UVa 642 - Word Amalgamation

题目：给你一个单词列表。再给你一些新的单词。输出列表中又一次排列能得到此新单词的词。

分析：字符串。对每一个字符串的字母排序生成新的传f（str）。总体排序，用二分来查找就可以。

说明：注意输出要满足字典序，先排序后查找。

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>

using namespace std;

typedef struct wnode
{
	char word[7];
	char abcd[7];
}words;
words W[101];

int cmp(words a, words b)
{
	int c = strcmp(a.abcd, b.abcd);
	if (c != 0) return c<0;
	return strcmp(a.word, b.word)<0;
}

int bs(char str[], int r)
{
	int l = 0,m,c;
	while (l < r) {
		m = (l+r)/2;
		c = strcmp(W[m].abcd, str);
		if (c < 0) 
			l = m+1;
		else r = m;
	}
	return l;
}

char buf[7];

int main()
{
	int count = 0;
	while (gets(W[count].word) && strcmp(W[count].word, "XXXXXX")) {
		strcpy(W[count].abcd, W[count].word);
		sort(W[count].abcd, W[count].abcd+strlen(W[count].abcd));
		count ++;
	}
	
	sort(W, W+count, cmp);
	
	while (gets(buf) && strcmp(buf, "XXXXXX")) {
		sort(buf, buf+strlen(buf));
		int s = bs(buf, count-1),flag = 0;
		while (!strcmp(buf, W[s].abcd)) {
			printf("%s\n",W[s ++].word);
			flag = 1;
		}
		if (!flag) 
			printf("NOT A VALID WORD\n");
		printf("******\n");
	}
	return 0;
}