康復計劃 Round 1: Codeforces Round #518 (Div. 1)

題目鏈接

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Codeforces - 1067A $Array Without Local Maximums$

參考Blog

看完轉移方程自己碼了出來，最後有個初始化的地方沒寫好卡了一下，還好給的時間和空間夠多，卡了過去。

代碼：

#include <bits/stdc++.h>
#define mst(a,b) memset(a,b,sizeof(a))
#define ALL(x) x.begin(),x.end()
#define pii pair<int,int>
#define eps 1e-6
#define debug(a) cout << #a": " << a << endl;
#define eularMod(a, b) a < b ? a : a % b + b
inline int lowbit(int x){ return x & -x; }
const int N = 1e5 + 10;
const int mod = (int) 998244353;
const int INF = 0x3f3f3f3f;
const long long LINF = (1LL << 62);
typedef long long LL;
typedef unsigned long long ULL;
const double PI = acos(-1.0);
using namespace std;

int n;
LL dp[N][205][3], sum[205][3], suf[205][3];
int v[N];

void getSum (int pos) {
    for (int i = 1; i <= 200; i++) {
        sum[i][0] = (sum[i - 1][0] + dp[pos][i][0]) % mod;
        sum[i][1] = (sum[i - 1][1] + dp[pos][i][1]) % mod;
        sum[i][2] = (sum[i - 1][2] + dp[pos][i][2]) % mod;
    }
}
void getSuf (int pos) {
    for (int i = 200; i >= 1; i--) {
        suf[i][0] = (suf[i + 1][0] + dp[pos][i][0]) % mod;
        suf[i][1] = (suf[i + 1][1] + dp[pos][i][1]) % mod;
        suf[i][2] = (suf[i + 1][2] + dp[pos][i][2]) % mod;
    }
}
void update (int i, int j) {
    dp[i][j][0] = (sum[j - 1][0] + sum[j - 1][1] + sum[j - 1][2]) % mod;
    dp[i][j][1] = (dp[i - 1][j][0] + dp[i - 1][j][1] + dp[i - 1][j][2]) % mod;
    dp[i][j][2] = (suf[j + 1][1] + suf[j + 1][2]) % mod;
}
// 0 表示 v[i - 1] < v[i]
// 1 表示 v[i - 1] == v[i]
// 2 表示 v[i - 1] > v[i]

int main() {
#ifdef LOCAL
    freopen("in.txt", "r", stdin);
//    freopen("out.txt","w",stdout);
#endif
    scanf("%d", &n);

    for (int i = 1; i <= n; i++) {
        scanf("%d", &v[i]);
    }

    if (v[1] == -1) {
        for (int i = 1; i <= 200; i++) {
            dp[1][i][0] = 1;
        }
    } else {
        dp[1][v[1]][0] = 1;
    }

    getSum(1);
    getSuf(1);

    for (int i = 2; i <= n; i++) {
        if (v[i] == -1) {
            for (int j = 1; j <= 200; j++) {
                update(i, j);
            }
        } else {
            update(i, v[i]);
        }
        mst(sum, 0);
        mst(suf, 0);
        getSum(i);
        getSuf(i);
    }

    LL ans = 0;
    for (int i = 1; i <= 200; i++) {
        ans = (ans + dp[n][i][2] + dp[n][i][1]) % mod;
    }

    printf("%lld\n", ans);
    return 0;
}

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Codeforces - 1067B $Multihedgehog$

題意有點繞，一開始讀不懂第二個條件，以爲是葉子節點的父節點至少有 3 個孩子就可以，後面才知道是一層一層的刪都要滿足葉子節點的父節點至少有 3 個孩子，我的做法是找直徑的終點，然後 dfs 一下檢查是否符合條件。

