Codeforces_723

A.取中间那个点即可。

#include<bits/stdc++.h>
using namespace std;
 
int a[];
 
int main()
{
    ios::sync_with_stdio(false);
    cin >> a[] >> a[] >> a[];
    sort(a,a+);
    cout << a[]-a[] << endl;
    return ;
 }

---

B.模拟，注意判断是否在括号内。

#include<bits/stdc++.h>
using namespace std;
 
int n;
string s;
 
int main()
{
    cin >> n;
    getchar();
    getline(cin,s);
    int maxx = ,cnt = ,now = ,flag = ;
    for(int i = ;i < n;i++)
    {
        if(s[i] == '(')
        {
            flag = ;
            now = ;
        }
        else if(s[i] == ')')
        {
            flag = ;
            if(now) cnt++;
            now = ;
        }
        else if(s[i] == '_')
        {
            if(flag ==  && now)    cnt++;
            now = ;
        }
        else
        {
            now++;
            if(flag == )   maxx = max(maxx,now);
        }
    }
    cout << maxx << " " << cnt << endl;
    return ;
 }

---

C.先算出ans，再把个数不到ans的数填满。

#include<bits/stdc++.h>
using namespace std;
 
int n,m,a[],b[] = {};
 
int main()
{
    ios::sync_with_stdio(false);
    cin >> n >> m;
    for(int i = ;i <= n;i++)
    {
        cin >> a[i];
        if(a[i] <= m)   b[a[i]]++;
    }
    int ans = n/m,cnt = ,now = ;
    for(int i = ;i <= n;i++)
    {
        if(a[i] <= m && b[a[i]] <= ans) continue;
        while(b[now] >= ans)    now++;
        if(now == m+)  break;
        if(a[i] <= m)   b[a[i]]--;
        a[i] = now;
        b[now]++;
        cnt++;
    }
    cout << ans << " " << cnt << endl;
    for(int i = ;i <= n;i++)   cout << a[i] << " ";
    cout << endl;
    return ;
 }

---

D.dfs找出每一个湖泊，按面积排序，把cnt-k个小的填满。

#include<bits/stdc++.h>
using namespace std;
 
int n,m,k,vis[][] = {},ok,sum;
int dx[] = {,,-,};
int dy[] = {-,,,};
string s[];
struct xxx
{
    int x,y,sum;
    friend bool operator <(xxx a,xxx b)
    {
        return a.sum < b.sum;
    }
}a[];
 
void dfs1(int x,int y)
{
    if(vis[x][y])   return;
    vis[x][y] = ;
    sum++;
    for(int i = ;i < ;i++)
    {
        int xx = x+dx[i],yy = y+dy[i];
        if(xx <  || xx > n || yy <  || yy > m)
        {
            ok = ;
            continue;
        }
        if(s[xx][yy] == '*')    continue;
        dfs1(xx,yy);
    }
}
 
void dfs2(int x,int y)
{
    s[x][y] = '*';
    for(int i = ;i < ;i++)
    {
        int xx = x+dx[i],yy = y+dy[i];
        if(xx <  || xx > n || yy <  || yy > m || s[xx][yy] == '*')    continue;
        dfs2(xx,yy);
    }
}
int main()
{
    ios::sync_with_stdio(false);
    cin >> n >> m >> k;
    int cnt = ;
    for(int i = ;i <= n;i++)
    {
        cin >> s[i];
        s[i] = " "+s[i];
    }
    for(int i = ;i <= n;i++)
    {
        for(int j = ;j <= m;j++)
        {
            if(s[i][j] == '*' || vis[i][j])   continue;
            ok = ;
            sum = ;
            dfs1(i,j);
            if(ok)
            {
                a[++cnt].x = i;
                a[cnt].y = j;
                a[cnt].sum = sum;
            }
        }
    }
    sort(a+,a++cnt);
    cnt -= k;
    int ans = ;
    for(int i = ;i <= cnt;i++)
    {
        ans += a[i].sum;
        dfs2(a[i].x,a[i].y);
    }
    cout << ans << endl;
    for(int i = ;i <= n;i++)
    {
        for(int j = ;j <= m;j++)   cout << s[i][j];
        cout << endl;
    }
    return ;
 }

