2013多校联合训练第四场

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原创最新推荐文章于 2026-04-05 02:31:56 发布 1.2k 阅读 AI 写同款· GEO 优化 ›

使用R语言中的NbClust包确定最佳聚类簇数聚类分析是一种常用的数据分析技术，它将相似的数据点分组成不同的簇。NbClust包是R语言中一个强大的工具，它提供了多种方法来帮助我们确定最佳的聚类簇数。综上所述，使用R语言中的NbClust包可以帮助我们确定最佳的聚类簇数。通过选择适当的聚类算法和评估指标，我们可以更好地理解数据集的结构和模式，并根据需要进行进一步的分析和解释。此外，NbClust包还提供了其他函数和参数，用于执行不同的聚类算法、选择不同的评估指标等。安装完毕后，我们需要加载NbClust包，以便使用其中的函数。函数来确定最佳聚类簇数。阅读详情

求一个字符串的回文子序列个数

很直观的DP但一开始看着这个题目愣了好久……

  scanf("%s" , s + 1);
  n = strlen(s + 1);
  memset(f , 0 , sizeof(f));
  for (i = 1 ; i <= n ; ++ i) f[i][i] = 1;
  for (k = 2 ; k <= n ; ++ k)
  {
    for (i = 1 ; i + k - 1 <= n ; ++ i)
    {
      j = i + k - 1;
      f[i][j] = f[i + 1][j] + f[i][j - 1];
      if (s[i] == s[j])
        f[i][j] ++;
      else f[i][j] += (Q - f[i + 1][j - 1]);
      f[i][j] %= Q;
    }
  }
  printf("Case %d: %d\n" , ++ ca , f[1][n]);

先预处理出从某个点出发往四个方向滑的终点位置以及路过能拿到的钥匙

然后就是一个BFS了，边权只有1或0，开deque。

可惜因为没有发现卡墙的问题比赛时没有过这个题TwT

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cstdlib>
#include <cmath>
#include <cctype>
#include <queue>
#include <stack>
#include <string>
#include <set>
#include <map>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
using namespace std;
#define N 205
#define ULL unsigned long long
int n , m , sx , sy , ex , ey;
char s[N][N];
int dx[] = {1 , -1 , 0 , 0} , dy[] = {0 , 0 , 1 , -1};

pair<int , int>  h[4][N][N] , null = make_pair(-1 , -1);
int kn[N][N], key[4][N][N] , ks , ans;

struct opt
{
  int x , y , p , w , r;
};
deque<opt> q;
int d[N][N][128];
bool f[N][N][128];

void update(int x , int y , int p , int v , int w , int r)
{
  if (!f[x][y][p])
  {
    if (w)
      q.push_back((opt){x , y , p , v + w , r});
    else q.push_front((opt){x , y , p , v + w , r});
  }
}
void BFS()
{
  int i , x , y , w , kk = (1 << ks) - 1;
  ans = 1 << 30;
  q.push_back((opt){sx , sy , 0 , 0 , 0});
  memset(f , 0 , sizeof(f));
  while (!q.empty())
  {
    opt a = q.front(); q.pop_front();
    if (f[a.x][a.y][a.p]) continue;
    f[a.x][a.y][a.p] = 1 , w = d[a.x][a.y][a.p] = a.w;
    x = a.x , y = a.y;
    cout << x <<' ' << y << ' ' << a.p << ' ' << w << endl;
    if (x == ex && y == ey)
    {
      if (a.p == kk)
        ans = min(ans , w);
      i = a.r;
      if (s[x + dx[i]][y + dy[i]] == '#' || h[i][x + dx[i]][y + dy[i]] == null) continue;
      update(h[i][x + dx[i]][y + dy[i]].fi , h[i][x + dx[i]][y + dy[i]] .se , a.p | key[i][x + dx[i]][y + dy[i]] , w , 0 , i);
      continue;
    }
    if (s[x][y] < 4)
    {
      i = s[x][y];
      if (s[x + dx[i]][y + dy[i]] == '#' || h[i][x + dx[i]][y + dy[i]] == null) continue;
      update(h[i][x + dx[i]][y + dy[i]].fi , h[i][x + dx[i]][y + dy[i]] .se , a.p | key[i][x + dx[i]][y + dy[i]] , w , 0 , i);
      continue;
    }
    for (i = 0 ; i < 4 ; ++ i)
      if (h[i][x][y] != null)
        update(h[i][x][y].fi , h[i][x][y].se , a.p | key[i][x][y] , w , 1 , i);
  }
  if (ans == 1 << 30)
    ans = -1;
  printf("%d\n" , ans);
}
void work()
{
  int i , j;
  memset(s , 0 , sizeof(s));
  memset(h , 0 , sizeof(h));
  memset(kn , -1 , sizeof(kn));
  memset(key , 0 , sizeof(key));
  ks = 0;

