An ascending sorted sequence of distinct values is one in which some form of a less-than operator is used to order the elements from smallest to largest. For example, the sorted sequence A, B, C, D implies that A < B, B < C and C < D. in this problem, we will give you a set of relations of the form A < B and ask you to determine whether a sorted order has been specified or not.
Input
Input consists of multiple problem instances. Each instance starts with a line containing two positive integers n and m. the first value indicated the number of objects to sort, where 2 <= n <= 26. The objects to be sorted will be the first n characters of the uppercase alphabet. The second value m indicates the number of relations of the form A < B which will be given in this problem instance. Next will be m lines, each containing one such relation consisting of three characters: an uppercase letter, the character "<" and a second uppercase letter. No letter will be outside the range of the first n letters of the alphabet. Values of n = m = 0 indicate end of input.
Output
For each problem instance, output consists of one line. This line should be one of the following three:
Sorted sequence determined after xxx relations: yyy...y.
Sorted sequence cannot be determined.
Inconsistency found after xxx relations.
where xxx is the number of relations processed at the time either a sorted sequence is determined or an inconsistency is found, whichever comes first, and yyy...y is the sorted, ascending sequence.
輸入樣例
4 6
A
A
A
輸出樣例
Sorted sequence determined after 4 relations: ABCD.
Inconsistency found after 2 relations.
Sorted sequence cannot be determined.
我們現(xiàn)在只看樣例：4和6的含義分別是：
4表示表達(dá)式中一共含有4個(gè)字符，也就是說我們需要排序的是4個(gè)樣例，6呢，就是需要用戶輸入6個(gè)表達(dá)式，我們根據(jù)這兩個(gè)數(shù)值和相應(yīng)的表達(dá)式，去推算字符的先后順序
，可能會(huì)有三種結(jié)果，一種當(dāng)然是排序成功，比如第一個(gè)輸入，一種是會(huì)有邏輯沖突，比如第二個(gè)輸入，一種就是條件不足，不能判斷
我的算法思想是這樣的，定義一個(gè)結(jié)構(gòu)體，該結(jié)構(gòu)體有三個(gè)變量，一個(gè)是key表示字符，一個(gè)是value，表示權(quán)重，一個(gè)就是布爾變量，表示該值是否已經(jīng)參加運(yùn)算
1.根據(jù)用戶輸入初始換結(jié)構(gòu)體數(shù)組變量（chain_head），將相應(yīng)的key值存到結(jié)構(gòu)體當(dāng)中，value則是UNSET，布爾類型變量則是false，表示還未參加到運(yùn)算當(dāng)中，
2.接下來，計(jì)算第n個(gè)表達(dá)式（n初始化成0），我們獲取第n個(gè)表達(dá)式的第一個(gè)和第三個(gè)字符，分別獲得該key在結(jié)構(gòu)體數(shù)組中的索引，index_a和index_b，這樣需要比較的就是
chain_head[index_a]和chain_head[index_b]了，首先看下chain_head[index_a].value是否小于chain_head[index_b].value,如果小于，則不進(jìn)行任何操作，繼續(xù)計(jì)算第n+1個(gè)
表達(dá)式，如果chain_head[index_a].value大于chain_head[index_b].value并且，chain_head[index_b].FirstApper=true(index_b的結(jié)構(gòu)體從未參加運(yùn)算)，則表示邏輯沖突，停
止循環(huán)，輸出結(jié)構(gòu)，如果chain_head[index_a].value==chain_head[index_b].value則chain_head[index_b].value=chain_head[index_a].value+1;即可，然后進(jìn)行第n+1個(gè)表達(dá)
式的計(jì)算，繼續(xù)進(jìn)行第二步的運(yùn)算
3.清理現(xiàn)場，輸出結(jié)果
源代碼如下：
// Sort_It.cpp : 定義控制臺(tái)應(yīng)用程序的入口點(diǎn)。
//
#include "stdafx.h"
#include
using namespace std;
/****************************************************************************************************************
***文件名稱：Sort_It.cpp
***文件描述：根據(jù)表達(dá)式對(duì)字符進(jìn)行排序
****************************************************************************************************************/
typedef struct {
char key;
int value;
bool FirstApper;
}Key_Value;
Key_Value* chain_head=NULL;
int nCount=0;//需要排序的字符個(gè)數(shù)
int nCur_Count=0;//當(dāng)前已經(jīng)賦值的個(gè)數(shù)
int result=0;
#define DONE 1
#define INCONSIST 2
#define UNENOUGH 3
#define UNSET 1
void operation(char key_a,char key_b);
/////////////////////////////////////////////////////////////////////////////////////////////////////////////////
///函數(shù)名稱：TestOver
///函數(shù)描述：檢測根據(jù)當(dāng)前獲得信息，檢測是否已經(jīng)可以完成排序（value的值疊加，如果等于1,2,3的和則表示排序完成）
/////////////////////////////////////////////////////////////////////////////////////////////////////////////////
bool TestOver(Key_Value* head)
{
int nSum_1=0,nSum=0;
for(int i=0;i