代碼：

#include <bits/stdc++.h>
#define mst(a,b) memset(a,b,sizeof(a))
#define ALL(x) x.begin(),x.end()
#define pii pair<int,int>
#define eps 1e-6
#define debug(a) cout << #a": " << a << endl;
#define eularMod(a, b) a < b ? a : a % b + b
inline int lowbit(int x){ return x & -x; }
const int N = 1e5 + 10;
const int mod = (int) 998244353;
const int INF = 0x3f3f3f3f;
const long long LINF = (1LL << 62);
typedef long long LL;
typedef unsigned long long ULL;
const double PI = acos(-1.0);
using namespace std;

int n, k;
int deg[N], head[N], cnt, tag[N];
int treeDiameter, point, tagPoint;
struct Edge {
    int nex, to;
}edge[N << 1];

inline void AddEdge (int u, int v) {
    edge[++cnt].nex = head[u];
    edge[cnt].to = v;
    head[u] = cnt;
    edge[++cnt].nex = head[v];
    edge[cnt].to = u;
    head[v] = cnt;
}
void dfs (int u, int fa, int len) {
    if (len > treeDiameter) {
        treeDiameter = len;
        point = u;
    }
    for (int i = head[u]; i; i = edge[i].nex) {
        int v = edge[i].to;
        if (v != fa) {
            dfs(v, u, len + 1);
        }
    }
}
bool dfs1 (int u, int fa, int len) {
    if (len == treeDiameter) {
        return false;
    }
    for (int i = head[u]; i; i = edge[i].nex) {
        int v = edge[i].to;
        if (v != fa) {
            if (!dfs1(v, u, len + 1)) {
                if (len == (treeDiameter >> 1)) {
                    tagPoint = u;
                }
                return false;
            }
        }
    }
    return true;
}
bool dfs2 (int u, int fa, int len) {
    if (fa && deg[u] == 1) {
        if (len == (treeDiameter >> 1)) return true;
        else return false;
    }
    int cnt = 0;
    for (int i = head[u]; i; i = edge[i].nex) {
        int v = edge[i].to;
        if (v != fa) {
            if (!dfs2(v, u, len + 1)) return false;
            cnt++;
        }
    }
    if (cnt >= 3) return true;
    else return false;
}

int main() {
#ifdef LOCAL
    freopen("in.txt", "r", stdin);
//    freopen("out.txt","w",stdout);
#endif
    scanf("%d%d", &n, &k);

    for (int i = 1; i <= n - 1; i++) {
        int u, v;
        scanf("%d%d", &u, &v);
        deg[u]++;
        deg[v]++;
        AddEdge(u, v);
    }

    for (int i = 1; i <= n; i++) if (deg[i] == 1) {
        tag[edge[head[i]].to]++;
    }

    for (int i = 1; i <= n; i++) if (tag[i] && tag[i] < 3) {
        printf("No\n");
        return 0;
    }

    dfs(1, 0, 0);
    dfs(point, 0, 0);

    if ((treeDiameter & 1) || k != (treeDiameter >> 1)) {
        printf("No\n");
        return 0;
    }

    dfs1(point, 0, 0);

//    printf("%d %d %d\n", point, tagPoint, treeDiameter);

    if (dfs2(tagPoint, 0, 0)) printf("Yes\n");
    else printf("No\n");
    return 0;
}

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Codeforces - 1067C $Knights$

神奇構造，參考題解：

代碼：

#include <bits/stdc++.h>
#define mst(a,b) memset(a,b,sizeof(a))
#define ALL(x) x.begin(),x.end()
#define pii pair<int,int>
#define eps 1e-6
#define debug(a) cout << #a": " << a << endl;
#define eularMod(a, b) a < b ? a : a % b + b
inline int lowbit(int x){ return x & -x; }
const int N = 1e5 + 10;
const int mod = (int) 998244353;
const int INF = 0x3f3f3f3f;
const long long LINF = (1LL << 62);
typedef long long LL;
typedef unsigned long long ULL;
const double PI = acos(-1.0);
using namespace std;

int main() {
#ifdef LOCAL
    freopen("in.txt", "r", stdin);
//    freopen("out.txt","w",stdout);
#endif
    int n;
    scanf("%d", &n);

    for (int i = 0; ; i += 2) {
        printf("%d %d\n", 0, i);
        if (!(--n)) break;
        printf("%d %d\n", 0, i + 1);
        if (!(--n)) break;
        printf("%d %d\n", 3, i + 1);
        if (!(--n)) break;
    }
    return 0;
}