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
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
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
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
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
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
现在的你,在干什么呢?
你是不是还记得,你说你想成为岩哥那样的人。