---

E.因为是无向图，所以有偶数个度为奇数的点，我们把他们与第n+1个点相连，则所有点的度都是偶数个，存在欧拉回路，输出这个回路的路径（不包括第n+1个点），符合要求。

#include<bits/stdc++.h>
using namespace std;
 
int n,m,d[];
set<int> s[];
 
void dfs(int now)
{
    while(s[now].size())
    {
        int t = *s[now].begin();
        s[now].erase(t);
        s[t].erase(now);
        if(now != n+ && t != n+)  cout << now << " " << t << endl;
        dfs(t);
    }
}
 
int main()
{
    int T;
    cin >> T;
    while(T--)
    {
        cin >> n >> m;
        memset(d,,sizeof(d));
        while(m--)
        {
            int x,y;
            cin >> x >> y;
            s[x].insert(y);
            s[y].insert(x);
            d[x]++;
            d[y]++;
        }
        int ans = ;
        for(int i = ;i <= n;i++)
        {
            if(d[i]%)
            {
                s[i].insert(n+);
                s[n+].insert(i);
            }
            else    ans++;
        }
        cout << ans << endl;
        for(int i = ;i <= n;i++)   dfs(i);
    }
    return ;
}

---

F.原图无向连通，可以把原图的除s，t外的点分为三类团。

①与s相连。②与t相连。③与s，t相连。

若不存在第③类团，则把相应团与s或t相连，最后把s与t相连。

若存在第③类团，则s与t不必相连，可以通过第③类团来把他们相连，这样其中某个点的度就少了1，为最优情况。

#include<bits/stdc++.h>
using namespace std;
 
struct xx
{
    int x,y;
    xx(int a,int b):x(a),y(b){};
};
int n,m,x,y,dx,dy,cnt = ,a[] = {},b[] = {},vis[] = {};
vector<int> v[];
vector<xx> ans;
 
void dfs(int now)
{
    vis[now] = ;
    for(int i = ;i < v[now].size();i++)
    {
        int t = v[now][i];
        if(t == x)  a[cnt] = now;
        else if(t == y) b[cnt] = now;
        else if(!vis[t])
        {
            ans.push_back(xx(now,t));
            dfs(t);
        }
    }
}
 
int main()
{
    ios::sync_with_stdio(false);
    cin >> n >> m;
    for(int i = ;i <= m;i++)
    {
        cin >> x >> y;
        v[x].push_back(y);
        v[y].push_back(x);
    }
    cin >> x >> y >> dx >> dy;
    for(int i = ;i <= n;i++)
    {
        if(vis[i] || i == x || i == y)  continue;
        ++cnt;
        dfs(i);
    }
    int z = ;
    for(int i = ;i <= cnt;i++)
    {
        if(a[i] == )
        {
            ans.push_back(xx(b[i],y));
            dy--;
        }
        else if(b[i] == )
        {
            ans.push_back(xx(a[i],x));
            dx--;
        }
        else z++;
    }
    if(dx <  || dy <  || dx+dy- < z)
    {
        cout << "No" << endl;
        return ;
    }
    if(z == )  ans.push_back(xx(x,y));
    else
    {
        int flag = ;
        for(int i = ;i <= cnt;i++)
        {
            if(a[i] ==  || b[i] == )  continue;
            if(flag)
            {
                if(dx)
                {
                    ans.push_back(xx(a[i],x));
                    dx--;
                }
                else
                {
                    ans.push_back(xx(b[i],y));
                    dy--;
                }
            }
            else
            {
                flag = ;
                ans.push_back(xx(a[i],x));
                dx--;
                ans.push_back(xx(b[i],y));
                dy--;
            }
        }
    }
    cout << "Yes" << endl;
    for(int i = ;i < ans.size();i++) cout << ans[i].x << " " << ans[i].y << endl;
    return ;
}