DIV1 250pt

题意：对于图G，有一些点和边，点中有一些点称为特殊点。问在所有特殊点最终不能处于同一个联通块的条件下，最多能给在图G中添加多少条边。

解法：首先，对于图G，处理出它有哪些联通块，然后，不含有特殊点的联通块要连接到某一个含有特殊点的联通块上。连接哪一个能使添加的边最多呢？当然是连接含有点数最多的含特殊点的联通块。

tag:graph, greedy

  1.  // BEGIN CUT HERE
  2.  /*
  3.   * Author:  plum rain
  4.   * score :
  5.   */
  6.  /*
  7.  
  8.   */
  9.  // END CUT HERE
  10.  #line 11 "AddElectricalWires.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-;
  55.  const double PI = atan(1.0)*;
  56.  const int inf =  / ;
  57.  
  58.  class AddElectricalWires
  59.  {
  60.      public:
  61.          int f[];
  62.          bool v[];
  63.  
  64.          int find(int x)
  65.          {
  66.              if (x != f[x]) f[x] = find(f[x]);
  67.              return f[x];
  68.          }
  69.  
  70.          int maxNewWires(vector <string> p, vector <int> num){
  71.              clr0 (v);
  72.              for (int i = ; i < sz(num); ++ i) v[num[i]] = ;
  73.              for (int i = ; i < sz(p); ++ i) f[i] = i;
  74.              for (int i = ; i < sz(p); ++ i)
  75.                  for (int j = ; j < sz(p); ++ j) if (p[i][j] == ''){
  76.                      int t1 = find(i), t2 = find(j);
  77.                      if ((v[t1] && v[t2]) || (t1 == t2)) continue;
  78.                      if (!v[t1]) f[t1] = t2;
  79.                      else f[t2] = t1;
  80.  
  81.                  }
  82.              int ans = , idx = num[];
  83.              for (int i = ; i < sz(num); ++ i){
  84.                  int tmp = ;
  85.                  for (int j = ; j < sz(p); ++ j)
  86.                      if (find(j) == num[i]) ++ tmp;
  87.                  if (tmp > ans)
  88.                      idx = num[i], ans = tmp;
  89.              }
  90.              int ret = ;
  91.              for (int i = ; i < sz(p); ++ i)
  92.                  for (int j = i+; j < sz(p); ++ j) if (p[i][j] == ''){
  93.                      int t1 = find(i), t2 = find(j);
  94.                      if (t1 == t2){
  95.                          ++ ret; continue;
  96.                      }
  97.                      if (v[t1] && v[t2]) continue;
  98.                      if (!v[t1] && !v[t2]){
  99.                          ++ ret; f[t1] = t2; continue;
  100.                      }
  101.                      if (t1 == idx || t2 == idx){
  102.                          ++ ret;
  103.                          if (t1 == idx) f[t2] = idx;
  104.                          else f[t1] = idx;
  105.                      }
  106.                  }
  107.              return ret;
  108.          }
  109.  
  110.  // BEGIN CUT HERE
  111.      public:
  112.      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(); }
  113.      private:
  114.      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(); }
  115.      void verify_case(int Case, const int &Expected, const int &Received) { cerr << "Test Case #" << Case << "..."; if (Expected == Received) cerr << "PASSED" << endl; else { cerr << "FAILED" << endl; cerr << "\tExpected: \"" << Expected << '\"' << endl; cerr << "\tReceived: \"" << Received << '\"' << endl; } }
  116.      void test_case_0() { string Arr0[] = {"","",""}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arr1[] = {}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); int Arg2 = ; verify_case(, Arg2, maxNewWires(Arg0, Arg1)); }
  117.      void test_case_1() { string Arr0[] = {"","",""}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arr1[] = {,}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); int Arg2 = ; verify_case(, Arg2, maxNewWires(Arg0, Arg1)); }
  118.      void test_case_2() { string Arr0[] = {"",""}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arr1[] = {}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); int Arg2 = ; verify_case(, Arg2, maxNewWires(Arg0, Arg1)); }
  119.      void test_case_3() { string Arr0[] = {"","","","",""}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arr1[] = {,,,,}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); int Arg2 = ; verify_case(, Arg2, maxNewWires(Arg0, Arg1)); }
  120.      void test_case_4() { string Arr0[] = {"","","","",""}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arr1[] = {,}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); int Arg2 = ; verify_case(, Arg2, maxNewWires(Arg0, Arg1)); }
  121.  
  122.  // END CUT HERE
  123.  
  124.  };
  125.  
  126.  // BEGIN CUT HERE
  127.  int main()
  128.  {
  129.  //    freopen( "a.out" , "w" , stdout );
  130.      AddElectricalWires ___test;
  131.      ___test.run_test(-);
  132.         return ;
  133.  }
  134.  // END CUT HERE

