//1706源码（AC，已测试）：

#include <stdio.h>
#include <string.h>
#include <ctype.h>

char line[100];							//当前处理的那一行
char ref[3002][83];						//记录Reference description
int orig[1002], now[1002], st[1002];	//分别是原始编号，现在编号，分配空间的起始行
int p, num, tt;

bool empty(char a[])					//判断是否是空行
{
	int i;
	for (i = 0; a[i] != 0; i++)
		if (a[i] != ' ')
			return false;
	return true;
}

int getnum(char a[], int &k)			//从字符数组中获取 数
{
	int tmp = 0;
	for (; isdigit(a[k]); k++)
		tmp = tmp * 10 + a[k] - '0';
	return tmp;
}

int find(int x)							//找原始编号为x的Reference的位置
{
	int i;
	for (i = 0; i < p; i++)
		if (orig[i] == x)
			return i;
	return -1;
}

void proc()								//处理正文
{
	int k, tmp, q;
	for (k = 0; line[k] != 0; k++)
		if (line[k] == '[')				//若引用Reference，则重新给它一个编号
		{
			k++;
			tmp = getnum(line, k);
			if ((q = find(tmp)) == -1)	//若没有记录过当前Reference，则分配新的编号
			{
				orig[p] = tmp; now[p] = num; st[p] = 0;
				q = p;
				p++; num++;
			}
			else
				if (now[q] == -1)		//否则若没有分配编号，则分配
				{
					now[q] = num;
					num++;
				}
			printf("[%d]", now[q] + 1);
		}
		else
			printf("%c", line[k]);		//其他数据正常输出
	printf("\n");
}

int main()
{
	bool rec, emptyline, cont;
	int k, tmp, q, i, j;

	tt = p = num = 0; rec = 0; emptyline = 1; cont = 1;
	while (gets(line))
	{
		if (!empty(line))				//非空行
		{
			if (line[0] == '[' && !cont)//若是Reference description的首行则记录
			{
				k = 1;
				tmp = getnum(line, k);
				if ((q = find(tmp)) == -1)
				{
					orig[p] = tmp;
					now[p] = -1;
					st[p] = tt;
					p++;
				}
				else
					st[q] = tt;
				strcpy(ref[tt], line);
				tt++;
				rec = 1;
			}
			else
				if (rec)				//是Reference description的接下去几行
				{
					strcpy(ref[tt], line);
					tt++;
				}
				else					//是正文
				{
					emptyline = 0;
					proc();
				}
		}
		else
		{
			rec = 0; cont = 0;			//若是空行，则多于一个空行只输出一个
			if (!emptyline)
			{
				printf("\n");
				emptyline = 1;
			}
		}
	}

	for (i = 0; i < p; i++)				//输出Reference description
	{
		for (j = 0; j < p; j++)			//按照顺序输出
			if (now[j] == i)
				break;

		j = st[j];
		printf("[%d]", i + 1);
		k = 1;
		getnum(ref[j], k);
		k++;
		for (; ref[j][k] != 0; k++)
			printf("%c", ref[j][k]);
		printf("\n");
		j++;
		while (j < tt && ref[j][0] != '[')
		{
			puts(ref[j]);
			j++;
		}
		if (i != p - 1)
			printf("\n");
	}
	return 0;	
}



用这个程序走题目中给的那个Sample得到的输出：
（和Sample Output不一样。
我觉得这个程序有问题，可为什么能AC呢？）

[1] Brownell, D, "Dynamic Reverse Address Resolution Protocol
    (DRARP)", Work in Progress.

The Reverse Address Resolution Protocol (RARP) [2] (through the
extensions defined in the Dynamic RARP (DRARP) [1]) explicitly
addresses the problem of network address discovery, and includes an
automatic IP address assignment mechanism.

The Trivial File Transfer Protocol (TFTP) [3] provides for transport
of a boot image from a boot server.  The Internet Control Message
Protocol (ICMP) [4] provides for informing hosts of additional routers
via "ICMP redirect" messages.

Works [2], [4] and [3] can be obtained via Internet.
[1] Finlayson, R., Mann, T., Mogul, J., and M. Theimer, "A Reverse
        Address Resolution Protocol", RFC 903, Stanford, June 1984.

[2] Finlayson, R., Mann, T., Mogul, J., and M. Theimer, "A Reverse
        Address Resolution Protocol", RFC 903, Stanford, June 1984.

[3] Sollins, K., "The TFTP Protocol (Revision 2)",  RFC 783, NIC,
     June 1981.

[4] Postel, J., "Internet Control Message Protocol", STD 5, RFC 792,
        USC/Information Sciences Institute, September 1981.

• Re:请教：可以AC，但过不了题目给的Sample，是什么原因？（附源码） (66) Grant 2008-03-13 05:22:21
  • Re:请教：可以AC，但过不了题目给的Sample，是什么原因？（附源码） (78) tiandi222 2010-09-30 21:13:28