拓扑排序——Codeforces Round #290 (Div. 2) C. Fox And Names

• 题目链接：http://codeforces.com/contest/510/problem/C

• 题意：见链接

• 分析：对于给出的N个字符串，我们按顺序判断相邻2个字符串，如果前者比后者长，肯定不构成字典序。否则按第一对不相等的字符建有向边。构图结束后，跑一遍拓扑排序，如果得到的字符数最后得到字符数小于26则不构成字典序。

• AC代码：

/*************************************************************************    > File Name: test.cpp    > Author: Akira     > Mail: [e-mail]  ************************************************************************/#include<bits/stdc++.h>typedef long long LL;typedef unsigned long long ULL;typedef long double LD;#define MST(a,b) memset(a,b,sizeof(a))#define CLR(a) MST(a,0)#define Sqr(a) ((a)*(a))using namespace std;#define MaxN 200#define MaxM 1000#define INF 0x3f3f3f3f#define PI 3.1415926535897932384626const int mod = 1E9+7;const double eps = 1e-6;#define bug cout<<88888888<<endl;#define debug(x) cout << #x" = " << x << endl;int n;char str[MaxN][110];struct Edge{    int u,v,next;}edge[MaxM];int cont,head[MaxN];void add(char u, char v){    edge[cont].u = u;    edge[cont].v = v;    edge[cont].next = head[u];    head[u] = cont++;}int index[MaxN];void init(){    cont = 0;    MST(head,-1);}bool judge(){    for(int i=0;i<n-1;i++)    {        for(int j=1;j<=strlen(str[i]+1);j++)        {            if( str[i][j]!=str[i+1][j])            {                if(str[i][j]==0) return false;                add(str[i][j], str[i+1][j]);                index[str[i+1][j]]++;                break;            }        }    }    return true;}char ans[26];bool topo(){    int cnt = 0;    for(int i='a';i<='z';i++) if(index[i]==0) ans[cnt++] = i;     int h = 0;    while(h<cnt)    {        int now = ans[h++];        for(int i=head[now];i!=-1;i=edge[i].next)        {            int v = edge[i].v;            index[v]--;            if(index[v]==0)ans[cnt++] = v;        }    }    if(cnt<26) return false;    printf("%s\n", ans);    return true;}int main(){    init();    //std::ios::sync_with_stdio(false);    scanf("%d", &n);    for(int i=0;i<n;i++) scanf("%s", str[i]+1);    if( judge() && topo());    else printf("Impossible\n");    //system("pause");}