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

  1 // BEGIN CUT HERE
  2 /*
  3  * Author:  plum rain
  4  * score :
  5  */
  6 /*
  7 
  8  */
  9 // END CUT HERE
 10 #line 11 "RandomSort.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 CLR(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 zero(x) (((x)>0?(x):-(x))<eps)
 42 #define out(x) cout<<#x<<":"<<(x)<<endl
 43 #define tst(a) cout<<#a<<endl
 44 #define CINBEQUICKER std::ios::sync_with_stdio(false)
 45 
 46 typedef vector<int> VI;
 47 typedef vector<string> VS;
 48 typedef vector<double> VD;
 49 typedef pair<int, int> pii;
 50 typedef long long int64;
 51 
 52 const double eps = 1e-8;
 53 const double PI = atan(1.0)*4;
 54 const int maxint = 2139062143;
 55 
 56 map<VI, double> mp;
 57 int a[20], sz;
 58 
 59 bool ok()
 60 {
 61     for (int i = 0; i < sz; ++ i)
 62         if (a[i] != i+1) return 0;
 63     return 1;
 64 }
 65 
 66 double dfs()
 67 {
 68     if (ok()) return 0;
 69     VI tmp; tmp.clear();
 70     for (int i = 0; i < sz; ++ i)
 71         tmp.PB (a[i]);
 72     if (mp.count(tmp)) return mp[tmp];
 73 
 74     int num = 0;
 75     for (int i = 0; i < sz; ++ i)
 76         for (int j = i+1; j < sz; ++ j)
 77             if (a[i] > a[j]) ++ num;
 78 
 79     double ret = 0;
 80     for (int i = 0; i < sz; ++ i)
 81         for (int j = i+1; j < sz; ++ j){
 82             if (a[i] < a[j]) continue; 
 83 
 84             swap(a[i], a[j]);
 85             ret += (dfs()+1) / num;
 86             swap(a[i], a[j]);
 87         }
 88     mp[tmp] = ret;
 89     return ret;
 90 }
 91 
 92 class RandomSort
 93 {
 94     public:
 95         double getExpected(vector <int> p){
 96             sz = p.size();
 97             mp.clear();
 98             for (int i = 0; i < sz; ++ i)
 99                 a[i] = p[i];
100             return dfs();
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     private:
107     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(); }
108     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; } }
109     void test_case_0() { int Arr0[] = {7, 2, 5, 4, 1, 3, 8, 6}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); double Arg1 = 1.0; verify_case(0, Arg1, getExpected(Arg0)); }
110     void test_case_1() { int Arr0[] = {4,3,2,1}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); double Arg1 = 4.066666666666666; verify_case(1, Arg1, getExpected(Arg0)); }
111     void test_case_2() { int Arr0[] = {1}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); double Arg1 = 0.0; verify_case(2, Arg1, getExpected(Arg0)); }
112     void test_case_3() { int Arr0[] = {2,5,1,6,3,4}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); double Arg1 = 5.666666666666666; verify_case(3, Arg1, getExpected(Arg0)); }
113 
114 // END CUT HERE
115 
116 };
117 
118 // BEGIN CUT HERE
119 int main()
120 {
121 //    freopen( "a.out" , "w" , stdout );    
122     RandomSort ___test;
123     ___test.run_test(-1);
124        return 0;
125 }
126 // END CUT HERE
View Code

 

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

  1 // BEGIN CUT HERE
  2 /*
  3  * Author:  plum rain
  4  * score :
  5  */
  6 /*
  7 
  8  */
  9 // END CUT HERE
 10 #line 11 "LargestGap.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 CLR(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 zero(x) (((x)>0?(x):-(x))<eps)
 42 #define out(x) cout<<#x<<":"<<(x)<<endl
 43 #define tst(a) cout<<#a<<endl
 44 #define CINBEQUICKER std::ios::sync_with_stdio(false)
 45 
 46 typedef vector<int> VI;
 47 typedef vector<string> VS;
 48 typedef vector<double> VD;
 49 typedef pair<int, int> pii;
 50 typedef long long int64;
 51 
 52 const double eps = 1e-8;
 53 const double PI = atan(1.0)*4;
 54 const int maxint = 2139062143;
 55 
 56 class LargestGap
 57 {
 58     public:
 59         int getLargest(vector <string> b){
 60             VI best, cur; int pos;
 61             string s = accumulate(b.begin(), b.end(), string(""));
 62             for (int i = 0; i < (int)s.size(); ++ i) if (s[i] == 'X'){
 63                 string s2 = s.substr(i) + s.substr(0,i) + "X";
 64                 cur.clear();
 65                 int len = 0;
 66                 for (int j = 0; j < (int)s2.size(); ++ j){
 67                     if (s2[j] == '.') ++ len;
 68                     else if (len) cur.PB (len), len = 0;
 69                 }
 70                 if (cur.size() > 1) cur[0] += cur.back(), cur.pop_back();
 71                 sort(cur.begin(), cur.end(), greater<int>());
 72                 if (cur > best) best = cur, pos = i;
 73             }
 74             return pos;
 75         }
 76         
 77 // BEGIN CUT HERE
 78     public:
 79     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();}
 80     private:
 81     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(); }
 82     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; } }
 83     //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)); }
 84     void test_case_0() { string Arr0[] = {"XXXX","....","XXXX","....","XXXX","...."}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 0; verify_case(1, Arg1, getLargest(Arg0)); }
 85     void test_case_1() { string Arr0[] = {"XXX.........XX...........XX..X"}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 12; verify_case(2, Arg1, getLargest(Arg0)); }
 86     void test_case_2() { string Arr0[] = {"XXX","X.....","....XX..XXXXXX","X........X..",".XXX."}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 32; verify_case(3, Arg1, getLargest(Arg0)); }
 87 
 88 // END CUT HERE
 89 
 90 };
 91 
 92 // BEGIN CUT HERE
 93 int main()
 94 {
 95 //    freopen( "a.out" , "w" , stdout );    
 96     LargestGap ___test;
 97     ___test.run_test(-1);
 98        return 0;
 99 }
100 // END CUT HERE
View Code

 

　　　

posted @ 2013-12-17 15:30  Plumrain  阅读(209)  评论(0)  编辑  收藏  举报