DIV1 250pt
题意：给定一个字符串s('a'-'z')，计其中出现次数最多和最少的字母分别出现c1次和c2次，若在s中去掉最多k个字母，求去掉以后c1 - c2的最小值。
解法：做题的时候，想到了用dfs暴力枚举，然后TLE了。然后想到了枚举c2，求c1的最小值，最后写了比较麻烦的代码，过了。然后看了题解才发现，枚举c1和c2。。。。。
真的是看到'a' - 'z'就应该想到这种方法。。。。
tag:think, brute-force
1 // BEGIN CUT HERE 2 /* 3 * Author: plum rain 4 * score : 5 */ 6 /* 7 8 */ 9 // END CUT HERE 10 #line 11 "RoughStrings.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 mp make_pair 41 #define sz(v) ((int)(v).size()) 42 #define all(t) t.begin(),t.end() 43 #define zero(x) (((x)>0?(x):-(x))<eps) 44 #define out(x) cout<<#x<<":"<<(x)<<endl 45 #define tst(a) cout<<a<<" " 46 #define tst1(a) cout<<#a<<endl 47 #define CINBEQUICKER std::ios::sync_with_stdio(false) 48 49 typedef vector<int> VI; 50 typedef vector<string> VS; 51 typedef vector<double> VD; 52 typedef pair<int, int> pii; 53 typedef long long int64; 54 55 const double eps = 1e-8; 56 const double PI = atan(1.0)*4; 57 const int inf = 2139062143 / 2; 58 59 int num[500], n2[500], n1[500]; 60 int idx[500], all; 61 pii an[100]; 62 63 int gao(int k) 64 { 65 clr0 (n2); 66 for (int i = 0; i < 26; ++ i) 67 ++ n2[num[i]]; 68 int pos = 50; 69 while (!n2[pos]) -- pos; 70 while (k > 0){ 71 if (pos < 0) return 0; 72 k -= n2[pos]; 73 n2[pos-1] += n2[pos]; 74 n2[pos--] = 0; 75 } 76 if (!k) return pos; 77 return pos + 1; 78 } 79 80 class RoughStrings 81 { 82 public: 83 int minRoughness(string s, int k){ 84 clr0 (num); 85 for (int i = 0; i < sz(s); ++ i) 86 ++ num[s[i] - 'a']; 87 all = 0; 88 for (int i = 0; i < 26; ++ i) 89 if (num[i]) n1[all++] = num[i]; 90 if (all == 1) return 0; 91 sort(n1, n1+all); 92 int ans = max(gao(k) - n1[0], 0), use = 0; 93 for (int i = 0; i < all; ++ i){ 94 use += n1[i]; 95 if (k < use) break; 96 if (i == all-1) ans = 0; 97 else ans = min(ans, max(gao(k-use) - n1[i+1], 0)); 98 } 99 return ans; 100 } 101 102 // BEGIN CUT HERE 103 public: 104 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(); if ((Case == -1) || (Case == 5)) test_case_5(); } 105 private: 106 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(); } 107 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 << " Expected: "" << Expected << '"' << endl; cerr << " Received: "" << Received << '"' << endl; } } 108 void test_case_0() { string Arg0 = "aaaaabbc"; int Arg1 = 1; int Arg2 = 3; verify_case(0, Arg2, minRoughness(Arg0, Arg1)); } 109 void test_case_1() { string Arg0 = "aaaabbbbc"; int Arg1 = 5; int Arg2 = 0; verify_case(1, Arg2, minRoughness(Arg0, Arg1)); } 110 void test_case_2() { string Arg0 = "veryeviltestcase"; int Arg1 = 1; int Arg2 = 2; verify_case(2, Arg2, minRoughness(Arg0, Arg1)); } 111 void test_case_3() { string Arg0 = "gggggggooooooodddddddllllllluuuuuuuccckkk"; int Arg1 = 5; int Arg2 = 3; verify_case(3, Arg2, minRoughness(Arg0, Arg1)); } 112 void test_case_4() { string Arg0 = "zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz"; int Arg1 = 17; int Arg2 = 0; verify_case(4, Arg2, minRoughness(Arg0, Arg1)); } 113 void test_case_5() { string Arg0 = "bbbccca"; int Arg1 = 2; int Arg2 = 0; verify_case(5, Arg2, minRoughness(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 RoughStrings ___test; 124 ___test.run_test(-1); 125 return 0; 126 } 127 // END CUT HERE
网上看到的代码：