  for (i = 1 ; i <= n ; ++ i)
    scanf("%s" , s[i] + 1);
  for (i = 1 ; i <= n ; ++ i)
    for (j = 1 ; j <= m ; ++ j)
    {
      if (s[i][j] == 'S')
        sx = i , sy = j;
      if (s[i][j] == 'E')
        ex = i , ey = j;
      if (s[i][j] == 'K')
        kn[i][j] = ks ++;
    }
  for (i = 1 ; i <= n ; ++ i)
    for (j = 1 ; j <= m ; ++ j)
    {
      if (s[i][j] == 'U')
        s[i][j] = 1;
      if (s[i][j] == 'D')
        s[i][j] = 0;
      if (s[i][j] == 'L')
        s[i][j] = 3;
      if (s[i][j] == 'R')
        s[i][j] = 2;
    }
  for (i = n + 1; i > 0 ; -- i)
    for (j = 1 ; j <= m ; ++ j)
    {
      if (i == n + 1) h[0][i][j] = null;
      else
      {
        if (s[i][j] < 4 || s[i][j] == 'E') h[0][i][j] = make_pair(i , j) , key[0][i][j] = 0;
        else if (s[i + 1][j] != '#')
        {
          h[0][i][j] = h[0][i + 1][j];
          key[0][i][j] = key[0][i + 1][j];
          key[0][i][j] |= ~kn[i][j] ? (1 << kn[i][j]) : 0;
        }
        else
        {
          h[0][i][j] = make_pair(i , j);
          key[0][i][j] = ~kn[i][j] ? (1 << kn[i][j]) : 0;
        }
      }
    }

  for (i = 0 ; i <= n ; ++ i)
    for (j = 1 ; j <= m ; ++ j)
    {
      if (i == 0) h[1][i][j] = null;
      else
      {
        if (s[i][j] < 4 || s[i][j] == 'E') h[1][i][j] = make_pair(i , j) , key[1][i][j] = 0;
        else if (s[i - 1][j] != '#')
        {
          h[1][i][j] = h[1][i - 1][j];
          key[1][i][j] = key[1][i - 1][j];
          key[1][i][j] |= ~kn[i][j] ? (1 << kn[i][j]) : 0;
        }
        else
        {
          h[1][i][j] = make_pair(i , j);
          key[1][i][j] = ~kn[i][j] ? (1 << kn[i][j]) : 0;
        }
      }
    }
  for (j = m + 1 ; j > 0 ; -- j)
    for (i = 1 ; i <= n ; ++ i)
    {
      if (j == m + 1) h[2][i][j] = null;
      else
      {
        if (s[i][j] < 4 || s[i][j] == 'E') h[2][i][j] = make_pair(i , j) , key[2][i][j] = 0;
        else if (s[i][j + 1] != '#')
        {
          h[2][i][j] = h[2][i][j + 1];
          key[2][i][j] = key[2][i][j + 1];
          key[2][i][j] |= ~kn[i][j] ? (1 << kn[i][j]) : 0;
        }
        else
        {
          h[2][i][j] = make_pair(i , j);
          key[2][i][j] = ~kn[i][j] ? (1 << kn[i][j]) : 0;
        }
      }
    }
  for (j = 0 ; j <= m ; ++ j)
    for (i = 1 ; i <= n ; ++ i)
    {
      if (j == 0) h[3][i][j] = null;
      else
      {
        if (s[i][j] < 4 || s[i][j] == 'E') h[3][i][j] = make_pair(i , j) , key[3][i][j] = 0;
        else if (s[i][j - 1] != '#')
        {
          h[3][i][j] = h[3][i][j - 1];
          key[3][i][j] = key[3][i][j - 1];
          key[3][i][j] |= ~kn[i][j] ? (1 << kn[i][j]) : 0;
        }
        else
        {
          h[3][i][j] = make_pair(i , j);
          key[3][i][j] = ~kn[i][j] ? (1 << kn[i][j]) : 0;
        }
      }
    }
  BFS();
}
int main()
{
  freopen("input.txt","r",stdin);
  while (~scanf("%d%d",&n,&m))
    work();
  return 0;
}

