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  • SRM 387(1-250pt)

    DIV1 300pt

    题意:有m种颜色的球若干个放在n个盒子里。每次操作可从一个盒子里拿出任意个球(不必同色),放进另一个盒子。要求终态为:1、最多有一个盒子里面装有不同色的球,该盒子成为joker box,也可以没有joker box;2、除了joker box,其他盒子要么为空,要么都为同色球;3、对于每种颜色的球,除了在joker box里的球,其余都在同一个盒子里,或者该颜色所有球都在joker box。

       给定初态时n个盒子里含有每种颜色的球各多少个,问转移到终态最少需要操作多少次。

    解法:贪心。有一种操作是,首先选出一个joker box,然后将其他盒子里的所有球全部拿到joker box里面,这样操作次数为n-1,每个盒子操作一次。而对于任意一个盒子,要使操作次数为0,有两种可能:1、初态时它为空;2、初态时它仅含一种颜色t的球,且之前没有盒子初态仅含颜色t的球。如果不满足则两个条件,直接将盒子拿空。

    tag:greedy, good

      1 // BEGIN CUT HERE
      2 /*
      3  * Author:  plum rain
      4  * score :
      5  */
      6 /*
      7 
      8  */
      9 // END CUT HERE
     10 #line 11 "MarblesRegroupingEasy.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 sz(v) ((int)(v).size())
     41 #define all(t) t.begin(),t.end()
     42 #define zero(x) (((x)>0?(x):-(x))<eps)
     43 #define out(x) cout<<#x<<":"<<(x)<<endl
     44 #define tst(a) cout<<a<<" "
     45 #define tst1(a) cout<<#a<<endl
     46 #define CINBEQUICKER std::ios::sync_with_stdio(false)
     47 
     48 typedef vector<int> vi;
     49 typedef vector<string> vs;
     50 typedef vector<double> vd;
     51 typedef pair<int, int> pii;
     52 typedef long long int64;
     53 
     54 const double eps = 1e-8;
     55 const double PI = atan(1.0)*4;
     56 const int inf = 2139062143 / 2;
     57 
     58 class MarblesRegroupingEasy
     59 {
     60     public:
     61         bool vis[100];
     62         int minMoves(vector <string> v){
     63             clr0 (vis);
     64             int cnt = 0;
     65             for (int i = 0; i < sz(v); ++ i){
     66                 string s = v[i];
     67                 int tmp = 0, idx = 0;
     68                 for (int j = 0; j < sz(s); ++ j)
     69                     if (s[j] != '0') ++ tmp, idx = j;
     70                 if (!tmp) continue;
     71                 if (tmp == 1 && !vis[idx]){
     72                     vis[idx] = 1; continue;
     73                 }
     74                 ++ cnt;
     75             }
     76             if (cnt) -- cnt;
     77             return cnt;
     78         }
     79         
     80 // BEGIN CUT HERE
     81     public:
     82     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(); }
     83     private:
     84     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(); }
     85     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; } }
     86     void test_case_0() { string Arr0[] = {"20",
     87  "11"}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 0; verify_case(0, Arg1, minMoves(Arg0)); }
     88     void test_case_1() { string Arr0[] = {"11",
     89  "11",
     90  "10"}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 1; verify_case(1, Arg1, minMoves(Arg0)); }
     91     void test_case_2() { string Arr0[] = {"10",
     92  "10",
     93  "01",
     94  "01"}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 1; verify_case(2, Arg1, minMoves(Arg0)); }
     95     void test_case_3() { string Arr0[] = {"11",
     96  "11",
     97  "11",
     98  "10",
     99  "10",
    100  "01"}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 3; verify_case(3, Arg1, minMoves(Arg0)); }
    101     void test_case_4() { string Arr0[] = {"020008000070",
    102  "000004000000",
    103  "060000600000",
    104  "006000000362",
    105  "000720000000",
    106  "000040000000", 
    107  "004009003000",
    108  "000800000000", 
    109  "020030003000",
    110  "000500200000",
    111  "000000300000"}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 6; verify_case(4, Arg1, minMoves(Arg0)); }
    112 
    113 // END CUT HERE
    114 
    115 };
    116 
    117 // BEGIN CUT HERE
    118 int main()
    119 {
    120 //    freopen( "a.out" , "w" , stdout );    
    121     MarblesRegroupingEasy ___test;
    122     ___test.run_test(-1);
    123        return 0;
    124 }
    125 // END CUT HERE
    View Code
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  • 原文地址:https://www.cnblogs.com/plumrain/p/srm_387.html
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