UVaLive 6950 && Gym 100299K Digraphs (DFS找环或者是找最长链)

题意：有n个只包含两个字母的字符串， 要求构造一个m*m的字母矩阵， 使得矩阵的每行每列都不包含所给的字符串， m要尽量大，

如果大于20的话构造20*20的矩阵就行了。

析：开始吧，并没有读对题意，后来才看懂什么意思，然后主要思想就是如果有环，那么一定是可以构造成20*20的，只要环一直重复就好，如果没有环，

那么就要找最长的链，然后矩阵长宽就是 （max+1）/2，然后和上面一样构造就好。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <tr1/unordered_map>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
using namespace std :: tr1;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 26 + 5;
const int mod = 1e9 + 7;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
int G[maxn][maxn];
char s[5];
int ans[maxn], a[maxn];
bool vis[maxn];
int cnt;

bool dfs(int u, int num){
    bool ok = false;
    for(int i = 0; i < 26; ++i){
        if(G[u][i]){
            a[num] = i;
            if(vis[i]){
                cnt = num;
                return true;
            }
            else{
                vis[i] = true;
                if(dfs(i, num+1))  return true;
                vis[i] = false;
            }
            ok = true;
        }
    }
    if(cnt < num && ok){
        cnt = num;
        memcpy(ans, a, sizeof ans);
    }
    return false;
}

int main(){
    int T;  cin >> T;
    while(T--){
        scanf("%d", &n);
        memset(G, -1, sizeof G);
        for(int i = 0; i < n; ++i){
            scanf("%s", s);
            G[s[0]-'a'][s[1]-'a'] = 0;
        }

        int anss = 0;cnt = 0;
        for(int i = 0; i < 26; ++i){
            memset(vis, false, sizeof vis);
            vis[i] = true;
            a[0] = i;
            if(dfs(i, 1)){
                bool ok = false;
                int id = 0;
                for(int j = 0; j < cnt; ++j)
                    if(ok)  ans[id++] = a[j];
                    else if(a[j] == a[cnt]){ ans[id++] = a[j]; ok = true; }
                anss = id;
                break;
            }
        }

        if(anss){
            for(int i = 0; i < 20; ++i){
                for(int j = 0; j < 20; ++j)
                    printf("%c", ans[(j+i)%anss]+'a');
                printf("\n");
            }
        }
        else{
            int x = (cnt+2)/2;
            for(int i = 0; i < x; ++i){
                for(int j = 0; j < x; ++j)
                    printf("%c", ans[i+j]+'a');
                printf("\n");
            }
        }

    }
    return 0;
}