noip模拟8

06-18-2021 约 1225 字预计阅读 3 分钟

T1

考场上想的是tarjan缩点，但是又忘了tarjan怎么打了。考完试发现tarjan也不是正解。所以tarjan：咕

正解是直接计算，统计自己连自己的和连别人的边数。如果边不联通就输出0

~~正解很短~~

CODE




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#include <bits/stdc++.h>
using namespace std;
const int maxn = 10010;

struct EDGE {
  int to, next;
} edge[maxn];
int tote, head[maxn];
void addEdge(int from, int to) {
  edge[++tote].next = head[from];
  edge[tote].to = to;
  head[from] = tote;
}
int m, n;
int visn;
bool vis[maxn];
void dfs(int id) {
  vis[id] = true;
  for (int i = head[id]; i; i = edge[i].next) {
    int to = edge[i].to;
    if (vis[to]) {
      visn++;
      continue;
    }
    dfs(to);
  }
}
int SC, CT, in[maxn];
int main() {
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    int f, t;
    cin >> f >> t;
    if (t == f) {
      SC++;
      addEdge(f, t);
    } else {
      CT++;
      addEdge(f, t);
      addEdge(t, f);
      in[t]++;
      in[f]++;
    }
  }
  for (int i = 1; i <= n; i++) {
    dfs(i);
    if (visn != 0) break;
    vis[i] = 0;
  }
  for (int i = 1; i <= n; i++) {
    if (!vis[i] && head[i] != 0) {
      cout << 0 << endl;
      return 0;
    }
  }
  int ans = 0;
  ans += SC * (SC - 1) / 2;
  ans += SC * CT;
  for (int i = 1; i <= n; i++) {
    if (in[i] != 0) ans += in[i] * (in[i] - 1) / 2;
  
  }cout << ans << endl;
  return 0;
}

# T2

~~正解更短~~

注意 r-l 优化否则会 TLE

彭师傅的解法LINK

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#include <bits/stdc++.h>
#define int long long 
using namespace std;
const int maxn = 100010;

int n, k;
int a[maxn];
int C;
signed main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    k += a[i];
  }
  int ans = 0;
  int r = 0;
  for (int l = 1; l <= k; l = r + 1) {
    r = k / (k / l);
    int d = 0;
    for (int i = 1; i <= n; i++) {
      d += ceil(1.0 * a[i] / r);
    }
    if (d * r <= k) ans = r;
  }
  cout << ans << endl;
  return 0;
}

T3

还不会

T4

大水题不提，快读+printf就能AC 写假了的码

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#include <bits/stdc++.h>
#define MOD 998244353

using namespace std;
const int maxn = 600010;

struct EDGE {
  int next, to;
} edge[maxn];
int tote, head[maxn];
void addEdge(int from, int to) {
  edge[++tote].to = to;
  edge[tote].next = head[from];
  head[from] = tote;
}
int size[maxn], father[maxn], dep[maxn], son[maxn], top[maxn], dfn[maxn],
    rk[maxn];
int cnt, root[maxn];

#ifdef __fread
inline char nc() {
  static char buf[100000], *p1 = buf, *p2 = buf;
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
inline int read() {
  char ch = nc();
  int sum = 0;
  while (!(ch >= '0' && ch <= '9')) ch = nc();
  while (ch >= '0' && ch <= '9') sum = sum * 10 + ch - 48, ch = nc();
  return sum;
}
#endif
#ifndef __fread
inline int read() {
  int ans = 0, f = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = -f;
    c = getchar();
  }
  while (isdigit(c)) {
    ans = (ans << 3) + (ans << 1) + (c ^ 48);
    c = getchar();
  }
  return ans * f;
}
#endif

int qpow(int base, int power, int mod) {
  int ans = 1;
  while (power) {
    if (power & 1) {
      ans = 1ll * ans * base % mod;
    }
    base = (1ll * base * base) % mod;
    power >>= 1;
  }
  return ans;
}
void dfs1(int id, int fa, int depth) {
  father[id] = fa;
  dep[id] = depth;
  size[id] = 1;
  for (int i = head[id]; i; i = edge[i].next) {
    int v = edge[i].to;
    if (v == fa) continue;
    dfs1(v, id, depth + 1);
    size[id] += size[v];
    if (size[v] > size[son[id]]) son[id] = v;
  }
}
void dfs2(int id, int tp) {
  top[id] = tp;
  dfn[id] = ++cnt;
  rk[cnt] = id;
  if (!son[id]) return;
  dfs2(son[id], tp);
  for (int i = head[id]; i; i = edge[i].next) {
    int v = edge[i].to;
    if (v != son[id] && v != father[id]) dfs2(v, v);
  }
}
int getLCA(int a, int b) {
  while (top[a] != top[b]) {
    if (dep[top[a]] > dep[top[b]]) {
      a = father[top[a]];
    } else {
      b = father[top[b]];
    }
  }
  if (dep[a] > dep[b]) {
    return b;
  } else {
    return a;
  }
}

int n, m;
int main() {
  n = read();
  for (int i = 1; i < n; i++) {
    int f, t;
    f = read();
    t = read();
    addEdge(f, t);
    addEdge(t, f);
  }
  dfs1(1, 1, 0);
  dfs2(1, 1);
  int m;
  cin >> m;
  for (int i = 1; i <= m; i++) {
    int x, y, k;
    x = read();
    y = read();
    k = read();
    int rtd = dep[getLCA(x, y)];
    int xpl = dep[x] - rtd;
    int ypl = dep[y] - rtd;
    long long ans = 0;
    while (xpl != 0) {
      ans += qpow(rtd + xpl, k, MOD);
      xpl--;
    }
    ans %= MOD;
    while (ypl != 0) {
      ans += qpow(rtd + ypl, k, MOD);
      ypl--;
    }
    ans %= MOD;
    ans += qpow(rtd, k, MOD);
    printf("%lld\n", ans % MOD);
  }
  return 0;
}

目录

noip模拟8

T1

注意 r-l 优化 否则会 TLE

T3

T4

注意 r-l 优化否则会 TLE