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:
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #include <iostream> 5 #include <string> 6 #include <vector> 7 #include <stack> 8 #include <bitset> 9 #include <cstdlib> 10 #include <cmath> 11 #include <set> 12 #include <list> 13 #include <deque> 14 #include <map> 15 #include <queue> 16 #include <fstream> 17 #include <cassert> 18 #include <unordered_map> 19 #include <unordered_set> 20 #include <cmath> 21 #include <sstream> 22 #include <time.h> 23 #include <complex> 24 #include <iomanip> 25 #include <tuple> 26 27 #define Max(a,b) ((a)>(b)?(a):(b)) 28 #define Min(a,b) ((a)<(b)?(a):(b)) 29 #define FOR(a,b,c) for (ll (a)=(b);(a)<(c);++(a)) 30 #define FORN(a,b,c) for (ll (a)=(b);(a)<=(c);++(a)) 31 #define DFOR(a,b,c) for (ll (a)=(b);(a)>=(c);--(a)) 32 #define FORSQ(a,b,c) for (ll (a)=(b);(a)*(a)<=(c);++(a)) 33 #define FORC(a,b,c) for (char (a)=(b);(a)<=(c);++(a)) 34 #define FOREACH(a,b) for (auto &(a) : (b)) 35 #define rep(i,n) FOR(i,0,n) 36 #define repn(i,n) FORN(i,1,n) 37 #define drep(i,n) DFOR(i,n-1,0) 38 #define drepn(i,n) DFOR(i,n,1) 39 #define MAX(a,b) a = Max(a,b) 40 #define MIN(a,b) a = Min(a,b) 41 #define SQR(x) ((LL)(x) * (x)) 42 #define Reset(a,b) memset(a,b,sizeof(a)) 43 #define fi first 44 #define se second 45 #define mp make_pair 46 #define pb push_back 47 #define all(v) v.begin(),v.end() 48 #define ALLA(arr,sz) arr,arr+sz 49 #define SIZE(v) (int)v.size() 50 #define SORT(v) sort(all(v)) 51 #define REVERSE(v) reverse(ALL(v)) 52 #define SORTA(arr,sz) sort(ALLA(arr,sz)) 53 #define REVERSEA(arr,sz) reverse(ALLA(arr,sz)) 54 #define PERMUTE next_permutation 55 #define TC(t) while(t--) 56 #define forever for(;;) 57 #define PINF 1000000000000 58 #define newline ' ' 59 60 #define test if(1)if(0)cerr 61 using namespace std; 62 using namespace std; 63 typedef vector<int> vi; 64 typedef vector<vi> vvi; 65 typedef pair<int,int> ii; 66 typedef pair<double,double> dd; 67 typedef pair<char,char> cc; 68 typedef vector<ii> vii; 69 typedef long long ll; 70 typedef unsigned long long ull; 71 typedef pair<ll, ll> l4; 72 73 74 75 76 77 int main() 78 { 79 ios::sync_with_stdio(false); 80 cin.tie(0); 81 int n; 82 cin >> n; 83 repn(i, n) 84 { 85 int x, y; cin >> x >> y; 86 cout << "Case #" << i << ": "; 87 if (y%2) cout << "BLACK "; 88 else cout << "WHITE "; 89 } 90 }
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：
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #include <iostream> 5 #include <string> 6 #include <vector> 7 #include <stack> 8 #include <bitset> 9 #include <cstdlib> 10 #include <cmath> 11 #include <set> 12 #include <list> 13 #include <deque> 14 #include <map> 15 #include <queue> 16 #include <fstream> 17 #include <cassert> 18 #include <unordered_map> 19 #include <unordered_set> 20 #include <cmath> 21 #include <sstream> 22 #include <time.h> 23 #include <complex> 24 #include <iomanip> 25 #include <tuple> 26 27 #define Max(a,b) ((a)>(b)?(a):(b)) 28 #define Min(a,b) ((a)<(b)?