题意：不同国家的人站在一起，同一个国家的人相邻站，给你一个数组，表示 这个位置属于的国家有几个人，0可以是任何数字，问你能不能确定这个数组（只有一个解）
解题思路：搜索
解题代码：
1 // BEGIN CUT HERE 2 /* 3 4 */ 5 // END CUT HERE 6 #line 7 "CountryGroupHard.cpp" 7 #include <cstdlib> 8 #include <cctype> 9 #include <cstring> 10 #include <cstdio> 11 #include <cmath> 12 #include <algorithm> 13 #include <vector> 14 #include <string> 15 #include <iostream> 16 #include <sstream> 17 #include <map> 18 #include <set> 19 #include <queue> 20 #include <stack> 21 #include <fstream> 22 #include <numeric> 23 #include <iomanip> 24 #include <bitset> 25 #include <list> 26 #include <stdexcept> 27 #include <functional> 28 #include <utility> 29 #include <ctime> 30 using namespace std; 31 32 #define PB push_back 33 #define MP make_pair 34 35 #define REP(i,n) for(i=0;i<(n);++i) 36 #define FOR(i,l,h) for(i=(l);i<=(h);++i) 37 #define FORD(i,h,l) for(i=(h);i>=(l);--i) 38 39 typedef vector<int> VI; 40 typedef vector<string> VS; 41 typedef vector<double> VD; 42 typedef long long LL; 43 typedef pair<int,int> PII; 44 45 int color[1000]; 46 int tmp[1000]; 47 int n ; 48 int ok = 0 ; 49 void isok() 50 { 51 int tok = 0 ; 52 for(int i = 0 ;i < n;i ++) 53 { 54 if(color[i] != 0 ) 55 tok = 1; 56 if(i >= 1 && color[i] == 0 && color[i-1] == 0 ) 57 ok = 2; 58 } 59 if(tok == 0 && n != 1) 60 return; 61 62 ok ++ ; 63 } 64 void dosolve(int k,int be, int c) 65 { 66 if(ok == 2) 67 return; 68 if(k == n) 69 { 70 isok(); 71 return; 72 } 73 if(tmp[k] == 0) 74 { 75 dosolve(k+1,be,c); 76 return; 77 } 78 if(be > k) 79 { 80 if(tmp[k] == c ) 81 { 82 dosolve(k+1,be,c); 83 } 84 else return; 85 }else{ 86 for(int j = max(be,k-tmp[k]+1); j <= k;j ++) 87 { 88 if(j + tmp[k] > n) 89 break; 90 for(int s = j ; s < j + tmp[k]; s ++) 91 color[s] = 1; 92 dosolve(k+1,j + tmp[k],tmp[k]); 93 for(int s = j ; s < j + tmp[k]; s ++) 94 color[s] = 0; 95 } 96 } 97 } 98 class CountryGroupHard 99 { 100 public: 101 string solve(vector <int> a) 102 { 103 ok = 0 ; 104 n = a.size(); 105 for(int i= 0 ;i< n;i ++) 106 tmp[i] = a[i]; 107 dosolve(0,0,-1); 108 //printf("%d ",ok); 109 if(ok == 1 ) 110 return "Sufficient"; 111 return "Insufficient"; 112 } 113 114 // BEGIN CUT HERE 115 public: 116 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(); } 117 private: 118 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(); } 119 void verify_case(int Case, const string &Expected, const string &Received) { cerr << "Test Case #" << Case << "..."; if (Expected == Received) cerr << "PASSED" << endl; else { cerr << "FAILED" << endl; cerr << " Expected: "" << Expected << '"' << endl; cerr << " Received: "" << Received << '"' << endl; } } 120 void test_case_0() { int Arr0[] = {0,2,3,0,0}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); string Arg1 = "Sufficient"; verify_case(0, Arg1, solve(Arg0)); } 121 void test_case_1() { int Arr0[] = {0,2,0}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); string Arg1 = "Insufficient"; verify_case(1, Arg1, solve(Arg0)); } 122 void test_case_2() { int Arr0[] = {0,3,0,0,3,0}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); string Arg1 = "Sufficient"; verify_case(2, Arg1, solve(Arg0)); } 123 void test_case_3() { int Arr0[] = {0,0,3,3,0,0}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); string Arg1 = "Insufficient"; verify_case(3, Arg1, solve(Arg0)); } 124 void test_case_4() { int Arr0[] = {2,2,0,2,2}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); string Arg1 = "Sufficient"; verify_case(4, Arg1, solve(Arg0)); } 125 126 // END CUT HERE 127 128 }; 129 130 // BEGIN CUT HERE 131 int main() 132 { 133 CountryGroupHard ___test; 134 ___test.run_test(-1); 135 return 0; 136 } 137 // END CUT HERE