SRM 402(1-250pt, 1-500pt)

DIV1 250pt

题意：对于任意一个由1～n组成的数列，其原始顺序为1,2,3..n。给出1~n的一个排列a[n]，要通过swp操作将其变回原始顺序。当i < j且a[i] > a[j]时，可以进行swp操作，即swap(a[i], a[j])。问要将给定排列变回原始顺序，所需要做的swp操作的次数期望。 n <= 8。

解法：概率dp。能用概率dp的原因是i < j 且 a[i] > a[j]时才能进行swp操作，这样就保证了不会出现环。

　　　这道题用递归的写法比较好写，但是用递归的同时一定要注意记忆化，记忆化以后时间复杂度就是O(8!)。否则会超时。

tag:概率dp

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

  */
 // END CUT HERE
 #line 11 "RandomSort.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 CLR(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 zero(x) (((x)>0?(x):-(x))<eps)
 #define out(x) cout<<#x<<":"<<(x)<<endl
 #define tst(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 maxint = ;

 map<VI, double> mp;
 int a[], sz;

 bool ok()
 {
     for (int i = ; i < sz; ++ i)
         if (a[i] != i+) return ;
     return ;
 }

 double dfs()
 {
     if (ok()) return ;
     VI tmp; tmp.clear();
     for (int i = ; i < sz; ++ i)
         tmp.PB (a[i]);
     if (mp.count(tmp)) return mp[tmp];

     int num = ;
     for (int i = ; i < sz; ++ i)
         for (int j = i+; j < sz; ++ j)
             if (a[i] > a[j]) ++ num;

     double ret = ;
     for (int i = ; i < sz; ++ i)
         for (int j = i+; j < sz; ++ j){
             if (a[i] < a[j]) continue; 

             swap(a[i], a[j]);
             ret += (dfs()+) / num;
             swap(a[i], a[j]);
         }
     mp[tmp] = ret;
     return ret;
 }

 class RandomSort
 {
     public:
         double getExpected(vector <int> p){
             sz = p.size();
             mp.clear();
             for (int i = ; i < sz; ++ i)
                 a[i] = p[i];
             return dfs();
         }

 // 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(); }
     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[]))); double Arg1 = 1.0; verify_case(, Arg1, getExpected(Arg0)); }
     void test_case_1() { int Arr0[] = {,,,}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); double Arg1 = 4.066666666666666; verify_case(, Arg1, getExpected(Arg0)); }
     void test_case_2() { int Arr0[] = {}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); double Arg1 = 0.0; verify_case(, Arg1, getExpected(Arg0)); }
     void test_case_3() { int Arr0[] = {,,,,,}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); double Arg1 = 5.666666666666666; verify_case(, Arg1, getExpected(Arg0)); }

 // END CUT HERE

 };

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

DIV1 450pt

题意：对于一个由.和X组成的环形字符串(即该字符串首尾相接)，连续的X(或者1个)组成blocks，连续的.(或者1个)组成gaps。比如..X.XX..由1个长度为4的gap(因为首尾相接)，1个长度为1的gap，1个长度为1的block，1个长度为2的block组成。定义gap array an[]为将所有gap的长度放到an[]中，从大到小做一个排序得到的数组即为gap array。现在，去掉某一个block，要使得到的an字典序最大，应该去掉哪个block？返回该block中下标最小的点的值，比如XX...X....XX.X，去掉长度为3的block，返回下标1，若去掉长度为2的block，返回10。若去掉某两个block后gap array相同，则返回下标最小值小的那一个。

　　　字符串长度 <= 2500。

解法：题意好复杂...

　　　还好这道题是可以暴力过掉的.....n*n*logn的复杂度能接受。但是一方面由于我编码能力太差，另一方面由于我对STL很多函数的不熟悉，导致我没有暴力出来......忧伤...看了官方题解提供的代码。点击打开。

　　　当然，这道题除了能暴力直接做，也是有线性解法的。若要比较去掉某两个block后生成的gap array的大小，设与第一个block相邻的两个gap长度分别为a,b，与第二个block相邻的gao长度分别为c,d，则只需要比较a+b和c+d，max(a,b)和max(c,d)，min(a,b)和min(c,d)即可。

　　　至于这个的原因嘛，想一下，很简单的- -....

tag:think

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

  */
 // END CUT HERE
 #line 11 "LargestGap.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 CLR(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 zero(x) (((x)>0?(x):-(x))<eps)
 #define out(x) cout<<#x<<":"<<(x)<<endl
 #define tst(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 maxint = ;

 class LargestGap
 {
     public:
         int getLargest(vector <string> b){
             VI best, cur; int pos;
             string s = accumulate(b.begin(), b.end(), string(""));
             for (int i = ; i < (int)s.size(); ++ i) if (s[i] == 'X'){
                 string s2 = s.substr(i) + s.substr(,i) + "X";
                 cur.clear();
                 int len = ;
                 for (int j = ; j < (int)s2.size(); ++ j){
                     if (s2[j] == '.') ++ len;
                     else if (len) cur.PB (len), len = ;
                 }
                 if (cur.size() > ) cur[] += cur.back(), cur.pop_back();
                 sort(cur.begin(), cur.end(), greater<int>());
                 if (cur > best) best = cur, pos = i;
             }
             return pos;
         }

 // 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();}
     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 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; } }
     //void test_case_0() { string Arr0[] = {; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 5; verify_case(0, Arg1, getLargest(Arg0)); }
     void test_case_0() { string Arr0[] = {"XXXX","....","XXXX","....","XXXX","...."}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; verify_case(, Arg1, getLargest(Arg0)); }
     void test_case_1() { string Arr0[] = {"XXX.........XX...........XX..X"}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; verify_case(, Arg1, getLargest(Arg0)); }
     void test_case_2() { string Arr0[] = {"XXX","X.....","....XX..XXXXXX","X........X..",".XXX."}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; verify_case(, Arg1, getLargest(Arg0)); }

 // END CUT HERE

 };

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