(a):(b)) 29 #define FOR(a,b,c) for (ll (a)=(b);(a)<(c);++(a)) 30 #define FORN(a,b,c) for (ll (a)=(b);(a)<=(c);++(a)) 31 #define DFOR(a,b,c) for (ll (a)=(b);(a)>=(c);--(a)) 32 #define FORSQ(a,b,c) for (ll (a)=(b);(a)*(a)<=(c);++(a)) 33 #define FORC(a,b,c) for (char (a)=(b);(a)<=(c);++(a)) 34 #define FOREACH(a,b) for (auto &(a) : (b)) 35 #define rep(i,n) FOR(i,0,n) 36 #define repn(i,n) FORN(i,1,n) 37 #define drep(i,n) DFOR(i,n-1,0) 38 #define drepn(i,n) DFOR(i,n,1) 39 #define MAX(a,b) a = Max(a,b) 40 #define MIN(a,b) a = Min(a,b) 41 #define SQR(x) ((LL)(x) * (x)) 42 #define Reset(a,b) memset(a,b,sizeof(a)) 43 #define fi first 44 #define se second 45 #define mp make_pair 46 #define pb push_back 47 #define all(v) v.begin(),v.end() 48 #define ALLA(arr,sz) arr,arr+sz 49 #define SIZE(v) (int)v.size() 50 #define SORT(v) sort(all(v)) 51 #define REVERSE(v) reverse(ALL(v)) 52 #define SORTA(arr,sz) sort(ALLA(arr,sz)) 53 #define REVERSEA(arr,sz) reverse(ALLA(arr,sz)) 54 #define PERMUTE next_permutation 55 #define TC(t) while(t--) 56 #define forever for(;;) 57 #define PINF 1000000000000 58 #define newline ' ' 59 60 #define test if(1)if(0)cerr 61 using namespace std; 62 using namespace std; 63 typedef vector<int> vi; 64 typedef vector<vi> vvi; 65 typedef pair<int,int> ii; 66 typedef pair<double,double> dd; 67 typedef pair<char,char> cc; 68 typedef vector<ii> vii; 69 typedef long long ll; 70 typedef unsigned long long ull; 71 typedef pair<ll, ll> l4; 72 73 74 int n, k; 75 int x[20], y[20]; 76 vector<int> bits; 77 int dp[(1<<15)]; 78 79 inline bool valid(int mid) 80 { 81 bits.clear(); 82 rep(i, n) rep(j, n) 83 { 84 int x1 = x[i], y1 = y[j]; 85 int x2 = x1+mid, y2 = y1+mid; 86 int ret = 0; 87 rep(k, n) if (x1 <= x[k] && x[k] <= x2 && y1 <= y[k] && y[k] <= y2) ret |= (1<<k); 88 bits.pb(ret); 89 } 90 Reset(dp, -1); 91 dp[0] = 0; 92 for (auto e : bits) 93 { 94 for (int cur = (1<<n)-1; cur >= 0; --cur) if (dp[cur] != -1) 95 { 96 int nxt = cur|e; 97 int nxtv = dp[cur]+1; 98 if (dp[nxt] == -1 || dp[nxt] > nxtv) dp[nxt] = nxtv; 99 } 100 } 101 return dp[(1<<n)-1] <= k; 102 } 103 int main() 104 { 105 ios::sync_with_stdio(false); 106 cin.tie(0); 107 int T; cin >> T; 108 repn(kase, T) 109 { 110 cin >> n >> k; 111 rep(i, n) cin >> x[i] >> y[i]; 112 int left = 0, right = 1e9, mid, ans; 113 while (left <= right) 114 { 115 mid = (left+right)>>1; 116 if (valid(mid)) 117 { 118 ans = mid; 119 right = mid-1; 120 } 121 else 122 left = mid+1; 123 } 124 cout << "Case #" << kase << ": " << ans << newline; 125 } 126 }
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:
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #include <iostream> 5 #include <string> 6 #include <vector> 7 #include <stack> 8 #include <bitset> 9 #include <cstdlib> 10 #include <cmath> 11 #include <set> 12 #include <list> 13 #include <deque> 14 #include <map> 15 #include <queue> 16 #include <fstream> 17 #include <cassert> 18 #include <unordered_map> 19 #include <unordered_set> 20 #include <cmath> 21 #include <sstream> 22 #include <time.h> 23 #include <complex> 24 #include <iomanip> 25 #include <tuple> 26 27 #define Max(a,b) ((a)>(b)?(a):(b)) 28 #define Min(a,b) ((a)<(b)?