SCC缩点之后再枚举割掉的所有入度为0或者出度为0的SCC，更新答案。

代码主要部分不是我写的

#pragma comment(linker,"/STACK:102400000,102400000")

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <cstring>
using namespace std;
#define N 100010
#define M 200010

int sum,first[N];
struct gtype {
       int x,y,next;
}g[M];

int cas,n,m,a,b,tot,top,cnt,pin[N],pout[N];
int low[N],dfn[N],con[N],stack[N],belong[N];
bool instack[N];

bool cmp(int a, int b) {
     return a>b;
}

void add(int x,int y) {
     sum++;
     g[sum].x = x;
     g[sum].y = y;
     g[sum].next = first[x];
     first[x] = sum;
}

void tarjan(int x) {
    int k;
    dfn[x] = low[x] = ++tot;
    instack[x] = true;
    stack[++top] = x;
    for (int t=first[x];t!=-1;t=g[t].next) {
        int y = g[t].y;
        if (!dfn[y]) {
            tarjan(y);
            low[x] = min(low[x],low[y]);
        }
        else if (instack[y]) low[x] = min(low[x],dfn[y]);
    }
    if (dfn[x]==low[x]) {
        cnt++;
        do {
            k = stack[top--];
            instack[k] = false;
            belong[k] = cnt;
            con[cnt]++;
        }while (k!=x);
    }
}

int main() {
    scanf("%d",&cas);
    for (int tt=1;tt<=cas;tt++) {
        sum = 0;
        memset(first,-1,sizeof(first));
        scanf("%d%d",&n,&m);
        for (int i=1;i<=m;i++) {
            scanf("%d%d",&a,&b);
            add(a,b);
        }

        tot = top = cnt = 0;
        memset(dfn,0,sizeof(dfn));
        memset(low,0,sizeof(low));
        memset(con,0,sizeof(con));
        memset(instack,0,sizeof(instack));
        for (int i=1;i<=n;i++)
            if (!dfn[i]) tarjan(i);

        printf("Case %d: ",tt);
        if (cnt==1) {
           printf("-1\n");
        }
        else {
             long long ans = 0;
             memset(pin,0,sizeof(pin));
             memset(pout,0,sizeof(pout));
             for (int i=1;i<=m;i++) {
                 if (belong[g[i].x] != belong[g[i].y]) {
                    pout[belong[g[i].x]]++;
                    pin[belong[g[i].y]]++;
                 }
             }
             for (int i=1;i<=cnt;i++)
             {
                 if (pin[i]==0)
                 {
                    ans = max(ans, (long long)n*(n-1)-m-(long long)(n-con[i])*con[i]);
                 }
                 if (pout[i]==0)
                 {
                    ans = max(ans, (long long)n*(n-1)-m-(long long)(n-con[i])*con[i]);
                 }
             }
             printf("%I64d\n",ans);
        }
    }
    return 0;
}

