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;

}

巴特西

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

最新文章

热门文章