SRM477

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：

 #line 7 "Islands.cpp"
 #include <cstdlib>
 #include <cctype>
 #include <cstring>
 #include <cstdio>
 #include <cmath>
 #include <algorithm>
 #include <vector>
 #include <string>
 #include <iostream>
 #include <sstream>
 #include <map>
 #include <set>
 #include <queue>
 #include <stack>
 #include <fstream>
 #include <numeric>
 #include <iomanip>
 #include <bitset>
 #include <list>
 #include <stdexcept>
 #include <functional>
 #include <utility>
 #include <ctime>
 using namespace std;

 #define PB push_back
 #define MP make_pair

 #define REP(i,n) for(i=0;i<(n);++i)
 #define FOR(i,l,h) for(i=(l);i<=(h);++i)
 #define FORD(i,h,l) for(i=(h);i>=(l);--i)

 typedef vector<int> VI;
 typedef vector<string> VS;
 typedef vector<double> VD;
 typedef long long LL;
 typedef pair<int,int> PII;

 class Islands
 {
         public:
         int beachLength(vector <string> S)
         {
              int ans = ;
              int n = S.size(), m = S[].size(), p;
              for (int i = ; i < n; ++i)
                 for (int j = ; j < m; ++j){
                      if (j >  && S[i][j] == '#' && S[i][j-] == '.') ++ans;
                      if (j >  && S[i][j] == '.' && S[i][j-] == '#') ++ans;
                      if (i == ) continue;
                      if (i & ) p = j;
                      else p = j - ;
                      if (p >=  && S[i][j] == '.' && S[i-][p] == '#') ++ans;
                      if (p >=  && S[i][j] == '#' && S[i-][p] == '.') ++ans;
                      if (p +  < m && S[i][j] == '.' && S[i-][p+] == '#') ++ans;
                      if (p +  < m && S[i][j] == '#' && S[i-][p+] == '.') ++ans;
                 }
              return ans;
         }
   };   

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：

 #line 7 "PythTriplets.cpp"
 #include <cstdlib>
 #include <cctype>
 #include <cstring>
 #include <cstdio>
 #include <cmath>
 #include <algorithm>
 #include <vector>
 #include <string>
 #include <iostream>
 #include <sstream>
 #include <map>
 #include <set>
 #include <queue>
 #include <stack>
 #include <fstream>
 #include <numeric>
 #include <iomanip>
 #include <bitset>
 #include <list>
 #include <stdexcept>
 #include <functional>
 #include <utility>
 #include <ctime>
 using namespace std;
 #define PB push_back
 #define MP make_pair
 #define REP(i,n) for(i=0;i<(n);++i)
 #define FOR(i,l,h) for(i=(l);i<=(h);++i)
 #define FORD(i,h,l) for(i=(h);i>=(l);--i)
 #define M0(a) memset(a, 0, sizeof(a))
 typedef vector<int> VI;
 typedef vector<string> VS;
 typedef vector<double> VD;
 typedef long long LL;
 typedef pair<int,int> PII;
 int match[], g[][];
 int n, m;
 bool vis[];
 class PythTriplets
 {
         public:
         vector<int> v1, v2;
         void makePoint(string &S){
              int tmp = ;
              for (int i = ; i < S.size(); ++i)
                 if (S[i] == ' '){
                      if (tmp == ) continue;
                      if (tmp & ) v1.push_back(tmp);
                      else v2.push_back(tmp);
                      tmp = ;
                 } else tmp = tmp *  + S[i] - ;
              if (tmp > ){
                      if (tmp & ) v1.push_back(tmp);
                      else v2.push_back(tmp);
              }
         }
         LL gcd(LL a, LL b){
               return b ? gcd(b, a % b) : a;
         }
         bool ok(long long a, long long b){
               if (gcd(a, b) > ) return false;
               long long c = floor(sqrt(a * a + b * b + .));
               if (c * c < a * a + b * b) ++c;
               if (a * a + b * b == c * c) return true;
               return false;
         }
         bool search_path(int u){
              for (int v = ; v < m; ++v) if (g[u][v] && !vis[v]){
                    vis[v] = true;
                    if (match[v] == - || search_path(match[v])){
                          match[v] = u;
                          return true;
                    }
              }
              return false;
         }
         int findMax(vector <string> stick)
         {
             v1.clear();
             v2.clear();
             string S = accumulate(stick.begin(), stick.end(), string(""));
             makePoint(S);
             n = v1.size();
             m = v2.size();
             M0(g);
             for (int i = ; i < n; ++i)
                for (int j = ; j < m; ++j)
                   if (ok(v1[i], v2[j])) g[i][j] = true;//, cout << v1[i] <<  " " << v2[j] << endl;
             int ans = ;
             memset(match, -, sizeof(match));
             for (int i = ; i < n; ++i){
                   M0(vis);
                   if (search_path(i)) ++ ans;
             }
             return ans;
         }
 };