a,b,c) for (ll (a)=(b);(a)<=(c);++(a))
#define DFOR(a,b,c) for (ll (a)=(b);(a)>=(c);--(a))
#define FORSQ(a,b,c) for (ll (a)=(b);(a)*(a)<=(c);++(a))
#define FORC(a,b,c) for (char (a)=(b);(a)<=(c);++(a))
#define FOREACH(a,b) for (auto &(a) : (b))
#define rep(i,n) FOR(i,0,n)
#define repn(i,n) FORN(i,1,n)
#define drep(i,n) DFOR(i,n-1,0)
#define drepn(i,n) DFOR(i,n,1)
#define MAX(a,b) a = Max(a,b)
#define MIN(a,b) a = Min(a,b)
#define SQR(x) ((LL)(x) * (x))
#define Reset(a,b) memset(a,b,sizeof(a))
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define all(v) v.begin(),v.end()
#define ALLA(arr,sz) arr,arr+sz
#define SIZE(v) (int)v.size()
#define SORT(v) sort(all(v))
#define REVERSE(v) reverse(ALL(v))
#define SORTA(arr,sz) sort(ALLA(arr,sz))
#define REVERSEA(arr,sz) reverse(ALLA(arr,sz))
#define PERMUTE next_permutation
#define TC(t) while(t--)
#define forever for(;;)
#define PINF 1000000000000
#define newline '
'

#define test if(1)if(0)cerr
using namespace std;
using namespace std;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int,int> ii;
typedef pair<double,double> dd;
typedef pair<char,char> cc;
typedef vector<ii> vii;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> l4;


int n, k;
int x[20], y[20];
vector<int> bits;
int dp[(1<<15)];

inline bool valid(int mid)
{
    bits.clear();
    rep(i, n) rep(j, n)
    {
        int x1 = x[i], y1 = y[j];
        int x2 = x1+mid, y2 = y1+mid;
        int ret = 0;
        rep(k, n) if (x1 <= x[k] && x[k] <= x2 && y1 <= y[k] && y[k] <= y2) ret |= (1<<k);
        bits.pb(ret);
    }
    Reset(dp, -1);
    dp[0] = 0;
    for (auto e : bits)
    {
        for (int cur = (1<<n)-1; cur >= 0; --cur) if (dp[cur] != -1)
        {
            int nxt = cur|e;
            int nxtv = dp[cur]+1;
            if (dp[nxt] == -1 || dp[nxt] > nxtv) dp[nxt] = nxtv;
        }
    }
    return dp[(1<<n)-1] <= k;
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int T;  cin >> T;
    repn(kase, T)
    {
        cin >> n >> k;
        rep(i, n) cin >> x[i] >> y[i];
        int left = 0, right = 1e9, mid, ans;
        while (left <= right)
        {
            mid = (left+right)>>1;
            if (valid(mid))
            {
                ans = mid;
                right = mid-1;
            }
            else
                left = mid+1;
        }
        cout << "Case #" << kase << ": " << ans << newline;
    }
}

C. Cycles

方法：

设禁止边集为K，则对K的每一个subset S，我们求出使用S中所有边构成的hamilton cycle有多少个，然后再利用容斥原理求解即可。

对于一个S，求出包含S中所有边的hamilton cycle的个数，可以这样做：首先，对于S产生的诱导子图，如果有一个点度数超过2，那么肯定不存在合理的hamilton 路，返回0个；同时，如果存在环，只有当只存在一个环而且该环本身就是hamilton路的时候，返回1，否则返回0。对于其他的情况，假设我们的到了path条路径，那么我们要先对这些路径定向，有2^path种方式，然后把这些路径看成点，再在所有的点中找出一条hamilton路；如果剩下有total个点（包含了path条路径），那么hamilton路有(total-1)!／2条。所以返回 2^path * (total-1)!/2。（cyclic permutation 的个数为(total-1)!， 同时要考虑方向去重）。

code:

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
#include <stack>
#include <bitset>
#include <cstdlib>
#include <cmath>
#include <set>
#include <list>
#include <deque>
#include <map>
#include <queue>
#include <fstream>
#include <cassert>
#include <unordered_map>
#include <unordered_set>
#include <cmath>
#include <sstream>
#include <time.h>
#include <complex>
#include <iomanip>
#include <tuple>

