SRM480

250pt:

题意：给定n个网站，以及n个网站的关键词，还有一个危险词库。如果一个网站的关键词中>=th的危险词，那么这个网站便是危险的。同时，他的所有关键词加入危险词库。问，有多少个危险网站。

思路：直接模拟。

code:

 #line 7 "InternetSecurity.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 InternetSecurity
 {
         public:
         vector<string> A[];
         bool v[];
         vector <string> determineWebsite(vector <string> wbs, vector <string> key, vector <string> dan, int th)
         {
                for (int i = ; i < key.size(); ++i){
                      string tmp;
                      stringstream ss;
                      A[i].clear();
                      ss << key[i];
                      while (ss >> tmp)  A[i].PB(tmp);
                   //   cout << A[i][A[i].size() - 1] << endl;
                }
                set<string> S;
                memset(v, , sizeof(v));
                for (int i = ; i < dan.size(); ++i) S.insert(dan[i]);
                bool keepLooking = true;
                while (keepLooking){
                      keepLooking = false;
                      for (int i = ; i < key.size(); ++i)
                         if  (!v[i]){
                               int cnt = ;
                               for (int j = ; j < A[i].size(); ++j)
                                  if (S.find(A[i][j]) != S.end()) ++cnt;
                               if (cnt >= th){
                                    keepLooking = true;
                                    v[i] = true;
                                    for (int j = ; j < A[i].size(); ++j)
                                        S.insert(A[i][j]);
                               }
                         }
                }
                vector<string> ans;
                ans.clear();
                for (int i = ; i < key.size(); ++i)
                    if (v[i]) ans.PB(wbs[i]);
                return ans;
         }
 };

450pt:

题意：给定一个DAG的客户机及服务器之间的关系，服务器只有入边。现在求在那些边上装一些安全设置，使得所有客户机到服务器至少都有经过一个安全设置的边。

思路：因为是DAG，那么直接对原图进行一边搜索，记录直接点的状态，转移到根状态

code:

 #line 7 "NetworkSecurity.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 NetworkSecurity
 {
         public:
         int n, m, ans;
         bool vis[];
         vector<string> S, C;
         long long S1[];
         void dfs(int u){
              if (vis[u]) return;
              vis[u] = true;
              S1[u] = ;
              for (int i = ; i < C[u].size(); ++i)
                   if (C[u][i] == 'Y'){
                         dfs(i);
                         S1[u] |= S1[i];
                   }
              for (int i = ; i < S[u].size(); ++i)
                  if (S[u][i] == 'Y'){
                       if (!((1LL << i) & S1[u])){
                           ++ans;
                           S1[u] |= (1LL << i);
                       }
                  }
         }
         int secureNetwork(vector <string> clientCable, vector <string> serverCable)
         {
                 ans = ;
                 S = serverCable;
                 C = clientCable;
                 memset(vis, , sizeof(vis));
                 for (int i = ; i < S.size(); ++i)
                      if(!vis[i]) dfs(i);
                 return ans;
         }
 };