Codeforces 907 矩阵编号不相邻构造 团操作状压DFS

A.

#include <bits/stdc++.h>
#define PI acos(-1.0)
#define mem(a,b) memset((a),b,sizeof(a))
#define TS printf("!!!
")
#define pb push_back
#define inf 1e9
//std::ios::sync_with_stdio(false);
using namespace std;
//priority_queue<int,vector<int>,greater<int>> que;
const double eps = 1.0e-8;
typedef pair<int, int> pairint;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 1e6 + 10;
const int  maxm = 300;
const int turn[4][2] = {{1, 0}, { -1, 0}, {0, 1}, {0, -1}};
//next_permutation
ll mod = 1e9 + 7;
int main()
{
        //freopen("out1.txt", "w", stdout);
        int v1, v2, v3, vm;
        int anser;
        cin >> v1 >> v2 >> v3 >> vm;
        int flag = 0;
        if (vm >= v2)
        {
                cout << -1 << endl;
                exit(0);
        }
        for (int i = v3; i <= 2 * v3; i++)
        {
                if (i >= vm && i <= 2 * vm)
                {
                        flag = 1;
                        anser = i;
                        break;
                }
        }
        if (flag)
        {
                cout << 2 * v1 << endl;
                cout << 2 * v2 << endl;
                cout << anser << endl;
        }
        else
        {
                cout << -1 << endl;
        }
}
View Code

B.

#include <bits/stdc++.h>
#define PI acos(-1.0)
#define mem(a,b) memset((a),b,sizeof(a))
#define TS printf("!!!
")
#define pb push_back
#define inf 1e9
//std::ios::sync_with_stdio(false);
using namespace std;
//priority_queue<int,vector<int>,greater<int>> que;
const double eps = 1.0e-8;
typedef pair<int, int> pairint;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 1e6 + 10;
const int  maxm = 300;
const int turn[4][2] = {{1, 0}, { -1, 0}, {0, 1}, {0, -1}};
//next_permutation
ll mod = 1e9 + 7;
char f[100][100];
bool judge(int x)
{
        int flag = 0;
        int cur = (x - 1) / 3;
        x = x % 3;
        if (x == 0)
        {
                x = 3;
        }
        int cur1 = x - 1;
//        cout << "y" << " " << cur1 * 3 + 1 << " " << cur1 * 3 + 3 << endl;
//        cout << "x" << " " << cur * 3 + 1 << " " << cur * 3 + 3 << endl;
        for (int i = cur * 3 + 1; i <= cur * 3 + 3; i++)
        {
                for (int j = cur1 * 3 + 1; j <= cur1 * 3 + 3; j++)
                {
                        if (f[i][j] == '.')
                        {
                                f[i][j] = '!';
                                flag = 1;
                        }
                }
        }
        if (flag)
        {
                return false;
        }
        else
        {
                for (int i = 1; i <= 9; i++)
                {
                        for (int j = 1; j <= 9; j++)
                        {
                                if (f[i][j] == '.')
                                {
                                        f[i][j] = '!';
                                }
                        }
                }
                return true;
        }
}
int main()
{
        //freopen("in.txt", "r", stdin);
        //freopen("out1.txt", "w", stdout);
        char now;
        for (int i = 1; i <= 9; i++)
        {
                for (int j = 1; j <= 9; j++)
                {
                        scanf("%c", &now);
                        if (now != '.' && now != 'x' && now != 'o')
                        {
                                j--;
                        }
                        else
                        {
                                f[i][j] = now;
                        }
                }
        }
        int x, y;
        cin >> x  >> y;
        x = x % 3;
        if (x == 0)
        {
                x = 3;
        }
        y = y % 3;
        if (y == 0)
        {
                y = 3;
        }
        int want = (x - 1) * 3 + y;
        //cout << want << endl;
        judge(want);
        for (int i = 1; i <= 9; i++)
        {
                for (int j = 1; j <= 9; j++)
                {
                        cout << f[i][j];
                        if (j % 3 == 0 && j % 9 != 0)
                        {
                                cout << " ";
                        }
                        if (j % 9 == 0)
                        {
                                cout << endl;
                        }
                }
                if (i % 3 == 0)
                {
                        cout << endl;
                }
        }
}
View Code

C.