#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define FOR(a,b,c) for (ll (a)=(b);(a)<(c);++(a))
#define FORN(a,b,c) for (ll (a)=(b);(a)<=(c);++(a))
#define DFOR(a,b,c) for (ll (a)=(b);(a)>=(c);--(a))
#define FORSQ(a,b,c) for (ll (a)=(b);(a)*(a)<=(c);++(a))
#define FORC(a,b,c) for (char (a)=(b);(a)<=(c);++(a))
#define FOREACH(a,b) for (auto &(a) : (b))
#define rep(i,n) FOR(i,0,n)
#define repn(i,n) FORN(i,1,n)
#define drep(i,n) DFOR(i,n-1,0)
#define drepn(i,n) DFOR(i,n,1)
#define MAX(a,b) a = Max(a,b)
#define MIN(a,b) a = Min(a,b)
#define SQR(x) ((LL)(x) * (x))
#define Reset(a,b) memset(a,b,sizeof(a))
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define all(v) v.begin(),v.end()
#define ALLA(arr,sz) arr,arr+sz
#define SIZE(v) (int)v.size()
#define SORT(v) sort(all(v))
#define REVERSE(v) reverse(ALL(v))
#define SORTA(arr,sz) sort(ALLA(arr,sz))
#define REVERSEA(arr,sz) reverse(ALLA(arr,sz))
#define PERMUTE next_permutation
#define TC(t) while(t--)
#define forever for(;;)
#define PINF 1000000000000
#define newline '
'

#define test if(1)if(0)cerr
using namespace std;
using namespace std;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int,int> ii;
typedef pair<double,double> dd;
typedef pair<char,char> cc;
typedef vector<ii> vii;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> l4;

const int mod = 9901;
int n, k;
vector<ii> E;
int pa[301];

ll fact[301];
int bcnt[(1<<15)];
int total = 0;
int sz[301];
int findpa(int id)
{
    return id==pa[id]?id:pa[id]=findpa(pa[id]);
}
bool merge(int x, int y)
{
    x = findpa(x), y = findpa(y);
    if (x != y)
    {
        pa[x] = y;
        sz[y] += sz[x];
        sz[x] = 0;
        --total;
        return true;
    }
    return false;
}
int deg[301];
void init(int n)
{
    repn(i, n) pa[i] = i;
    repn(i, n) sz[i] = 1;
    total = n;
    Reset(deg, 0);
}

ll power(ll base, ll p)
{
    if (!base) return 0;
    ll ret = 1;
    while (p)
    {
        if (p&1) ret = ret * base % mod;
        base = base * base % mod;
        p >>= 1;
    }
    return ret;
}
ll solve(int bit)
{
    int error = 0;
    init(n);
    rep(i, k) if ((1<<i)&bit)
    {
        deg[E[i].first] += 1;
        deg[E[i].second] += 1;
        if (deg[E[i].first] > 2 || deg[E[i].second] > 2) return 0;
        if (!merge(E[i].first, E[i].second))
        {
            ++error;
        }
    }
    if (error == 1 && total == 1) return 2;
    if (error) return 0;
    int path = 0;
    repn(i, n) if (sz[i] > 1) path += 1;
    return power(2, path) % mod * fact[total-1] % mod;
}
int main()
{
    fact[0] = 1;
    repn(i, 300)
    fact[i] = fact[i-1] * i % mod;
    bcnt[0] = 0;
    repn(i, (1<<15)-1)
    bcnt[i] = bcnt[i>>1] + (i&1);
    ios::sync_with_stdio(false);
    cin.tie(0);
    int T;  cin >> T;
    repn(kase, T)
    {
        cin >> n >> k;
        E.resize(k);
        rep(i, k)
        cin >> E[i].first >> E[i].second;
        ll ret = 0;
        rep(bit, (1<<k))
        {
            int sign = 1;
            if (bcnt[bit] % 2) sign = -1;
            ll tmp = solve(bit);
            ret = (ret + sign * tmp) % mod;
        }
        cout << "Case #" << kase << ": " << (ret*9902/2%mod+mod)%mod << newline;
        
    }
}