Shawn's Blog

// Reference: https://leetcode.cn/problems/minimum-score-after-removals-on-a-tree/solution/dfs-shi-jian-chuo-chu-li-shu-shang-wen-t-x1kk/
class Solution {
public:
    int minimumScore(vector<int>& nums, vector<vector<int>>& edges) {
        // 建图。
        int n = nums.size();
        vector<vector<int>> g(n);
        for (auto &e : edges) {
            int u = e[0], v = e[1];
            g[u].push_back(v);
            g[v].push_back(u);
        }
        int xr[n], in[n], out[n], clock = 0;
        function<void(int, int)> dfs = [&](int u, int f) {
            in[u] = ++clock;
            xr[u] = nums[u];
            for (int v : g[u])
                if (v != f) {
                    dfs(v, u);
                    xr[u] ^= xr[v];
                }
            out[u] = clock;
        };
        // 以 0 为根遍历并记录节点访问次序。
        dfs(0, -1);
        // 根据访问次序判断节点父子关系。
        auto is_ancestor = [&](int x, int y) -> bool { return in[x] < in[y] && in[y] <= out[x]; };
        int ans = INT_MAX;
        // 枚举删除的两条边所对应的端点（四个点中的两个子点，因此 0 除外）。
        for (int i = 2, x, y, z; i < n; ++i) // 枚举被删除的点 i。
            for (int j = 1; j < i; ++j) { // 枚举被删除的另一个点 j。
                if (is_ancestor(i, j)) x = xr[j], y = xr[i] ^ x, z = xr[0] ^ xr[i]; // i 是 j 的父节点。
                else if (is_ancestor(j, i)) x = xr[i], y = xr[j] ^ x, z = xr[0] ^ xr[j]; // j 是 i 的父节点。
                else x = xr[i], y = xr[j], z = xr[0] ^ x ^ y; // 删除的两条边分属于两个子树。
                ans = min(ans, max({x, y, z}) - min({x, y, z}));
                if (ans == 0) return 0; // 剪枝。
            }
        return ans;
    }
};

// Reference: https://leetcode.cn/problems/number-of-distinct-roll-sequences/solution/by-endlesscheng-tgkn/
const int MOD = 1e9 + 7, MX = 1e4;
int f[MX + 1][6][6];
int init = []() {
    for (int i = 0; i < 6; ++i)
        for (int j = 0; j < 6; ++j)
            if (j != i && gcd(j + 1, i + 1) == 1)
                f[2][i][j] = 1;
    for (int i = 2; i < MX; ++i)
        for (int j = 0; j < 6; ++j)
            for (int last = 0; last < 6; ++last)
                if (last != j && gcd(last + 1, j + 1) == 1)
                    for (int last2 = 0; last2 < 6; ++last2)
                        if (last2 != j && last2 != last)
                            f[i + 1][j][last] = (f[i + 1][j][last] + f[i][last][last2]) % MOD;
    return 0;
}();

class Solution {
public:
    int distinctSequences(int n) {
        if (n == 1) return 6;
        int ans = 0;
        for (int i = 0; i < 6; ++i)
            for (int j = 0; j < 6; ++j)
                ans = (ans + f[n][i][j]) % MOD;
        return ans;
    }
};

// Reference: https://leetcode.cn/problems/number-of-distinct-roll-sequences/solution/by-endlesscheng-tgkn/
const int MOD = 1e9 + 7, MX = 1e4;
int f[MX + 1][6];
int init = []() {
    for (int i = 0; i < 6; ++i)
        f[1][i] = 1;
    for (int i = 2; i <= MX; ++i)
        for (int j = 0; j < 6; ++j) {
            long s = 0L;
            for (int k = 0; k < 6; ++k)
                if (k != j && gcd(k + 1, j + 1) == 1)
                    s += f[i - 1][k] - f[i - 2][j];
            if (i > 3) s += f[i - 2][j];
            f[i][j] = s % MOD;
        }
    return 0;
}();

class Solution {
public:
    int distinctSequences(int n) {
        long ans = 0L;
        for (int v : f[n])
            ans += v;
        return (ans % MOD + MOD) % MOD;
    }
};

#include <bits/stdc++.h>
using namespace std;
using PII = pair<int, int>;
const int N = 100001;
int n, m, x, y, dist[N];
vector<PII> g[N];
priority_queue<PII, vector<PII>, greater<PII>> q;
int main() {
    scanf("%d%d", &n, &m);
    while (m--) {
        scanf("%d%d", &x, &y);
        g[x].emplace_back(y, 0);
        g[y].emplace_back(x, 1);
    }
    memset(dist, 0x3f, sizeof(dist));
    q.emplace(0, 1);
    dist[1] = 0;
    while (q.size()) {
        auto p = q.top(); q.pop();
        int d = p.first, u = p.second;
        for (auto &x : g[u]) {
            int v = x.first, w = x.second;
            if (dist[v] > d + w) {
                dist[v] = d + w;
                q.emplace(dist[v], v);
            }
        }
    }
    printf("%d\n", dist[n] == 0x3f3f3f3f ? -1 : dist[n]);
    return 0;
}

#include <bits/stdc++.h>
using namespace std;
using PII = pair<int, int>;
const int N = 100001;
int n, m, x, y, dist[N];
vector<PII> g[N];
queue<int> q;
bool vis[N];
int main() {
    scanf("%d%d", &n, &m);
    while (m--) {
        scanf("%d%d", &x, &y);
        g[x].emplace_back(y, 0);
        g[y].emplace_back(x, 1);
    }
    memset(dist, 0x3f, sizeof(dist));
    dist[1] = 0;
    q.emplace(1);
    while (q.size()) {
        int u = q.front(); q.pop();
        vis[u] = false;
        for (auto &x : g[u]) {
            int v = x.first, w = x.second;
            if (dist[v] > dist[u] + w) {
                dist[v] = dist[u] + w;
                if (!vis[v])
                    q.emplace(v), vis[v] = true;
            }
        }
    }
    printf("%d\n", dist[n] == 0x3f3f3f3f ? -1 : dist[n]);
    return 0;
}

class Solution {
    using TII = tuple<int, int, int>;
public:
    int minCost(vector<vector<int>>& grid) {
        // 1->right, 2->left, 3->down, 4->up.
        int m = grid.size(), n = grid[0].size(), ds[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
        vector<vector<int>> dist(m, vector<int>(n, 0x3f3f3f3f));
        vector<vector<bool>> vis(m, vector<bool>(n));
        dist[0][0] = 0;
        priority_queue<TII, vector<TII>, greater<TII>> q;
        q.emplace(0, 0, 0);
        while (q.size()) {
            auto [d, x, y] = q.top(); q.pop();
            if (vis[x][y]) continue;
            vis[x][y] = true;
            for (int i = 0; i < 4; ++i) {
                int nx = x + ds[i][0], ny = y + ds[i][1];
                if (nx < 0 || nx >= m || ny < 0 || ny >= n) continue;
                int w = grid[x][y] != i + 1;
                if (dist[nx][ny] > dist[x][y] + w) {
                    dist[nx][ny] = dist[x][y] + w;
                    q.emplace(dist[nx][ny], nx, ny);
                }
            }
        }
        return dist[m - 1][n - 1];
    }
};