• 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

      1 // BEGIN CUT HERE
      2 /*
      3  * Author:  plum rain
      4  * score :
      5  */
      6 /*
      7 
      8  */
      9 // END CUT HERE
     10 #line 11 "CorporationSalary.cpp"
     11 #include <sstream>
     12 #include <stdexcept>
     13 #include <functional>
     14 #include <iomanip>
     15 #include <numeric>
     16 #include <fstream>
     17 #include <cctype>
     18 #include <iostream>
     19 #include <cstdio>
     20 #include <vector>
     21 #include <cstring>
     22 #include <cmath>
     23 #include <algorithm>
     24 #include <cstdlib>
     25 #include <set>
     26 #include <queue>
     27 #include <bitset>
     28 #include <list>
     29 #include <string>
     30 #include <utility>
     31 #include <map>
     32 #include <ctime>
     33 #include <stack>
     34 
     35 using namespace std;
     36 
     37 #define clr0(x) memset(x, 0, sizeof(x))
     38 #define clr1(x) memset(x, -1, sizeof(x))
     39 #define pb push_back
     40 #define sz(v) ((int)(v).size())
     41 #define all(t) t.begin(),t.end()
     42 #define zero(x) (((x)>0?(x):-(x))<eps)
     43 #define out(x) cout<<#x<<":"<<(x)<<endl
     44 #define tst(a) cout<<a<<" "
     45 #define tst1(a) cout<<#a<<endl
     46 #define CINBEQUICKER std::ios::sync_with_stdio(false)
     47 
     48 typedef vector<int> vi;
     49 typedef vector<string> VS;
     50 typedef vector<double> VD;
     51 typedef pair<int, int> pii;
     52 typedef long long int64;
     53 
     54 const double eps = 1e-8;
     55 const double PI = atan(1.0)*4;
     56 const int inf = 2139062143 / 2;
     57 
     58 vi pat[100];
     59 bool son[100];
     60 int64 an[100];
     61 
     62 void dfs (int x)
     63 {
     64     int64 &ret = an[x];
     65     if (ret) return;
     66 
     67     int n = sz(pat[x]);
     68     if (!n){
     69         an[x] = 1; return;
     70     }
     71 
     72     for (int i = 0; i < n; ++ i){
     73         dfs (pat[x][i]);
     74         ret += an[pat[x][i]];
     75     }
     76 }
     77 
     78 class CorporationSalary
     79 {
     80     public:
     81         long long totalSalary(vector <string> v){
     82             clr0 (an); clr0 (son); 
     83             int n = sz(v);
     84             for (int i = 0; i < n; ++ i)
     85                 pat[i].clear();
     86 
     87             for (int i = 0; i < n; ++ i)
     88                 for (int j = 0; j < n; ++ j)
     89                     if (v[i][j] == 'Y')
     90                         pat[i].pb (j), son[j] = 1;
     91 
     92             for (int i = 0; i < n; ++ i) 
     93                 if (!son[i]) dfs (i); 
     94             int64 ans = 0;
     95             for (int i = 0; i < n; ++ i)
     96                 ans += an[i];
     97             return ans;
     98         }
     99         
    100 // BEGIN CUT HERE
    101     public:
    102     void run_test(int Case) { if ((Case == -1) || (Case == 0)) test_case_0(); if ((Case == -1) || (Case == 1)) test_case_1(); if ((Case == -1) || (Case == 2)) test_case_2(); if ((Case == -1) || (Case == 3)) test_case_3(); if ((Case == -1) || (Case == 4)) test_case_4(); }
    103     private:
    104     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(); }
    105     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 << "	Expected: "" << Expected << '"' << endl; cerr << "	Received: "" << Received << '"' << endl; } }
    106     void test_case_0() { string Arr0[] = {"N"}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); long long Arg1 = 1LL; verify_case(0, Arg1, totalSalary(Arg0)); }
    107     void test_case_1() { string Arr0[] = {"NNYN",
    108  "NNYN",
    109  "NNNN",
    110  "NYYN"}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); long long Arg1 = 5LL; verify_case(1, Arg1, totalSalary(Arg0)); }
    111     void test_case_2() { string Arr0[] = {"NNNNNN",
    112  "YNYNNY",
    113  "YNNNNY",
    114  "NNNNNN",
    115  "YNYNNN",
    116  "YNNYNN"}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); long long Arg1 = 17LL; verify_case(2, Arg1, totalSalary(Arg0)); }
    117     void test_case_3() { string Arr0[] = {"NYNNYN",
    118  "NNNNNN",
    119  "NNNNNN",
    120  "NNYNNN",
    121  "NNNNNN",
    122  "NNNYYN"}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); long long Arg1 = 8LL; verify_case(3, Arg1, totalSalary(Arg0)); }
    123     void test_case_4() { string Arr0[] = {"NNNN",
    124  "NNNN",
    125  "NNNN",
    126  "NNNN"}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); long long Arg1 = 4LL; verify_case(4, Arg1, totalSalary(Arg0)); }
    127 
    128 // END CUT HERE
    129 
    130 };
    131 
    132 // BEGIN CUT HERE
    133 int main()
    134 {
    135 //    freopen( "a.out" , "w" , stdout );    
    136     CorporationSalary ___test;
    137     ___test.run_test(-1);
    138        return 0;
    139 }
    140 // END CUT HERE
    View Code

