poj1094 Sorting It All Out (拓扑排序)

题　意：给你一定的关系，当给定的其中若干关系（按照时间顺序）已经形成回路，对剩下的关系不处理，直接判定矛盾；当 　　　　　其中若干（按照时间顺序）关系已经能够确定一种唯一的拓扑排序时，强调唯一，对剩下的关系不处理；否则判定　　　　　　不能确定。

思　路：每对关系依次进行拓扑排序。

学习点：拓扑排序。

源代码：

#include <stdio.h>#include <string.h>#define MAXN 27//#define OOint nNum, nPair;int nMap[MAXN][MAXN];int nIndegree[MAXN];char nOrder[MAXN];int topoSort();int main(){int i;int nFrom, nTo;int nFlag, nIndex;char nStr[5];#ifdef OOfreopen("poj1094.in", "r", stdin);#endifwhile (2 == scanf("%d %d", &nNum, &nPair)){if (!nNum && !nPair) break;nFlag = 0; memset(nMap, 0, sizeof(nMap));for (i = 0; i != nPair; ++i){scanf("%s", nStr);if (nFlag) continue;nFrom = nStr[0] - 'A'; nTo = nStr[2] - 'A';nMap[nFrom][nTo] = 1;nIndex = topoSort();if (nIndex == 1){printf("Sorted sequence determined after %d relations: %s./n", i + 1, nOrder);nFlag = 1;}else if (nIndex == 2){printf("Inconsistency found after %d relations./n", i + 1);nFlag = 1;}}if (nFlag == 0) printf("Sorted sequence cannot be determined./n");}return 0;}int topoSort(){int i, j, k;int nCnt, isOnly, nIndex;memset(nIndegree, 0, sizeof(nIndegree));memset(nOrder, '/0', sizeof(nOrder));for (i = 0; i != nNum; ++i){for (j = 0; j != nNum; ++j){if (nMap[i][j]) nIndegree[j] ++;}}isOnly = 1; k = 0;for (i = 0; i != nNum; ++i){nCnt = 0; nFlag = 1;for (j = 0; j != nNum; ++j){if (nIndegree[j]) continue;nCnt ++;nIndex = j;}if (nCnt == 0) return 2; // k = 0,表示所有点的都有入度，形成环，矛盾。if (nCnt > 1) {isOnly = 0;}nIndegree[nIndex] --;nOrder[k++] = nIndex + 'A';for (j = 0; j != nNum; ++j){if (nMap[nIndex][j]) nIndegree[j]--;}}if (isOnly) return 1;else return 3;}