比赛结束了才被DS提醒是被水了过去……

之前用线段树想的太麻烦，后来才发现可以用数组直接模拟分为了多少段

#pragma comment(linker,"/STACK:102400000,102400000")
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cctype>
#include<cmath>
#include<algorithm>
#define maxn 100010
using namespace std;
typedef long long LL;
int pos[maxn],color[maxn],n,m;
void RD(int& x)
{
  x = 0; char c;
  for (c = getchar() ;!isdigit(c) ; c = getchar()); x = c - '0';
  for (c = getchar() ; isdigit(c) ; c = getchar()) x = x * 10 + c - '0';
}
struct Question
{
    int l,r,id;
    int a;
    friend bool operator < (const Question &a,const Question &b)
    {
        return pos[a.l]<pos[b.l]||(pos[a.l]==pos[b.l]&&a.r<b.r);
    }
}ask[maxn];
bool cmp_id(const Question &a,const Question &b)
{
    return a.id<b.id;
}
bool ff[maxn];
int anss;
void read()
{
    RD(n) , RD(m);
    memset(ff , 0 , sizeof(ff));
    for(int i=1;i<=n;i++)
        RD(color[i]);
    int limit=(int)sqrt(n+0.5);
    for(int j=1;j<=n;j++)
        pos[j]=(j-1)/limit+1;
    for(int i=1;i<=m;i++)
    {
        RD(ask[i].l) , RD(ask[i].r);
        ask[i].id=i;
    }
    sort(ask+1,ask+m+1);
}

void Modify(int i,int add)
{
  if (add == 1)
  {
        int x = color[i];
        ff[x] = 1;
        if (ff[x - 1] && ff[x + 1]) -- anss;
        else if (!ff[x - 1] && !ff[x + 1]) ++ anss;
  }
  else
  {
        int x = color[i];
        ff[x] = 0;
        if (ff[x - 1] && ff[x + 1]) ++ anss;
        else if (!ff[x - 1] && !ff[x + 1]) -- anss;
  }
}
void work()
{
    anss=0;
    for(int i=1,l=1,r=0;i<=m;i++)
    {
        if(r<ask[i].r)
        {
            for(r=r+1;r<=ask[i].r;r++)
                Modify(r,1);
            r--;
        }
        if(r>ask[i].r)
            for(;r>ask[i].r;r--)
                Modify(r,-1);
        if(l<ask[i].l)
            for(;l<ask[i].l;l++)
                Modify(l,-1);
        else if(l>ask[i].l)
        {
            for(l=l-1;l>=ask[i].l;l--)
                Modify(l,1);
            l++;
        }
        ask[i].a=anss;
    }
}
void print()
{
    sort(ask+1,ask+m+1,cmp_id);
    for(int i=1;i<=m;i++)
        printf("%d\n",ask[i].a);
}
int main()
{
  int _; RD(_); while (_--)
    read(),
    work(),
    print();
  return 0;
}

每一组hehehehe明显可以分开处理，一段长度为n的hehe....he的方案就是斐波那契的[n/2+1]项，乘起来就ok

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cstdlib>
#include <cmath>
#include <cctype>
#include <queue>
#include <stack>
#include <string>
#include <set>
#include <map>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
using namespace std;
#define N 15005
#define M 2000005
typedef long long LL;
using namespace std;
char s[N];
int n , m , a[N] , Q = 10007 , f[N] , ca;
void work()
{
  int i , j , ans = 1 , x;
  scanf("%s", s);
  n = strlen(s);
  m = 0;
  for (i = 0 ; i + 3 < n ; ++ i)
    if (s[i] == 'h' && s[i + 1] == 'e' && s[i + 2] == 'h' && s[i + 3] == 'e')
      a[m ++] = i;
  for (i = 0 ; i < m ; i = j)
  {
    j = i + 1 , x = 1;
    while (a[j] - a[j - 1] == 2)
      ++ j , ++ x;
    ans *= f[x] , ans %= Q;
  }
  printf("Case %d: %d\n" , ++ ca , ans);
}

int main()
{
  f[0] = 1 , f[1] = 2;
  for (int i = 2 ; i <= 10086 ; ++ i)
    f[i] = f[i - 1] + f[i - 2] , f[i] %= Q;
  int _; cin >> _ ; while (_ --)
    work();
  return 0;
}