SRM 407(1-250pt, 1-500pt)

DIV1 250pt

题意：每个员工可以有几个直系上司，也可以有几个直系下属。没有直系下属的人工资为1，有直系下属的人工资为所有直系下属工资之和。求所有人工资之和。人数 <= 50。

解法：直接dfs一遍求出所有人工资就好。(官方题解说该问题满足拓补序，用dfs做拓补序再求)

tag:dfs

 // BEGIN CUT HERE
 /*
  * Author:  plum rain
  * score :
  */
 /*

  */
 // END CUT HERE
 #line 11 "CorporationSalary.cpp"
 #include <sstream>
 #include <stdexcept>
 #include <functional>
 #include <iomanip>
 #include <numeric>
 #include <fstream>
 #include <cctype>
 #include <iostream>
 #include <cstdio>
 #include <vector>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <cstdlib>
 #include <set>
 #include <queue>
 #include <bitset>
 #include <list>
 #include <string>
 #include <utility>
 #include <map>
 #include <ctime>
 #include <stack>

 using namespace std;

 #define clr0(x) memset(x, 0, sizeof(x))
 #define clr1(x) memset(x, -1, sizeof(x))
 #define pb push_back
 #define sz(v) ((int)(v).size())
 #define all(t) t.begin(),t.end()
 #define zero(x) (((x)>0?(x):-(x))<eps)
 #define out(x) cout<<#x<<":"<<(x)<<endl
 #define tst(a) cout<<a<<" "
 #define tst1(a) cout<<#a<<endl
 #define CINBEQUICKER std::ios::sync_with_stdio(false)

 typedef vector<int> vi;
 typedef vector<string> VS;
 typedef vector<double> VD;
 typedef pair<int, int> pii;
 typedef long long int64;

 const double eps = 1e-;
 const double PI = atan(1.0)*;
 const int inf =  / ;

 vi pat[];
 bool son[];
 int64 an[];

 void dfs (int x)
 {
     int64 &ret = an[x];
     if (ret) return;

     int n = sz(pat[x]);
     if (!n){
         an[x] = ; return;
     }

     for (int i = ; i < n; ++ i){
         dfs (pat[x][i]);
         ret += an[pat[x][i]];
     }
 }

 class CorporationSalary
 {
     public:
         long long totalSalary(vector <string> v){
             clr0 (an); clr0 (son);
             int n = sz(v);
             for (int i = ; i < n; ++ i)
                 pat[i].clear();

             for (int i = ; i < n; ++ i)
                 for (int j = ; j < n; ++ j)
                     if (v[i][j] == 'Y')
                         pat[i].pb (j), son[j] = ;

             for (int i = ; i < n; ++ i)
                 if (!son[i]) dfs (i);
             int64 ans = ;
             for (int i = ; i < n; ++ i)
                 ans += an[i];
             return ans;
         }

 // BEGIN CUT HERE
     public:
     void run_test(int Case) { if ((Case == -) || (Case == )) test_case_0(); if ((Case == -) || (Case == )) test_case_1(); if ((Case == -) || (Case == )) test_case_2(); if ((Case == -) || (Case == )) test_case_3(); if ((Case == -) || (Case == )) test_case_4(); }
     private:
     template <typename T> string print_array(const vector<T> &V) { ostringstream os; os << "{ "; for (typename vector<T>::const_iterator iter = V.begin(); iter != V.end(); ++iter) os << '\"' << *iter << "\","; os << " }"; return os.str(); }
     void verify_case(int Case, const long long &Expected, const long long &Received) { cerr << "Test Case #" << Case << "..."; if (Expected == Received) cerr << "PASSED" << endl; else { cerr << "FAILED" << endl; cerr << "\tExpected: \"" << Expected << '\"' << endl; cerr << "\tReceived: \"" << Received << '\"' << endl; } }
     void test_case_0() { string Arr0[] = {"N"}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); long long Arg1 = 1LL; verify_case(, Arg1, totalSalary(Arg0)); }
     void test_case_1() { string Arr0[] = {"NNYN",
  "NNYN",
  "NNNN",
  "NYYN"}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); long long Arg1 = 5LL; verify_case(, Arg1, totalSalary(Arg0)); }
     void test_case_2() { string Arr0[] = {"NNNNNN",
  "YNYNNY",
  "YNNNNY",
  "NNNNNN",
  "YNYNNN",
  "YNNYNN"}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); long long Arg1 = 17LL; verify_case(, Arg1, totalSalary(Arg0)); }
     void test_case_3() { string Arr0[] = {"NYNNYN",
  "NNNNNN",
  "NNNNNN",
  "NNYNNN",
  "NNNNNN",
  "NNNYYN"}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); long long Arg1 = 8LL; verify_case(, Arg1, totalSalary(Arg0)); }
     void test_case_4() { string Arr0[] = {"NNNN",
  "NNNN",
  "NNNN",
  "NNNN"}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); long long Arg1 = 4LL; verify_case(, Arg1, totalSalary(Arg0)); }

 // END CUT HERE

 };

 // BEGIN CUT HERE
 int main()
 {
 //    freopen( "a.out" , "w" , stdout );
     CorporationSalary ___test;
     ___test.run_test(-);
        return ;
 }
 // END CUT HERE