#include <bits/stdc++.h>
#define PI acos(-1.0)
#define mem(a,b) memset((a),b,sizeof(a))
#define TS printf("!!!
")
#define pb push_back
#define inf 1e9
//std::ios::sync_with_stdio(false);
using namespace std;
//priority_queue<int,vector<int>,greater<int>> que;
const double eps = 1.0e-8;
typedef pair<int, int> pairint;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 1e6 + 10;
const int  maxm = 300;
const int turn[4][2] = {{1, 0}, { -1, 0}, {0, 1}, {0, -1}};
//next_permutation
ll mod = 1e9 + 7;
int flag = 0;
int cur = 0;
char letter = '.';
int num[30];
int test[30];
unordered_set<char> ans;
int main()
{
        //freopen("out1.txt", "w", stdout);
        ios::sync_with_stdio(0);
        cin.tie(0);
        cout.tie(0);


        int n;
        int anser = 0;
        cin >> n;
        char a;
        string b;
        for (int i = 0; i < 26; i++)
        {
                ans.insert('a' + i);
        }
        for (int i = 1; i <= n; i++)
        {
                cin >> a >> b;
                if (ans.size() == 1)
                {
                        if (a == '!')
                        {
                                anser++;
                        }
                        else if (a == '?')
                        {
                                if (b[0] == letter)
                                {
                                        break;
                                }
                                else if(!ans.count(b[0]))
                                {
                                        anser++;
                                }
                        }
                        continue;
                }
                if (a == '!')
                {

                        unordered_set<char> now;
                        for (char j : b)
                        {
                                now.insert(j);
                        }
                        for (auto it = ans.begin(); it != ans.end();)
                        {
                                if (!now.count(*it))
                                {
                                        ans.erase(it++);
                                }
                                else
                                {
                                        it++;
                                }
                        }
                }
                else
                {
                        for (char j : b)
                        {
                                ans.erase(j);
                        }
                }
        }
        cout << anser << endl;
}
View Code

D.给你一个N*M的矩阵 每位的编号为(i-1)*m+j  要求你重新排列 使得重新排列后每个数与原来相邻的数互不相邻 不存在则输出-1

#include <bits/stdc++.h>
#define PI acos(-1.0)
#define mem(a,b) memset((a),b,sizeof(a))
#define TS printf("!!!
")
#define pb push_back
#define inf 1e9
//std::ios::sync_with_stdio(false);
using namespace std;
//priority_queue<int,vector<int>,greater<int>> que;
const double eps = 1.0e-8;
typedef pair<int, int> pairint;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 1e6 + 10;
const int  maxm = 300;
const int turn[4][2] = {{1, 0}, { -1, 0}, {0, 1}, {0, -1}};
//next_permutation
ll mod = 1e9 + 7;
int n, m;
int ans[100005];
int pop = 1;
int now[100005];
int f[30][30];
int num[30][30];
int cur[30][30];
int flagf = 0;
stack<int> s;
void dfs(int x, int y)
{
        if (x == n * m + 1)
        {
                flagf = 1;
                for (int i = n; i >= 1; i--)
                {
                        for (int j = m; j >= 1; j--)
                        {
                                s.push(cur[i][j]);
                        }
                }
                return ;
        }
        if (flagf)
        {
                return ;
        }
        int aimx = (x - 1) / m + 1;
        int aimy = (x - 1) % m + 1;
        for (int i = 1; i <= n * m; i++)
        {
                if ((y & (1 << i)) == 0)
                {
                        if ((!num[cur[aimx - 1][aimy]][i]) && (!num[cur[aimx][aimy - 1]][i]))
                        {
                                int numcur = y | (1 << i);
                                //cout << aimx << " " << aimy << " " << i << endl;
                                cur[aimx][aimy] = i;
                                dfs(x + 1, numcur);
                        }
                }
        }
}
void solve()
{
        for (int i = 1; i <= n; i++)
        {
                for (int j = 1; j <= m; j++)
                {
                        f[i][j] = (i - 1) * m + j;
                }
        }
        for (int i = 1; i <= n; i++)
        {
                for (int j = 1; j <= m; j++)
                {
                        for (int k = 0; k <= 3; k++)
                        {
                                int dx = i + turn[k][0];
                                int dy = j + turn[k][1];
                                if (f[dx][dy] != 0)
                                {
                                        num[f[i][j]][f[dx][dy]] = num[f[dx][dy]][f[i][j]] = 1;
                                }
                        }

