CodeForces#290 div.2 题解

A题：

题意：按要求输出图形(详见题目样例)

思路：模拟。循环随便搞

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <vector>
#include <deque>
#include <list>
#include <cctype>
#include <algorithm>
#include <climits>
#include <queue>
#include <stack>
#include <cmath>
#include <map>
#include <set>
#include <iomanip>
#include <cstdlib>
#include <ctime>
#define ll long long
#define ull unsigned long long
#define all(x) (x).begin(), (x).end()
#define clr(a, v) memset( a , v , sizeof(a) )
#define pb push_back
#define mp make_pair
#define read(f) freopen(f, "r", stdin)
#define write(f) freopen(f, "w", stdout)
using namespace std;
const double pi = acos(-1);

int main()
{
    ios::sync_with_stdio( false );

    int n, m;
    while ( cin >> n >> m ){
        for ( int i = 1; i <= n; i ++ ){
            if ( i & 1 ){
                for ( int j = 0; j < m; j ++ ){
                    cout << '#';
                }
            }
            else if ( i % 4 ){
                for ( int j = 0; j < m - 1; j ++ ){
                    cout << '.';
                }
                cout << '#';
            }
            else{
                cout << '#';
                for ( int i = 1; i < m; i ++ ){
                    cout << '.';
                }
            }
            cout << endl;
        }
    }
    
    return 0;
} 

B题：

题意：给出一张图，判断是否存在一个环，环上所有点的颜色相同，且环上至少有四个点

思路：深搜，判断是否有四点以上的环

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <vector>
#include <deque>
#include <list>
#include <cctype>
#include <algorithm>
#include <climits>
#include <queue>
#include <stack>
#include <cmath>
#include <map>
#include <set>
#include <iomanip>
#include <cstdlib>
#include <ctime>
#define ll long long
#define ull unsigned long long
#define all(x) (x).begin(), (x).end()
#define clr(a, v) memset( a , v , sizeof(a) )
#define pb push_back
#define mp make_pair
#define read(f) freopen(f, "r", stdin)
#define write(f) freopen(f, "w", stdout)
using namespace std;
const double pi = acos(-1);

int flag = 0;
string s[105];
bool vis[105][105];
int dir[4][2] = { 1,0, -1,0, 0,1, 0,-1 };
int n, m;
int be, en;
void init(){
    memset ( vis, 0, sizeof ( vis ) );
    string tmp = "";
    for ( int i = 0; i < 105; i ++ ){
        s[i] = tmp;
    }
    flag = 0;
}
void dfs ( int x, int y, int cnt ){
    for ( int i = 0; i < 4; i ++ ){
        int xi = x + dir[i][0];
        int yi = y + dir[i][1];
        if ( xi >= 0 && xi < n && yi >= 0 && yi < m ){
            if ( xi == be && yi == en && cnt >= 4 ){
                flag = 1;
                return;
            }
            if ( s[xi][yi] == s[x][y] && ( ! vis[xi][yi] ) ){
                vis[xi][yi] = 1;
                dfs ( xi, yi, cnt + 1 );
            }
            if ( flag ) return;
        }
    }
}
int main()
{
    while ( cin >> n >> m ){
        init();
        for ( int i = 0; i < n; i ++ ){
            cin.ignore();
            cin >> s[i];
        }
        for ( int i = 0; i < n; i ++ ){
            for ( int j = 0; j < m; j ++ ){
                clr ( vis, 0 );
                be = i;
                en = j;
                vis[be][en] = 1;
                dfs ( i, j, 1 );
                if ( flag )break;
            }
            if ( flag ) break;
        }
        if ( flag ){
            cout << "Yes" << endl;
        }
        else{
            cout << "No" << endl;
        }
    }
    return 0;
}

C题：

题意：给出若干个字符串，并且这些字符串已经按“字典序"进行排序了，输出这种”字典序“，若不存在，输出-1

思路：拓扑排序，按拓扑排序建图，输出拓扑序即可，若存在环，输出-1

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <vector>
#include <deque>
#include <list>
#include <cctype>
#include <algorithm>
#include <climits>
#include <queue>
#include <stack>
#include <cmath>
#include <map>
#include <set>
#include <iomanip>
#include <cstdlib>
#include <ctime>
#define ll long long
#define ull unsigned long long
#define all(x) (x).begin(), (x).end()
#define clr(a, v) memset( a , v , sizeof(a) )
#define pb push_back
#define mp make_pair
#define read(f) freopen(f, "r", stdin)
#define write(f) freopen(f, "w", stdout)
using namespace std;
const double pi = acos(-1);
//Problem C
string s[105];
ll id[30];
ll adj[30][30];
int main()
{
    ios::sync_with_stdio( false );
    int n;
    while ( cin >> n ){
        int flag = 0;
        clr ( id, 0 );
        clr ( adj, 0 );
        for ( int i = 0; i < n; i ++ ){
            cin.ignore();
            cin >> s[i];
            for ( int j = 0; j < i; j ++ ){
                int k;
                for ( k = 0; k < s[j].size() && k < s[i].size(); k ++ ){
                    if ( s[j][k] != s[i][k] ){
                        id[s[i][k]-'a'] ++;
                        adj[s[j][k]-'a'][s[i][k]-'a'] ++;
                        break;
                    }
                }
                if ( k == s[i].size() && k < s[j].size() ){
                    flag = 1;
                }
            }
        }
        if ( flag == 1 ){
            cout << "Impossible" << endl;
            continue;
        }
        string ans = "";
        int vis[30] = { 0 };
        for ( int i = 0; i < 26; i ++ ){
            for ( int j = 0; j < 26; j ++ ){
                if ( vis[j] == 0 && id[j] == 0 ){
                    char ch = 'a';
                    ch += j;
                    string tmp = "a";
                    tmp[0] += j;
                    ans += tmp;
                    vis[j] = 1;
                    for ( int k = 0; k < 26; k ++ ){
                        id[k] -= adj[j][k];
                    }
                    break;
                }
            }
        }
        if ( ans.size() == 26 ){
            cout << ans << endl;
        }
        else{
            cout << "Impossible" << endl;
        }
    }
    return 0;
}

D题：

题意：给出n个步长，每种步长需要花费x，求出最少消耗使得能够走到距离

思路：可能是dp，没做

E题：

(未完待续)