(a):(b)) 29 #define FOR(a,b,c) for (ll (a)=(b);(a)<(c);++(a)) 30 #define FORN(a,b,c) for (ll (a)=(b);(a)<=(c);++(a)) 31 #define DFOR(a,b,c) for (ll (a)=(b);(a)>=(c);--(a)) 32 #define FORSQ(a,b,c) for (ll (a)=(b);(a)*(a)<=(c);++(a)) 33 #define FORC(a,b,c) for (char (a)=(b);(a)<=(c);++(a)) 34 #define FOREACH(a,b) for (auto &(a) : (b)) 35 #define rep(i,n) FOR(i,0,n) 36 #define repn(i,n) FORN(i,1,n) 37 #define drep(i,n) DFOR(i,n-1,0) 38 #define drepn(i,n) DFOR(i,n,1) 39 #define MAX(a,b) a = Max(a,b) 40 #define MIN(a,b) a = Min(a,b) 41 #define SQR(x) ((LL)(x) * (x)) 42 #define Reset(a,b) memset(a,b,sizeof(a)) 43 #define fi first 44 #define se second 45 #define mp make_pair 46 #define pb push_back 47 #define all(v) v.begin(),v.end() 48 #define ALLA(arr,sz) arr,arr+sz 49 #define SIZE(v) (int)v.size() 50 #define SORT(v) sort(all(v)) 51 #define REVERSE(v) reverse(ALL(v)) 52 #define SORTA(arr,sz) sort(ALLA(arr,sz)) 53 #define REVERSEA(arr,sz) reverse(ALLA(arr,sz)) 54 #define PERMUTE next_permutation 55 #define TC(t) while(t--) 56 #define forever for(;;) 57 #define PINF 1000000000000 58 #define newline ' ' 59 60 #define test if(1)if(0)cerr 61 using namespace std; 62 using namespace std; 63 typedef vector<int> vi; 64 typedef vector<vi> vvi; 65 typedef pair<int,int> ii; 66 typedef pair<double,double> dd; 67 typedef pair<char,char> cc; 68 typedef vector<ii> vii; 69 typedef long long ll; 70 typedef unsigned long long ull; 71 typedef pair<ll, ll> l4; 72 73 const int mod = 9901; 74 int n, k; 75 vector<ii> E; 76 int pa[301]; 77 78 ll fact[301]; 79 int bcnt[(1<<15)]; 80 int total = 0; 81 int sz[301]; 82 int findpa(int id) 83 { 84 return id==pa[id]?id:pa[id]=findpa(pa[id]); 85 } 86 bool merge(int x, int y) 87 { 88 x = findpa(x), y = findpa(y); 89 if (x != y) 90 { 91 pa[x] = y; 92 sz[y] += sz[x]; 93 sz[x] = 0; 94 --total; 95 return true; 96 } 97 return false; 98 } 99 int deg[301]; 100 void init(int n) 101 { 102 repn(i, n) pa[i] = i; 103 repn(i, n) sz[i] = 1; 104 total = n; 105 Reset(deg, 0); 106 } 107 108 ll power(ll base, ll p) 109 { 110 if (!base) return 0; 111 ll ret = 1; 112 while (p) 113 { 114 if (p&1) ret = ret * base % mod; 115 base = base * base % mod; 116 p >>= 1; 117 } 118 return ret; 119 } 120 ll solve(int bit) 121 { 122 int error = 0; 123 init(n); 124 rep(i, k) if ((1<<i)&bit) 125 { 126 deg[E[i].first] += 1; 127 deg[E[i].second] += 1; 128 if (deg[E[i].first] > 2 || deg[E[i].second] > 2) return 0; 129 if (!merge(E[i].first, E[i].second)) 130 { 131 ++error; 132 } 133 } 134 if (error == 1 && total == 1) return 2; 135 if (error) return 0; 136 int path = 0; 137 repn(i, n) if (sz[i] > 1) path += 1; 138 return power(2, path) % mod * fact[total-1] % mod; 139 } 140 int main() 141 { 142 fact[0] = 1; 143 repn(i, 300) 144 fact[i] = fact[i-1] * i % mod; 145 bcnt[0] = 0; 146 repn(i, (1<<15)-1) 147 bcnt[i] = bcnt[i>>1] + (i&1); 148 ios::sync_with_stdio(false); 149 cin.tie(0); 150 int T; cin >> T; 151 repn(kase, T) 152 { 153 cin >> n >> k; 154 E.resize(k); 155 rep(i, k) 156 cin >> E[i].first >> E[i].second; 157 ll ret = 0; 158 rep(bit, (1<<k)) 159 { 160 int sign = 1; 161 if (bcnt[bit] % 2) sign = -1; 162 ll tmp = solve(bit); 163 ret = (ret + sign * tmp) % mod; 164 } 165 cout << "Case #" << kase << ": " << (ret*9902/2%mod+mod)%mod << newline; 166 167 } 168 }