Codeforces Round #290 (Div. 2)C. Fox And Names(拓扑排序)
题目链接： 题目链接
Fox Ciel is going to publish a paper on FOCS (Foxes Operated Computer Systems, pronounce: "Fox"). She heard a rumor: the authors list on the paper is always sorted in the lexicographical order.
After checking some examples, she found out that sometimes it wasn't true. On some papers authors' names weren't sorted inlexicographical order in normal sense. But it was always true that after some modification of the order of letters in alphabet, the order of authors becomes lexicographical!
She wants to know, if there exists an order of letters in Latin alphabet such that the names on the paper she is submitting are following in the lexicographical order. If so, you should find out any such order.
Lexicographical order is defined in following way. When we compare s and t, first we find the leftmost position with differing characters: si ≠ ti. If there is no such position (i. e. s is a prefix of t or vice versa) the shortest string is less. Otherwise, we compare characters si and tiaccording to their order in alphabet.
The first line contains an integer n (1 ≤ n ≤ 100): number of names.
Each of the following n lines contain one string namei (1 ≤ |namei| ≤ 100), the i-th name. Each name contains only lowercase Latin letters. All names are different.
If there exists such order of letters that the given names are sorted lexicographically, output any such order as a permutation of characters 'a'–'z' (i. e. first output the first letter of the modified alphabet, then the second, and so on).
Otherwise output a single word "Impossible" (without quotes).
3 rivest shamir adleman
bcdefghijklmnopqrsatuvwxyz
10 tourist petr wjmzbmr yeputons vepifanov scottwu oooooooooooooooo subscriber rowdark tankengineer
Impossible
10 petr egor endagorion feferivan ilovetanyaromanova kostka dmitriyh maratsnowbear bredorjaguarturnik cgyforever
aghjlnopefikdmbcqrstuvwxyz
7 car care careful carefully becarefuldontforgetsomething otherwiseyouwillbehacked goodluck
acbdefhijklmnogpqrstuvwxyz
题意：给你n个字符串，问你该怎么自定义字符表里的大小关系，来使得这n个字符串的大小是升序的。
比如：
abc
aad
如果它们两个是升序的的话，那么自定义的b肯定大于a。
然后输出任意一个符合条件的字母表。
题解：
其实题目比较简单，利用拓扑排序就可以实现该操作。
但是要注意字符串的比较是如何的。
遇到不相同的就返回其比较值，后面的无需比较
如果不能输出拓扑排序的序列，那么就无法实现。
代码：
#include<cstdio>
#include<cstring>
#include<iostream>
#include<sstream>
#include<algorithm>
#include<vector>
#include<bitset>
#include<set>
#include<queue>
#include<stack>
#include<map>
#include<cstdlib>
#include<cmath>
#define PI 2*asin(1.0)
#define LL __int64
const int MOD = 1e9 + 7;
const int N = 1e2 + 15;
const int INF = (1 << 30) - 1;
const int letter = 130;
using namespace std;
int n;
char str[N][N];
int mp[N][N];
bool dis[N][N];
int in[N], out[N];
bool vis[N];
queue<int>q;
vector<char>vs;
int value(char aim)
{
if(aim == ' ') return 0;
else return aim - 'a' + 1;
}
int main()
{
while(scanf("%d", &n) != EOF)
{
memset(vis, 0, sizeof(vis));
memset(mp, 0, sizeof(mp));
memset(in, 0, sizeof(in));
memset(out, 0, sizeof(out));
memset(dis, 0, sizeof(dis));
getchar();
int max1 = -1, len;
for(int i = 0; i < n; i++)
{
gets(str[i]);
len = strlen(str[i]);
if(len > max1) max1 = len;
}
for(int i = 0; i < n; i++)
{
for(int j = 0; j < max1; j++)
{
if(!(str[i][j] >= 'a' && str[i][j] <= 'z')) str[i][j] = ' ';
}
str[i][max1] = '\0';
}
int a, b;
int t = 0;
for(int j = 0; j < max1; j++)
{
for(int i = 0; i < n - 1; i++)
{
if(!dis[i][i + 1]) ///last
{
a = value(str[i][j]);
b = value(str[i + 1][j]);
if(a != b)
{
dis[i][i + 1] = 1;
t++;
// in[b]++;
mp[a][b] = 1;
/// printf("%d %d\n", a, b);
}
if(t == n - 1) break; ///bijiao n-1ci
}
}
if(t == n - 1) break;
}
for(int i = 0;i <= 26;i++)
for(int j = 0;j <= 26;j++)
{
if(mp[i][j]) in[j]++;
}
while(!q.empty()) q.pop();
vs.clear();
for(int i = 00; i <= 26; i++) if(in[i] == 0) q.push(i);
if(in[0])
{
puts("Impossible");
continue;
}
int flag = 1;
while(!q.empty())
{
int x = q.front();
q.pop();
vs.push_back(x);
for(int i = 0; i <= 26; i++)
{
if(mp[x][i])
{
in[i]--;
if(in[i] == 0 && i != 0) q.push(i);
out[x]--;
}
}
}
for(int i = 1; i <= 26; i++)
{
if(in[i])
{
flag = 0;
break;
}
}
if(flag == 0) puts("Impossible");
else
{
for(int i = 0; i < vs.size(); i++)
{
if(vs[i] != 0)
{
printf("%c", vs[i] + 'a' - 1);
vis[vs[i]] = 1;
}
}
for(int i = 1; i <= 26; i++)
{
if(!vis[i]) printf("%c", i + 'a' - 1);
}
printf("\n");
}
}
return 0;
}