SRM 394(1-250pt)

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

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

 */
// END CUT HERE
#line 11 "RoughStrings.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 clr0(x) memset(x, 0, sizeof(x))
#define clr1(x) memset(x, -1, sizeof(x))
#define pb push_back
#define mp make_pair
#define sz(v) ((int)(v).size())
#define all(t) t.begin(),t.end()
#define zero(x) (((x)>0?(x):-(x))<eps)
#define out(x) cout<<#x<<":"<<(x)<<endl
#define tst(a) cout<<a<<" "
#define tst1(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-8;
const double PI = atan(1.0)*4;
const int inf = 2139062143 / 2;

int num[500], n2[500], n1[500];
int idx[500], all;
pii an[100];

int gao(int k)
{
    clr0 (n2);
    for (int i = 0; i < 26; ++ i)
        ++ n2[num[i]];
    int pos = 50;
    while (!n2[pos]) -- pos;
    while (k > 0){
        if (pos < 0) return 0;
        k -= n2[pos];
        n2[pos-1] += n2[pos];
        n2[pos--] = 0;
    }
    if (!k) return pos;
    return pos + 1;
}

class RoughStrings
{
    public:
        int minRoughness(string s, int k){
            clr0 (num);
            for (int i = 0; i < sz(s); ++ i)
                ++ num[s[i] - 'a'];
            all = 0;
            for (int i = 0; i < 26; ++ i)
                if (num[i]) n1[all++] = num[i];
            if (all == 1) return 0;
            sort(n1, n1+all);
            int ans = max(gao(k) - n1[0], 0), use = 0;
            for (int i = 0; i < all; ++ i){
                use += n1[i];
                if (k < use) break;
                if (i == all-1) ans = 0;
                else ans = min(ans, max(gao(k-use) - n1[i+1], 0));
            }
            return ans;
        }
        
// BEGIN CUT HERE
    public:
    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(); }
    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 << "	Expected: "" << Expected << '"' << endl; cerr << "	Received: "" << Received << '"' << endl; } }
    void test_case_0() { string Arg0 = "aaaaabbc"; int Arg1 = 1; int Arg2 = 3; verify_case(0, Arg2, minRoughness(Arg0, Arg1)); }
    void test_case_1() { string Arg0 = "aaaabbbbc"; int Arg1 = 5; int Arg2 = 0; verify_case(1, Arg2, minRoughness(Arg0, Arg1)); }
    void test_case_2() { string Arg0 = "veryeviltestcase"; int Arg1 = 1; int Arg2 = 2; verify_case(2, Arg2, minRoughness(Arg0, Arg1)); }
    void test_case_3() { string Arg0 = "gggggggooooooodddddddllllllluuuuuuuccckkk"; int Arg1 = 5; int Arg2 = 3; verify_case(3, Arg2, minRoughness(Arg0, Arg1)); }
    void test_case_4() { string Arg0 = "zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz"; int Arg1 = 17; int Arg2 = 0; verify_case(4, Arg2, minRoughness(Arg0, Arg1)); }
    void test_case_5() { string Arg0 = "bbbccca"; int Arg1 = 2; int Arg2 = 0; verify_case(5, Arg2, minRoughness(Arg0, Arg1)); }

// END CUT HERE

};

// BEGIN CUT HERE
int main()
{
//    freopen( "a.out" , "w" , stdout );    
    RoughStrings ___test;
    ___test.run_test(-1);
       return 0;
}
// END CUT HERE

 网上看到的代码：