JOI 2019 Final

A

统计后缀和

#include<bits/stdc++.h>
#define cs const
#define pb push_back
using namespace std;
typedef long long ll;
cs int N = 3e3 + 50;
int n, m; char mp[N][N];
int a[N][N], b[N][N];
int main(){
  #ifdef FSYo
  freopen("1.in", "r", stdin);
  #endif
  cin >> n >> m;
  for(int i = 0; i < n; i++)
  scanf("%s", mp[i]);
  for(int i = 0; i < n; i++)
  for(int j = m - 1; ~j; j--)
  a[i][j] = a[i][j + 1] + (mp[i][j] == 'O');
  for(int i = n - 1; ~i; i--)
  for(int j = 0; j < m; j++)
  b[i][j] = b[i + 1][j] + (mp[i][j] == 'I');
  ll ans = 0; 
  for(int i = 0; i < n; i++)
  for(int j = 0; j < m; j++)
  if(mp[i][j] == 'J')
  ans += a[i][j] * b[i][j];
  cout << ans; return 0;  
}

B

贪心，先按 $V_i$ 排序，假设选了 $k$ 个，那么一定会用最大的 $k$ 个 $c_i$
从后向前 $\mathcal{DP}$，求出到 $i$ 最多选几个即可，复杂度 $\mathcal{O}(n\log n)$

#include<bits/stdc++.h>
#define cs const
#define pb push_back
#define fi first
#define se second
using namespace std;
cs int N = 1e5 + 50;
typedef pair<int, int> pi;
int n, m;
pi a[N]; int c[N];
int main(){
  #ifdef FSYo
  freopen("1.in", "r", stdin);
  #endif
  cin >> n >> m;
  for(int i = 1; i <= n; i++)
  scanf("%d%d", &a[i].se, &a[i].fi);
  sort(a + 1, a + n + 1);
  for(int i = 1; i <= m; i++)
  scanf("%d", &c[i]);
  sort(c + 1, c + m + 1);
  int mx = 0; 
  for(int i = n; i; i--){
    int r = lower_bound(c + 1, c + m + 1, a[i].se) - c; 
    int z = min(mx + 1, m - r + 1);
    mx = max(mx, z);
  } cout << mx; return 0;
}

C

注意到相同颜色之间的顺序不会变
设计 $dp_{i,j,k,0/1/2}$ 表示已经放了 $i,j,k$ 个的逆序对数
复杂度 $\mathcal{O}(n^3)$

#include<bits/stdc++.h>
#define cs const
#define pb push_back
using namespace std;
cs int N = 405;
int n, p, q, r; 
char S[N]; int dp[N][N][N][3], sm[3][N], a[3][N];
void chkmn(int &a, int b){ if(a > b) a = b; }
int main(){
  #ifdef FSYo
  freopen("1.in", "r", stdin);
  #endif
  scanf("%d%s", &n, S + 1);
  for(int i = 1; i <= n; i++){
    for(int j = 0; j < 3; j++)
    sm[j][i] = sm[j][i - 1];
    if(S[i] == 'R') a[0][++p] = i, ++sm[0][i]; 
    if(S[i] == 'G') a[1][++q] = i, ++sm[1][i];
    if(S[i] == 'Y') a[2][++r] = i, ++sm[2][i];
  }
  memset(dp, 0x3f, sizeof(dp));
  dp[0][0][0][0] = 
  dp[0][0][0][1] = 
  dp[0][0][0][2] = 0; 
  static int v[3], w[3];
  for(int i = 0; i <= p; i++)
  for(int j = 0; j <= q; j++)
  for(int k = 0; k <= r; k++){
    v[0] = i, v[1] = j, v[2] = k;
    for(int x = 0; x < 3; x++) if(v[x]){
      for(int z = 0; z < 3; z++)
      w[z] = v[z] - (x == z);
      int c = 0, vl = 1e9; 
      for(int z = 0; z < 3; z++) if(z != x){
        c += max(0, v[z] - sm[z][a[x][v[x]]]);
        vl = min(vl, dp[w[0]][w[1]][w[2]][z]);
      } if(vl < 1e9) dp[i][j][k][x] = vl + c;
    } 
  } 
  int ans = 1e9;
  for(int i = 0; i < 3; i++)
  chkmn(ans, dp[p][q][r][i]);
  ans = (ans == 1e9) ? -1 : ans;
  cout << ans << '\n';
  return 0; 
}

