250pt:
题意：给定一块蜂巢状的N*M矩阵，每块六边形和周围6个六边形相邻，现在告诉你哪些是陆地，哪些是水，问水陆交界处的长度。
思路：直接模拟
code：
1 #line 7 "Islands.cpp" 2 #include <cstdlib> 3 #include <cctype> 4 #include <cstring> 5 #include <cstdio> 6 #include <cmath> 7 #include <algorithm> 8 #include <vector> 9 #include <string> 10 #include <iostream> 11 #include <sstream> 12 #include <map> 13 #include <set> 14 #include <queue> 15 #include <stack> 16 #include <fstream> 17 #include <numeric> 18 #include <iomanip> 19 #include <bitset> 20 #include <list> 21 #include <stdexcept> 22 #include <functional> 23 #include <utility> 24 #include <ctime> 25 using namespace std; 26 27 #define PB push_back 28 #define MP make_pair 29 30 #define REP(i,n) for(i=0;i<(n);++i) 31 #define FOR(i,l,h) for(i=(l);i<=(h);++i) 32 #define FORD(i,h,l) for(i=(h);i>=(l);--i) 33 34 typedef vector<int> VI; 35 typedef vector<string> VS; 36 typedef vector<double> VD; 37 typedef long long LL; 38 typedef pair<int,int> PII; 39 40 41 class Islands 42 { 43 public: 44 int beachLength(vector <string> S) 45 { 46 int ans = 0; 47 int n = S.size(), m = S[0].size(), p; 48 for (int i = 0; i < n; ++i) 49 for (int j = 0; j < m; ++j){ 50 if (j > 0 && S[i][j] == '#' && S[i][j-1] == '.') ++ans; 51 if (j > 0 && S[i][j] == '.' && S[i][j-1] == '#') ++ans; 52 if (i == 0) continue; 53 if (i & 1) p = j; 54 else p = j - 1; 55 if (p >= 0 && S[i][j] == '.' && S[i-1][p] == '#') ++ans; 56 if (p >= 0 && S[i][j] == '#' && S[i-1][p] == '.') ++ans; 57 if (p + 1 < m && S[i][j] == '.' && S[i-1][p+1] == '#') ++ans; 58 if (p + 1 < m && S[i][j] == '#' && S[i-1][p+1] == '.') ++ans; 59 } 60 return ans; 61 } 62 };
500pt:
题意:给定最多200个正整数，问最多能组成多少个勾股数对。勾股数对定义：对于互质数对(a, b)，存在c,使得a*a+b*b=c*c,
思路：隐蔽的二分图。
很明显先求出勾股数对，那么接下来便是一个最大匹配了。
不过，是二分图吗？
对于奇数a与奇数b，由于互质，那么a^2+b^2必定是2的倍数，必然不存在c
对于偶数a与偶数b,不互质显然不存在。
所以是个二分图，直接匹配即可
ｃｏｄｅ：
1 #line 7 "PythTriplets.cpp" 2 #include <cstdlib> 3 #include <cctype> 4 #include <cstring> 5 #include <cstdio> 6 #include <cmath> 7 #include <algorithm> 8 #include <vector> 9 #include <string> 10 #include <iostream> 11 #include <sstream> 12 #include <map> 13 #include <set> 14 #include <queue> 15 #include <stack> 16 #include <fstream> 17 #include <numeric> 18 #include <iomanip> 19 #include <bitset> 20 #include <list> 21 #include <stdexcept> 22 #include <functional> 23 #include <utility> 24 #include <ctime> 25 using namespace std; 26 #define PB push_back 27 #define MP make_pair 28 #define REP(i,n) for(i=0;i<(n);++i) 29 #define FOR(i,l,h) for(i=(l);i<=(h);++i) 30 #define FORD(i,h,l) for(i=(h);i>=(l);--i) 31 #define M0(a) memset(a, 0, sizeof(a)) 32 typedef vector<int> VI; 33 typedef vector<string> VS; 34 typedef vector<double> VD; 35 typedef long long LL; 36 typedef pair<int,int> PII; 37 int match[210], g[210][210]; 38 int n, m; 39 bool vis[210]; 40 class PythTriplets 41 { 42 public: 43 vector<int> v1, v2; 44 void makePoint(string &S){ 45 int tmp = 0; 46 for (int i = 0; i < S.size(); ++i) 47 if (S[i] == ' '){ 48 if (tmp == 0) continue; 49 if (tmp & 1) v1.push_back(tmp); 50 else v2.push_back(tmp); 51 tmp = 0; 52 } else tmp = tmp * 10 + S[i] - 48; 53 if (tmp > 0){ 54 if (tmp & 1) v1.push_back(tmp); 55 else v2.push_back(tmp); 56 } 57 } 58 LL gcd(LL a, LL b){ 59 return b ? gcd(b, a % b) : a; 60 } 61 bool ok(long long a, long long b){ 62 if (gcd(a, b) > 1) return false; 63 long long c = floor(sqrt(a * a + b * b + .0)); 64 if (c * c < a * a + b * b) ++c; 65 if (a * a + b * b == c * c) return true; 66 return false; 67 } 68 bool search_path(int u){ 69 for (int v = 0; v < m; ++v) if (g[u][v] && !vis[v]){ 70 vis[v] = true; 71 if (match[v] == -1 || search_path(match[v])){ 72 match[v] = u; 73 return true; 74 } 75 } 76 return false; 77 } 78 int findMax(vector <string> stick) 79 { 80 v1.clear(); 81 v2.clear(); 82 string S = accumulate(stick.begin(), stick.end(), string("")); 83 makePoint(S); 84 n = v1.size(); 85 m = v2.size(); 86 M0(g); 87 for (int i = 0; i < n; ++i) 88 for (int j = 0; j < m; ++j) 89 if (ok(v1[i], v2[j])) g[i][j] = true;//, cout << v1[i] << " " << v2[j] << endl; 90 int ans = 0; 91 memset(match, -1, sizeof(match)); 92 for (int i = 0; i < n; ++i){ 93 M0(vis); 94 if (search_path(i)) ++ ans; 95 } 96 return ans; 97 } 98 };