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 << "\tExpected: \"" << Expected << '\"' << endl; cerr << "\tReceived: \"" << 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

 

 

 

posted @ 2013-12-30 12:21  Plumrain  阅读(194)  评论(0)  编辑  收藏  举报