SRM387 div1

250pt：

题目：有一些盒子（不大于50个），每个盒子里有一些大理石（最多50种颜色），然后给定每个盒子里每种颜色大理石的个数（没有为0），求最少操作几步满足：

1：最多只能一个盒子里有多种颜色，叫做jaker

2：每种颜色最多位于一个一个非jaker的盒子里，且每个非jaker的盒子最多只含有一种颜色：

移动一步可以移动一个盒子任意个石头到另外一个。。

思路：如果我们枚举哪个是jaker，那么对于剩下来的盒子，就剩下要不要移动到jaker里的抉择了。。而且：

1： 如果该盒子有多种颜色，那么一定要移动，全部移动到jaker里就行

2： 如果该盒子有只有一种颜色，那么该颜色用过（就是在前面也有一个只有该颜色的）则必须移动，否者不移动

3：空盒子不移动

code：

 // BEGIN CUT HERE
 /*

 */
 // END CUT HERE
 #line 7 "MarblesRegroupingEasy.cpp"
 #include <cstdlib>
 #include <cctype>
 #include <cstring>
 #include <cstdio>
 #include <cmath>
 #include <algorithm>
 #include <vector>
 #include <string>
 #include <iostream>
 #include <sstream>
 #include <map>
 #include <set>
 #include <queue>
 #include <stack>
 #include <fstream>
 #include <numeric>
 #include <iomanip>
 #include <bitset>
 #include <list>
 #include <stdexcept>
 #include <functional>
 #include <utility>
 #include <ctime>
 #define M0(a) memset(a, 0, sizeof(a))
 using namespace std;

 #define PB push_back
 #define MP make_pair

 #define REP(i,n) for(i=0;i<(n);++i)
 #define FOR(i,l,h) for(i=(l);i<=(h);++i)
 #define FORD(i,h,l) for(i=(h);i>=(l);--i)

 typedef vector<int> VI;
 typedef vector<string> VS;
 typedef vector<double> VD;
 typedef long long LL;
 typedef pair<int,int> PII;

 int vis[], n, m;
 vector<string> a;
 int work(int nt){
      M0(vis);
      int ret = ;
      for (int i = ; i < n; ++i)
        if (i != nt){
          int cnt = , p = -;
          for (int j = ; j < m; ++j)
             if (a[i][j] != '') ++cnt, p = j;
          if (cnt == ) continue;
          else if (cnt > ) ret ++;
          else {
              if (!vis[p]) vis[p] = ;
              else ++ret;
          }
      }
      return ret;
 }

 class MarblesRegroupingEasy
 {
         public:
         int minMoves(vector <string> b)
         {
              n = b.size();
              m = b[].size();
              a = b;
              int ret = ;
              for (int i = ; i < n; ++i)
                 ret = min(ret, work(i));
              return ret;
         }

 // BEGIN CUT HERE
     public:
     void run_test(int Case) { if ((Case == -) || (Case == )) test_case_0(); if ((Case == -) || (Case == )) test_case_1(); if ((Case == -) || (Case == )) test_case_2(); if ((Case == -) || (Case == )) test_case_3(); if ((Case == -) || (Case == )) test_case_4(); }
     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 << "\tExpected: \"" << Expected << '\"' << endl; cerr << "\tReceived: \"" << Received << '\"' << endl; } }
     void test_case_0() { string Arr0[] = {"",
  ""}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; verify_case(, Arg1, minMoves(Arg0)); }
     void test_case_1() { string Arr0[] = {"",
  "",
  ""}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; verify_case(, Arg1, minMoves(Arg0)); }
     void test_case_2() { string Arr0[] = {"",
  "",
  "",
  ""}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; verify_case(, Arg1, minMoves(Arg0)); }
     void test_case_3() { string Arr0[] = {"",
  "",
  "",
  "",
  "",
  ""}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; verify_case(, Arg1, minMoves(Arg0)); }
     void test_case_4() { string Arr0[] = {"",
  "",
  "",
  "",
  "",
  "",
  "",
  "",
  "",
  "",
  ""}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; verify_case(, Arg1, minMoves(Arg0)); }