DIV1 500pt

题意：给定整数k,n和一个数组A[]，要按顺序访问范围为0-(n-1)的数轴上编号为A[i]的点，访问的方法如下。首先，找到一段长度为k的连续的点，(比如编号为x, x+1, x+2...x+k-1)，然后查看缓存中的点，有三种可能：某点在缓存中但这次不需要访问，则移出缓存；某点不在缓存中但这次要访问，访问该点并将它加入缓存；某点在缓存中且要访问，不访问该点但将其留在缓存中。要按顺序访问A[i]中的点，最少需要进行多少次访问操作。

　　n <= 10^9, k <= n, A.size() <= 50。

　　注意，如果A[] = {2, 3, 10}，k = 3，n = 1000则只需访问2, 3, 4, 8, 9, 10即可。

解法：本题的关键点在于要注意到一点，要访问A[i]，最优解访问的连续k个点只有两类可能：A[j], A[j]+1...A[j]+k-1和A[j]-k+1, A[j]-k+2...A[j]。(*)

　　　所以，先预处理出可能从哪些点开始访问。然后，由于要按顺序访问A[i]，所以只需要做一遍dp即可。

tag:dp, think

  1.  // BEGIN CUT HERE
  2.  /*
  3.   * Author:  plum rain
  4.   * score :
  5.   */
  6.  /*
  7.  
  8.   */
  9.  // END CUT HERE
  10.  #line 11 "ContiguousCache.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-;
  55.  const double PI = atan(1.0)*;
  56.  const int inf =  / ;
  57.  const int64 llinf = 1LL<<;
  58.  
  59.  class ContiguousCache
  60.  {
  61.      public:
  62.          int64 d[][];
  63.  
  64.          int64 over(int a, int b, int c, int d, int k)
  65.          {
  66.              if (d < a || c > b) return k;
  67.              return max(b, d) - min(a, c) +  - k;
  68.          }
  69.  
  70.          long long minimumReads(int len, int k, vector <int> pos){
  71.              vi x;
  72.              for (int i = ; i < sz(pos); ++ i){
  73.                  x.pb (min(pos[i], len-k));
  74.                  x.pb (max(, pos[i]-k+));
  75.              }
  76.              x.erase(unique(x.begin(), x.end()), x.end());
  77.  
  78.              for (int i = ; i < sz(pos); ++ i)
  79.                  for (int j = ; j < sz(x); ++ j) d[i][j] = llinf;
  80.  
  81.              for (int i = ; i < sz(x); ++ i) if (x[i] <= pos[] && x[i]+k- >= pos[])
  82.                  d[][i] = k;
  83.  
  84.              for (int i = ; i < sz(pos); ++ i)
  85.                  for (int j = ; j < sz(x); ++ j) if (x[j] <= pos[i] && x[j]+k- >= pos[i])
  86.                      for (int t = ; t < sz(x); ++ t) if (d[i-][t] != llinf)
  87.                          d[i][j] = min(d[i][j], d[i-][t] + over(x[t], x[t]+k-, x[j], x[j]+k-, k));
  88.  
  89.              int64 ret = llinf;
  90.              for (int i = ; i < sz(x); ++ i) ret = min(ret, d[sz(pos)-][i]);
  91.              return ret;
  92.          }
  93.  
  94.  // BEGIN CUT HERE
  95.      public:
  96.      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(); }
  97.      private:
  98.      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(); }
  99.      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; } }
  100.      void test_case_0() { int Arg0 = ; int Arg1 = ; int Arr2[] = {, , , }; vector <int> Arg2(Arr2, Arr2 + (sizeof(Arr2) / sizeof(Arr2[]))); long long Arg3 = 7LL; verify_case(, Arg3, minimumReads(Arg0, Arg1, Arg2)); }
  101.      void test_case_1() { int Arg0 = ; int Arg1 = ; int Arr2[] = {,,,,,}; vector <int> Arg2(Arr2, Arr2 + (sizeof(Arr2) / sizeof(Arr2[]))); long long Arg3 = 29LL; verify_case(, Arg3, minimumReads(Arg0, Arg1, Arg2)); }
  102.      void test_case_2() { int Arg0 = ; int Arg1 = ; int Arr2[] = {, , , , , }; vector <int> Arg2(Arr2, Arr2 + (sizeof(Arr2) / sizeof(Arr2[]))); long long Arg3 = 1987654320LL; verify_case(, Arg3, minimumReads(Arg0, Arg1, Arg2)); }
  103.      void test_case_3() { int Arg0 = ; int Arg1 = ; int Arr2[] = {}; vector <int> Arg2(Arr2, Arr2 + (sizeof(Arr2) / sizeof(Arr2[]))); long long Arg3 = 2LL; verify_case(, Arg3, minimumReads(Arg0, Arg1, Arg2)); }
  104.  
  105.  // END CUT HERE
  106.  
  107.  };
  108.  
  109.  // BEGIN CUT HERE
  110.  int main()
  111.  {
  112.  //    freopen( "a.out" , "w" , stdout );
  113.      ContiguousCache ___test;
  114.      ___test.run_test(-);
  115.         return ;
  116.  }
  117.  // END CUT HERE

　　　　

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