                }
        }
        dfs(1, 0);
        if (flagf)
        {
                cout << "YES" << endl;
                for (int i = 1; i <= n; i++)
                {
                        for (int j = 1; j <= m; j++)
                        {
                                cout << s.top() << " ";
                                s.pop();
                        }
                        cout << endl;
                }
        }
        else
        {
                cout << "NO" << endl;
        }
}
int main()
{
        //freopen("in.txt", "r", stdin);
        //freopen("out1.txt", "w", stdout);
        while (cin >> n >> m)
        {
                if (n <= 4 && m <= 4)
                {
                        solve();
                        exit(0);
                }
                if (n < m)
                {
                        printf("YES
");
                        for (int i = 1; i <= n; i++)
                        {
                                if (i & 1)
                                {
                                        for (int j = (i - 1) * m + 1; j <= i * m; j += 2)
                                        {
                                                cout << j << " ";
                                        }
                                        for (int j = (i - 1) * m + 2; j <= i * m; j += 2)
                                        {
                                                cout << j << " ";
                                        }
                                        cout << endl;
                                }
                                else
                                {
                                        for (int j = (i - 1) * m + 2; j <= i * m; j += 2)
                                        {
                                                cout << j << " ";
                                        }
                                        for (int j = (i - 1) * m + 1; j <= i * m; j += 2)
                                        {
                                                cout << j << " ";
                                        }
                                        cout << endl;
                                }
                        }
                }
                else
                {
                        printf("YES
");
                        for (int i = 1; i <= m; i++)
                        {
                                if (i & 1)
                                {
                                        now[i] = 1;
                                }
                                else
                                {
                                        now[i] = 2;
                                }
                        }
                        for (int i = 1; i <= n; i++)
                        {
                                for (int j = 1; j <= m; j++)
                                {
                                        ans[pop++] = (now[j] - 1) * m + j;
                                        now[j] += 2;
                                        if (now[j] > n)
                                        {
                                                now[j] = (j & 1) + 1;
                                        }
                                        //cout<<now[j]<<endl;
                                }
                        }
                        for (int i = 1; i < pop; i++)
                        {
                                printf("%d", ans[i]);
                                if (i % m == 0)
                                {
                                        printf("
");
                                }
                                else
                                {
                                        printf(" ");
                                }
                        }
                }
        }
}
View Code

 E.状压DFS

#include <bits/stdc++.h>
#define PI acos(-1.0)
#define mem(a,b) memset((a),b,sizeof(a))
#define TS printf("!!!
")
#define pb push_back
#define inf 1e9
//std::ios::sync_with_stdio(false);
using namespace std;
//priority_queue<int,vector<int>,greater<int>> que;
const double eps = 1.0e-8;
typedef pair<int, int> pairint;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 1e6 + 10;
const int  maxm = 300;
const int turn[4][2] = {{1, 0}, { -1, 0}, {0, 1}, {0, -1}};
//next_permutation
ll mod = 1e9 + 7;
int n, m;
int num[30][30];
stack<int> cur, ans;
int anser = 30;
void dfs(int now)
{
        if (cur.size() >= ans.size())
        {
                return ;
        }
        if (now == n)
        {
                for (int i = 0; i < n; i++)
                {
                        if (num[n][i] != (1 << n) - 1)
                        {
                                return ;
                        }
                }
                ans = cur;
                return ;
        }
        memcpy(num[now + 1], num[now], sizeof(num[now]));
        dfs(now + 1);
        for (int i = 0; i < n; i++)
        {
                if ((1 << i) & num[now][now])
                {
                        num[now + 1][i] |= num[now][now];
                }
        }
        cur.push(now);
        dfs(now + 1);
        cur.pop();
}
int main()
{
        //freopen("in.txt", "r", stdin);
        //freopen("out1.txt", "w", stdout);
        cin >> n >> m;
        int pa, pb;
        for (int i = 0; i < n; i++)
        {
                num[0][i] = 1 << i;
                //cout<<p[i]<<endl;
                ans.push(i);
        }
        for (int i = 1; i <= m; i++)
        {
                scanf("%d %d", &pa, &pb);
                pa--, pb--;
                num[0][pa] |= 1 << pb;
                num[0][pb] |= 1 << pa;
        }
        //pf();
        dfs(0);
        cout << ans.size() << endl;
        while (!ans.empty())
        {
                cout << ans.top() + 1 << " ";
                ans.pop();
        }
        cout << endl;
}
View Code
原文地址:https://www.cnblogs.com/Aragaki/p/8099330.html