 // END CUT HERE

 };

 // BEGIN CUT HERE
 int main()
 {
         MarblesRegroupingEasy ___test;
         ___test.run_test(-);
     //    system("pause");
         return ;
 }
 // END CUT HERE

500pt:

题目：给定一些一维坐标里的线段，问把他们可以分成多少个子集，每个，子集符合下面情况：

1：任意一个不再子集里的线段与子集相交

2.子集元素都不相交

思路：

因为坐标最大才100，所以以坐标进行统计（类似dp思想）

　  直接看代码吗，

不然不好说清楚，代码还是很短的。。

 // BEGIN CUT HERE
 /*

 */
 // END CUT HERE
 #line 7 "IntervalSubsets.cpp"
 #include <cstdlib>
 #include <cctype>
 #include <cstring>
 #include <cstdio>
 #include <cmath>
 #include <algorithm>
 #include <vector>
 #include <string>
 #include <iostream>
 #include <sstream>
 #include <map>
 #include <set>
 #include <queue>
 #include <stack>
 #include <fstream>
 #include <numeric>
 #include <iomanip>
 #include <bitset>
 #include <list>
 #include <stdexcept>
 #include <functional>
 #include <utility>
 #include <ctime>
 using namespace std;

 #define PB push_back
 #define MP make_pair
 #define M0(a) memset(a, 0, sizeof(a))
 #define REP(i,n) for(i=0;i<(n);++i)
 #define FOR(i,l,h) for(i=(l);i<=(h);++i)
 #define FORD(i,h,l) for(i=(h);i>=(l);--i)

 typedef vector<int> VI;
 typedef vector<string> VS;
 typedef vector<double> VD;
 typedef long long LL;
 typedef pair<int,int> PII;
 int f[];

 class IntervalSubsets
 {
         public:
         int numberOfSubsets(vector <int> start, vector <int> finish)
         {
              int n = start.size();
              M0(f);
              f[] = ;
              for (int i = ; i <= ; ++i){
                  int L = -;
                  for (int j = ; j < n; ++j)
                      if (finish[j] <= i) L = max(start[j], L);
                  if (L == -)
                         f[i] = f[i - ];
                  else
                     for (int j = ; j < n; ++j)
                         if (finish[j] >= L && finish[j] <= i) f[i] += f[start[j] - ];
              }
              return f[];
         }

 // BEGIN CUT HERE
     public:
     void run_test(int Case) { if ((Case == -) || (Case == )) test_case_0(); if ((Case == -) || (Case == )) test_case_1(); if ((Case == -) || (Case == )) test_case_2(); if ((Case == -) || (Case == )) test_case_3(); if ((Case == -) || (Case == )) test_case_4(); }
     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 << "\tExpected: \"" << Expected << '\"' << endl; cerr << "\tReceived: \"" << Received << '\"' << endl; } }
     void test_case_0() { int Arr0[] = {,}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arr1[] = {,}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); int Arg2 = ; verify_case(, Arg2, numberOfSubsets(Arg0, Arg1)); }
     void test_case_1() { int Arr0[] = {,,}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arr1[] = {,,}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); int Arg2 = ; verify_case(, Arg2, numberOfSubsets(Arg0, Arg1)); }
     void test_case_2() { int Arr0[] = {,,,,}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arr1[] = {,,,,}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); int Arg2 = ; verify_case(, Arg2, numberOfSubsets(Arg0, Arg1)); }
     void test_case_3() { int Arr0[] = {,,,,,,,,}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arr1[] = {,,,,,,,,}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); int Arg2 = ; verify_case(, Arg2, numberOfSubsets(Arg0, Arg1)); }
     void test_case_4() { int Arr0[] = {, , , }; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arr1[] = {, , , }; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); int Arg2 = ; verify_case(, Arg2, numberOfSubsets(Arg0, Arg1)); }

 // END CUT HERE

 };

 // BEGIN CUT HERE
 int main()
 {
         IntervalSubsets ___test;
         ___test.run_test(-);
      //   system("pause");
         return ;
 }
 // END CUT HERE