    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

      1 // BEGIN CUT HERE
      2 /*
      3  * Author:  plum rain
      4  * score :
      5  */
      6 /*
      7 
      8  */
      9 // END CUT HERE
     10 #line 11 "PointsGame.cpp"
     11 #include <sstream>
     12 #include <stdexcept>
     13 #include <functional>
     14 #include <iomanip>
     15 #include <numeric>
     16 #include <fstream>
     17 #include <cctype>
     18 #include <iostream>
     19 #include <cstdio>
     20 #include <vector>
     21 #include <cstring>
     22 #include <cmath>
     23 #include <algorithm>
     24 #include <cstdlib>
     25 #include <set>
     26 #include <queue>
     27 #include <bitset>
     28 #include <list>
     29 #include <string>
     30 #include <utility>
     31 #include <map>
     32 #include <ctime>
     33 #include <stack>
     34 
     35 using namespace std;
     36 
     37 #define clr0(x) memset(x, 0, sizeof(x))
     38 #define clr1(x) memset(x, -1, sizeof(x))
     39 #define pb push_back
     40 #define sz(v) ((int)(v).size())
     41 #define all(t) t.begin(),t.end()
     42 #define zero(x) (((x)>0?(x):-(x))<eps)
     43 #define out(x) cout<<#x<<":"<<(x)<<endl
     44 #define tst(a) cout<<a<<" "
     45 #define tst1(a) cout<<#a<<endl
     46 #define CINBEQUICKER std::ios::sync_with_stdio(false)
     47 
     48 typedef vector<int> vi;
     49 typedef vector<string> vs;
     50 typedef vector<double> vd;
     51 typedef pair<int, int> pii;
     52 typedef long long int64;
     53 
     54 const double eps = 1e-8;
     55 const double PI = atan(1.0)*4;
     56 const int inf = 2139062143 / 2;
     57 
     58 int n;
     59 bool flag[50];
     60 vi an, px, py;
     61 map<vi, double> mp;
     62 
     63 double gao(int x)
     64 {
     65     //out (x);
     66     //for (int i = 0; i < n; ++ i)
     67         //tst(i), out (an[i]);
     68     if (mp.count(an)) return mp[an];
     69 
     70     if(x == n){
     71         double ret = 0;
     72         for (int i = 0; i < n; ++ i) if (an[i] == 1)
     73             for (int j = 0; j < n; ++ j) if (an[j] == 2)
     74                 ret += sqrt((px[i]-px[j])*(px[i]-px[j]) + (py[i]-py[j])*(py[i]-py[j]) + 0.0);
     75         mp[an] = ret;
     76         return ret;
     77     }
     78     
     79     double ret = x & 1 ? inf : 0;
     80     for (int i = 0; i < n; ++ i) if (!an[i]){
     81         an[i] = (x & 1) + 1;
     82         double tmp = gao(x + 1);
     83         //out (tmp);
     84         if (x & 1) ret = min(tmp, ret);
     85         else ret = max(tmp, ret);
     86         an[i] = 0;
     87     }
     88     mp[an] = ret;
     89     return ret;
     90 }
     91 
     92 class PointsGame
     93 {
     94     public:
     95         double gameValue(vector <int> PX, vector <int> PY){
     96             px = PX; py = PY;
     97             n = sz(px);
     98             for (int i = 0; i < n; ++ i) an.pb (0);
     99             clr0 (flag); mp.clear();
    100             return gao(0);
    101         }
    102         
    103 // BEGIN CUT HERE
    104     public:
    105     void run_test(int Case) { if ((Case == -1) || (Case == 0)) test_case_0(); if ((Case == -1) || (Case == 1)) test_case_1(); if ((Case == -1) || (Case == 2)) test_case_2(); if ((Case == -1) || (Case == 3)) test_case_3(); }
    106     //void run_test(int Case) { if ((Case == -1) || (Case == 0)) test_case_0();}
    107     private:
    108     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(); }
    109     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 << "	Expected: "" << Expected << '"' << endl; cerr << "	Received: "" << Received << '"' << endl; } }
    110     void test_case_0() { int Arr0[] = {0,0}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arr1[] = {0,1}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); double Arg2 = 1.0; verify_case(0, Arg2, gameValue(Arg0, Arg1)); }
    111     void test_case_1() { int Arr0[] = {0,0,1,1}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arr1[] = {0,5,0,5}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); double Arg2 = 12.198039027185569; verify_case(1, Arg2, gameValue(Arg0, Arg1)); }
    112     void test_case_2() { int Arr0[] = {0,0,0,0}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arr1[] = {0,1,5,6}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); double Arg2 = 12.0; verify_case(2, Arg2, gameValue(Arg0, Arg1)); }
    113     void test_case_3() { int Arr0[] = {0,1,2,3}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arr1[] = {0,0,0,0}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); double Arg2 = 6.0; verify_case(3, Arg2, gameValue(Arg0, Arg1)); }
    114 
    115 // END CUT HERE
    116 
    117 };
    118 
    119 // BEGIN CUT HERE
    120 int main()
    121 {
    122 //    freopen( "a.out" , "w" , stdout );    
    123     PointsGame ___test;
    124     ___test.run_test(-1);
    125        return 0;
    126 }
    127 // END CUT HERE
    View Code
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  • 原文地址:https://www.cnblogs.com/plumrain/p/srm_407.html
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