DIV1 500pt

题意：平面上有很多点，A和B轮流将点染色，A将点染为红色，B将点染为蓝色。染完所有颜色后最终的score为所有两端颜色不一样的边的长度和(一端为红色，一端为蓝色)。B想使score尽量小，A想使score尽量大，两人都采用最优策略，A先手，问最终score的为多少。点数 <= 12。

解法：这道题最关键的是注意到两点：1、最终score之和染色结果有关，和染色顺序有关；2、在忽略染色顺序的情况下，每个点只有三种状态，无色，红色，蓝色。O(12^3)的复杂度完全能接受。也就是说，只需要dfs加记忆化就好。

tag：dfs, memorization

 // BEGIN CUT HERE
 /*
  * Author:  plum rain
  * score :
  */
 /*

  */
 // END CUT HERE
 #line 11 "PointsGame.cpp"
 #include <sstream>
 #include <stdexcept>
 #include <functional>
 #include <iomanip>
 #include <numeric>
 #include <fstream>
 #include <cctype>
 #include <iostream>
 #include <cstdio>
 #include <vector>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <cstdlib>
 #include <set>
 #include <queue>
 #include <bitset>
 #include <list>
 #include <string>
 #include <utility>
 #include <map>
 #include <ctime>
 #include <stack>

 using namespace std;

 #define clr0(x) memset(x, 0, sizeof(x))
 #define clr1(x) memset(x, -1, sizeof(x))
 #define pb push_back
 #define sz(v) ((int)(v).size())
 #define all(t) t.begin(),t.end()
 #define zero(x) (((x)>0?(x):-(x))<eps)
 #define out(x) cout<<#x<<":"<<(x)<<endl
 #define tst(a) cout<<a<<" "
 #define tst1(a) cout<<#a<<endl
 #define CINBEQUICKER std::ios::sync_with_stdio(false)

 typedef vector<int> vi;
 typedef vector<string> vs;
 typedef vector<double> vd;
 typedef pair<int, int> pii;
 typedef long long int64;

 const double eps = 1e-;
 const double PI = atan(1.0)*;
 const int inf =  / ;

 int n;
 bool flag[];
 vi an, px, py;
 map<vi, double> mp;

 double gao(int x)
 {
     //out (x);
     //for (int i = 0; i < n; ++ i)
         //tst(i), out (an[i]);
     if (mp.count(an)) return mp[an];

     if(x == n){
         double ret = ;
         for (int i = ; i < n; ++ i) if (an[i] == )
             for (int j = ; j < n; ++ j) if (an[j] == )
                 ret += sqrt((px[i]-px[j])*(px[i]-px[j]) + (py[i]-py[j])*(py[i]-py[j]) + 0.0);
         mp[an] = ret;
         return ret;
     }

     double ret = x &  ? inf : ;
     for (int i = ; i < n; ++ i) if (!an[i]){
         an[i] = (x & ) + ;
         double tmp = gao(x + );
         //out (tmp);
         if (x & ) ret = min(tmp, ret);
         else ret = max(tmp, ret);
         an[i] = ;
     }
     mp[an] = ret;
     return ret;
 }

 class PointsGame
 {
     public:
         double gameValue(vector <int> PX, vector <int> PY){
             px = PX; py = PY;
             n = sz(px);
             for (int i = ; i < n; ++ i) an.pb ();
             clr0 (flag); mp.clear();
             return gao();
         }

 // BEGIN CUT HERE
     public:
     void run_test(int Case) { if ((Case == -) || (Case == )) test_case_0(); if ((Case == -) || (Case == )) test_case_1(); if ((Case == -) || (Case == )) test_case_2(); if ((Case == -) || (Case == )) test_case_3(); }
     //void run_test(int Case) { if ((Case == -1) || (Case == 0)) test_case_0();}
     private:
     template <typename T> string print_array(const vector<T> &V) { ostringstream os; os << "{ "; for (typename vector<T>::const_iterator iter = V.begin(); iter != V.end(); ++iter) os << '\"' << *iter << "\","; os << " }"; return os.str(); }
     void verify_case(int Case, const double &Expected, const double &Received) { cerr << "Test Case #" << Case << "..."; if (Expected == Received) cerr << "PASSED" << endl; else { cerr << "FAILED" << endl; cerr << "\tExpected: \"" << Expected << '\"' << endl; cerr << "\tReceived: \"" << Received << '\"' << endl; } }
     void test_case_0() { int Arr0[] = {,}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arr1[] = {,}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); double Arg2 = 1.0; verify_case(, Arg2, gameValue(Arg0, Arg1)); }
     void test_case_1() { int Arr0[] = {,,,}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arr1[] = {,,,}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); double Arg2 = 12.198039027185569; verify_case(, Arg2, gameValue(Arg0, Arg1)); }
     void test_case_2() { int Arr0[] = {,,,}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arr1[] = {,,,}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); double Arg2 = 12.0; verify_case(, Arg2, gameValue(Arg0, Arg1)); }
     void test_case_3() { int Arr0[] = {,,,}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arr1[] = {,,,}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); double Arg2 = 6.0; verify_case(, Arg2, gameValue(Arg0, Arg1)); }

 // END CUT HERE

 };

 // BEGIN CUT HERE
 int main()
 {
 //    freopen( "a.out" , "w" , stdout );
     PointsGame ___test;
     ___test.run_test(-);
        return ;
 }
 // END CUT HERE