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/*
拓扑序列变形 之 poj 1094 Sorting It All Out
变形:
在每消去唯一一个入度为0的点后,只剩下唯一一个入度为0的点。
这样获得的n个点才是排序好的。
*/
1 #include <iostream> 2 #include <cstdlib> 3 #include <cstdio> 4 #include <cstddef> 5 #include <iterator> 6 #include <algorithm> 7 #include <string> 8 #include <locale> 9 #include <cmath> 10 #include <vector> 11 #include <cstring> 12 #include <map> 13 #include <utility> 14 #include <queue> 15 #include <stack> 16 #include <set> 17 #include <functional> 18 using namespace std; 19 typedef pair<int, int> PII; 20 typedef long long int64; 21 const int INF = 0x3f3f3f3f; 22 const int modPrime = 3046721; 23 const double eps = 1e-9; 24 const int MaxN = 30; 25 const int MaxM = 10010; 26 27 int n, m; 28 bool G[MaxN][MaxN]; 29 int ndCnt[MaxN]; 30 char chList[MaxN]; 31 32 enum emFlag 33 { 34 fgSUCCESS, fgFAIL, fgWAIT 35 }; 36 37 38 emFlag Solve() 39 { 40 stack<int> stk; 41 42 int ndCntTmp[MaxN]; 43 for (int i = 0; i < n; ++i) 44 { 45 ndCntTmp[i] = ndCnt[i]; 46 if (0 == ndCnt[i]) 47 { 48 stk.push(i); 49 } 50 } 51 52 bool sign = true; 53 int chLen = 0; 54 55 while (!stk.empty()) 56 { 57 // 下面的if语句,就是需要注意的地方 58 if (stk.size() > 1) 59 { 60 sign = false; 61 } 62 int nd = stk.top(); 63 stk.pop(); 64 chList[chLen++] = static_cast<char> (‘A‘ + nd); 65 for (int i = 0; i < n; ++i) 66 { 67 if (G[nd][i]) 68 { 69 --ndCntTmp[i]; 70 if (0 == ndCntTmp[i]) 71 { 72 stk.push(i); 73 } 74 } 75 } 76 } 77 78 if (n == chLen) 79 { 80 if (sign) 81 { 82 return fgSUCCESS; 83 } 84 else 85 { 86 return fgWAIT; 87 } 88 } 89 else 90 { 91 return fgFAIL; 92 } 93 } 94 95 void ini() 96 { 97 for (int i = 0; i < n; ++i) 98 { 99 ndCnt[i] = 0; 100 for (int j = 0; j < n; ++j) 101 { 102 G[i][j] = false; 103 } 104 } 105 } 106 107 int main() 108 { 109 #ifdef HOME 110 freopen("in", "r", stdin); 111 //freopen("out", "w", stdout); 112 #endif 113 string str; 114 while ((cin >> n >> m) && (n || m)) 115 { 116 ini(); 117 118 bool sign = false; 119 int pos; 120 emFlag fg = fgWAIT; 121 for (int i = 1; i <= m; ++i) 122 { 123 cin >> str; 124 if (!sign) 125 { 126 int a = str[0] - ‘A‘, b = str[2] - ‘A‘; 127 if (!G[a][b]) 128 { 129 G[a][b] = true; 130 ++ndCnt[b]; 131 fg = Solve(); 132 if (fg != fgWAIT) 133 { 134 pos = i; 135 sign = true; 136 } 137 } 138 } 139 } 140 if (sign) 141 { 142 if (fgSUCCESS == fg) 143 { 144 cout << "Sorted sequence determined after " << pos << " relations: "; 145 for (int j = 0; j < n; ++j) 146 { 147 cout << chList[j]; 148 } 149 cout << "." << endl; 150 } 151 else 152 { 153 cout << "Inconsistency found after " << pos << " relations." << endl; 154 } 155 } 156 else 157 { 158 cout << "Sorted sequence cannot be determined." << endl; 159 } 160 } 161 162 #ifdef HOME 163 cerr << "Time elapsed: " << clock() / CLOCKS_PER_SEC << " ms" << endl; 164 _CrtDumpMemoryLeaks(); 165 #endif 166 return 0; 167 }
拓扑序列变形 之 poj 1094 Sorting It All Out
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原文地址:http://www.cnblogs.com/shijianming/p/5055923.html