C
Klee in Solitary Confinement
思维题
先预处理出每个数的个数，若k为0，直接输出最大的个数。若不为0，对于每个数，对自己x贡献-1，并且当贡献≤0时，重新置0，意味着不对前面的数+k。然后对x+k贡献+1,并更新最大值，最后输出最大值即可
Code
#include <cstdio>
#include <algorithm>
#define py 2000000
using namespace std;
int n, k, dp[4000010], ans, sum[4000010], a[4000010];
int main()
{
// freopen("_in.txt", "r", stdin);
scanf("%d%d", &n, &k);
int x;
for (int i = 1; i <= n; i++)
{
scanf("%d", &a[i]);
sum[a[i] + py]++;
ans = max(ans, sum[a[i] + py]);
}
if(k==0)
{
printf("%d\n",ans);
return 0;
}
for (int i = 1; i <= n; i++)
{
x = a[i];
dp[x + py]--;
if (dp[x + py] < 0)
dp[x + py] = 0;
dp[x + py + k]++;
ans = max(ans, sum[x + py + k] + dp[x + py + k]);
}
printf("%d\n", ans);
return 0;
}
D
Paimon Sorting
因为对于序列的每一个数都要询问一次，且该数后面的数对结果没有影响，故考虑插入法
如果插入的数小于当前最大值，则直到最后一轮之前该数都不会对结果有贡献，最后一轮的贡献则为前面比它大的数的个数（去重后），这里用两个树状数组维护。
如果相等，则始终不会产生贡献。
如果插入的数大于当前最大值，经过手模，观察出为2+当前最大值第二次出现后的数的个数。
Code
#include <bits/stdc++.h>
#define DEBUG freopen("_in.txt", "r", stdin);
// #define DEBUG freopen("_in.txt", "r", stdin), freopen("_out.txt", "w", stdout);
typedef long long ll;
using namespace std;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll maxn = 2e5 + 10;
const ll maxm = 2e7 + 10;
const ll mod = 1e9 + 7;
const double pi = acos(-1);
const double eps = 1e-8;
ll T, n;
ll c1[maxn], vis[maxn],c2[maxn];
ll lowbit(ll i)
{
return i & (-i);
}
void update1(ll x)
{
for (ll i = x; i <= n; i += lowbit(i))
{
c1[i]++;
}
}
ll query1(ll x)
{
ll ans = 0;
for (ll i = x; i; i -= lowbit(i))
{
ans += c1[i];
}
return ans;
}
void update2(ll x)
{
for (ll i = x; i <= n; i += lowbit(i))
{
c2[i]++;
}
}
ll query2(ll x)
{
ll ans = 0;
for (ll i = x; i; i -= lowbit(i))
{
ans += c2[i];
}
return ans;
}
int main()
{
// DEBUG;
scanf("%lld", &T);
while (T--)
{
scanf("%lld", &n);
for (ll i = 1; i <= n; i++)
{
c1[i] = 0;
c2[i] = 0;
vis[i]=0;
}
ll v;
scanf("%lld", &v);
ll maxx = v, cnt1 = 1, cnt2 = 0, ans = 0, cnt3 = 0;
update1(v);
vis[v] = 1;
printf("%lld", ans);
for (ll i = 2; i <= n; i++)
{
ll v;
scanf("%lld", &v);
if (v < maxx)
{
ans += i - query1(v)-1-(cnt3 - query2(v));
printf(" %lld", ans);
if (cnt1 >= 2)
{
cnt2++;
}
}
else if (v == maxx)
{
printf(" %lld", ans);
cnt1++;
if (cnt1 >= 2)
{
cnt2++;
}
}
else
{
ans += 2;
ans += cnt2;
printf(" %lld", ans);
cnt1 = 1;
cnt2 = 0;
maxx = v;
}
if(vis[v])
{
cnt3++;
update2(v);
}
update1(v);
vis[v] = 1;
}
printf("\n");
}
return 0;
}
H
Crystalfly
树上DP，dp[i][2],表示点i的子树所贡献的最大值，dp[i][0]，该点的直连子节点的dp[i][2]的和，对于一个点
1、存在1个3s子节点，则为一个1s点并且损失其全部直连子节点+1个3s子节点
2、存在1个以上3s子节点，则为一个1s点或3s子节点并且损失其全部直连子节点+1个3s子节点
3、不存在3s子节点，则优先选择其中点权最大的子节点
Code
#include <cstdio>
#include <algorithm>
using namespace std;
long long T, n, val[100010], t[100010], head[100010], k, fa[100010], dp[100010][3];
struct edge
{
long long to, next;
} ed[200010];
void adde(long long u, long long v)
{
ed[++k].to = v;
ed[k].next = head[u];
head[u] = k;
}
void dfs(long long u)
{
long long sum_dp0 = 0;
long long max_i = 0;
long long max_it3 = 0;
long long max_02 = -0x3f3f3f3f3f3f3f3f;
long long temp = 0;
long long lans = 0;
for (long long i = head[u]; i; i = ed[i].next)
{
long long v = ed[i].to;
if (v == fa[u])
continue;
fa[v] = u;
dfs(v);
sum_dp0 += dp[v][0];
max_i = max(max_i, val[v]);
if (t[v] == 3)
{
if (max_it3 <= val[v])
{
max_it3 = val[v];
temp = v;
}
}
}
for (long long i = head[u]; i; i = ed[i].next)
{
long long v = ed[i].to;
if (v == fa[u])
continue;
if (v == temp)
continue;
max_02 = max(max_02, dp[v][2] - dp[v][0] + val[v]);
}
lans = max_it3 + max_02;//第一种情况
max_it3 = 0;
for (long long i = head[u]; i; i = ed[i].next)
{
long long v = ed[i].to;
if (v == fa[u])
continue;
if (v == temp)
continue;
if (t[v] == 3)
max_it3 = max(max_it3, val[v]);
}
lans = max(lans, max_it3 + dp[temp][2] - dp[temp][0] + val[temp]);//第二种情况
dp[u][0] = max(sum_dp0 + max_i, sum_dp0 + lans);//第三种情况
dp[u][2] = sum_dp0;
return;
}
int main()
{
scanf("%lld", &T);
while (T--)
{
scanf("%lld", &n);
k = 0;
for (int i = 1; i <= n; i++)
{
head[i] = 0;
fa[i] = 0;
dp[i][0] = 0;
dp[i][2] = 0;
}
for (long long i = 1; i <= n; i++)
scanf("%lld", &val[i]);
for (long long i = 1; i <= n; i++)
scanf("%lld", &t[i]);
for (long long i = 1; i <= n - 1; i++)
{
long long u, v;
scanf("%lld%lld", &u, &v);
adde(u, v);
adde(v, u);
}
dfs(1);
printf("%lld\n", dp[1][0] + val[1]);
}
return 0;
}