POJ 1094 Sorting It All Out

/*
*POJ 1094 Sorting It All Out
*参考http://hi.baidu.com/bnjyjncwbdbjnzr/item/a4ffa006defc47c42f4c6bd5
*以邻接表为图的存储结构的算法:
*a)扫描顶点表,将入度为零的顶点入栈;
    (ps:这里栈可以使优先队列 and so on)
*b)当栈非空时:
*   输出栈顶元素v:出栈;
*   检查v的出边,将每条出边的终端顶点的入度减1,若该顶点入度为0,入栈;
*c)当栈空时,若输出的顶点小于顶点数,则说明AOV网有回路,否则拓扑排序完成.
*/

/*
本题:
    首先是用邻接表存储 各个点的先后顺序;
    再用cnt[]数组存储每个点的入度;
    每输入一条边就需判断一下,是否是矛盾(产生环),还是顺序已经确定;
    如果输入完后,仍然不能判断就是"Sorted sequence cannot be determined."
*/

#include <cstdio>
#include <cstring>

#define MAXN 29
int n, m;
bool adj[MAXN][MAXN];
int cnt[MAXN];
char order[MAXN];

int topoSort()
{
	int i, j, k;
	memset(cnt, 0, sizeof(cnt));
	memset(order, ‘\0‘, sizeof(order));
	bool flag = true;
	/*统计入度*/
	for (i = 1; i <= n; i++) {
		for (j = 1; j <= n; j++) {
			if (adj[i][j]) {
				cnt[j]++;
			}
		}
	}
	for ( i = 1; i <= n; i++) {
		k = 0;
		for (j = 1; j <= n; j++) {
			if (cnt[j] == 0) {
				if (k == 0) {
					k = j;
				} else {
					flag = false; //入度为零的点不唯一
				}
			}
		}
		if (k == 0) {
			return 0; //存在环
		}
		cnt[k] = -1;
		order[i - 1] = k + ‘A‘ - 1;
		for (j = 1; j <= n; j++) {
			if (adj[k][j]) {
				cnt[j]--;   //k指向的所有节点入度-1
			}
		}
	}
	if (flag) {
		return 1;
	} else {
		return 2;
	}
}

int main()
{
	int u, v;
	int result;
	char s[10];
	int i;
	while (scanf("%d %d", &n, &m) && n + m) {
		memset(adj, false, sizeof(adj));
		bool flag = false;
		for (i = 1; i <= m; i++) {
			scanf("%s", s);
			if (flag) {
				continue;
			}
			u = s[0] - ‘A‘ + 1;
			v = s[2] - ‘A‘ + 1;
			adj[u][v] = true;
			result = topoSort();
			if (result == 0) {
				printf("Inconsistency found after %d relations.\n", i);
				flag = true;
			} else if (result == 1) {
				printf("Sorted sequence determined after %d relations: %s.\n", i, order);
				flag = true;
			}
		}
		if (!flag) {
			printf("Sorted sequence cannot be determined.\n");
		}
	}
	return 0;
}