D

将每个点移到最近的点
统计出 $c_{i,j}$ 表示还缺的个数
那么我们可以贪心从左边扫到右边，复杂度 $\mathcal{O}(n)$

#include<bits/stdc++.h>
#define cs const
#define pb push_back
using namespace std;
typedef long long ll;
cs int N = 1e5 + 50;
int n, c[N][2]; ll ans; 
int main(){
  #ifdef FSYo
  freopen("1.in", "r", stdin);
  #endif
  cin >> n; 
  for(int i = 1, x, y; i <= n + n; i++){
    scanf("%d%d", &x, &y); int _x, _y;
    if(y <= 1) ans += 1 - y, _y = 0;
    else ans += y - 2, _y = 1;
    if(x < 1) _x = 1, ans += 1 - x; 
    else if(x > n) _x = n, ans += x - n;
    else _x = x; ++c[_x][_y];
  }
  for(int i = 1; i <= n; i++)
  for(int j = 0; j < 2; j++)
  c[i][j] = 1 - c[i][j];
  for(int i = 1; i <= n; i++){
    if(c[i][0] > 0 && c[i][1] < 0){
      int d = min(c[i][0], -c[i][1]);
      c[i][0] -= d, c[i][1] += d, ans += d;  
    }
    if(c[i][0] < 0 && c[i][1] > 0){
      int d = min(-c[i][0], c[i][1]);
      c[i][0] += d, c[i][1] -= d, ans += d;
    } 
    ans += abs(c[i][0]) + abs(c[i][1]);
    c[i + 1][0] += c[i][0];
    c[i + 1][1] += c[i][1];
  } cout << ans; return 0; 
}

E

产生贡献的只能是 $x$ 到直径的较远的一个端点的某一段
这启示我们从两个直径开始 $dfs$，统计每个点的贡献
用栈维护根到当前点产生贡献的点以及个数
向下走一步要将其余儿子中最深的覆盖的那些点 $ban$ 掉
那么我们只需要先 $dfs$ 重儿子，复杂度 $\mathcal{O}(n)$

#include<bits/stdc++.h>
#define cs const
#define pb push_back
using namespace std;
cs int N = 2e5 + 50;
int n, m, c[N];
vector<int> G[N];
int S, T, rt, mx;
void dfs(int u, int f, int d){
  if(d > mx) mx = d, rt = u;
  for(int v : G[u]) if(v != f)
    dfs(v, u, d + 1);
}
int dep[N], mxd[N], son[N], se[N], ans[N];
void pre_dfs(int u, int f){
  dep[u] = dep[f] + 1; 
  mxd[u] = se[u] = son[u] = 0; 
  for(int v : G[u]) if(v != f){
    pre_dfs(v, u);
    mxd[u] = max(mxd[u], mxd[v] + 1);
    if(mxd[son[u]] <= mxd[v]) son[u] = v; 
  } 
  for(int v : G[u]) if(v != f && v != son[u])
  se[u] = max(se[u], mxd[v] + 1);
}
int s[N], tp, bin[N], Sm;
void ins(int x){ Sm += !bin[x], ++bin[x]; }
void del(int x){ --bin[x], Sm -= !bin[x]; }
void sub_dfs(int u, int f){
  while(tp && dep[u] - dep[s[tp]] <= se[u]) 
    del(c[s[tp--]]);
  ins(c[s[++tp] = u]);
  if(son[u]) sub_dfs(son[u], u);
  while(tp && dep[u] - dep[s[tp]] <= mxd[u])
    del(c[s[tp--]]);
  ans[u] = max(ans[u], Sm);
  for(int v : G[u]) if(v != f && v != son[u]){
    if(!tp || s[tp] != u) 
    ins(c[s[++tp] = u]); sub_dfs(v, u);
  } if(tp && s[tp] == u) 
  del(c[s[tp--]]);
} 
int main(){
  #ifdef FSYo
  freopen("1.in", "r", stdin);
  #endif
  cin >> n >> m;
  for(int i = 1, u, v; i < n; i++)
  scanf("%d%d", &u, &v), 
  G[u].pb(v), G[v].pb(u);
  for(int i = 1; i <= n; i++)
  scanf("%d", &c[i]);
  dfs(1, 0, 0), S = rt; rt = mx = 0; 
  dfs(S, 0, 0), T = rt; 
  tp = 0, pre_dfs(S, 0), sub_dfs(S, 0);
  tp = 0, pre_dfs(T, 0), sub_dfs(T, 0);
  for(int i = 1; i <= n; i++)
  cout << ans[i] << '\n';
  return 0; 
}