2019 ICPC NanChang Regional Online Contest

有点毒的比赛。

题目链接：https://www.jisuanke.com/contest/3870?view=challenges

---

B：

添加一个点连接所有救火点，路径设为0，就变成了跑两次dijkstra的题。

/* Contest nanchang_2019_online
 * Problem B
 * Team: Make One For Us
 */
#include <bits/stdc++.h>

using namespace std;

const long long int INF = 10000000000000L;

struct edge {
    int u, v;
    long long int w;
};

struct node {
    int u;
    long long int d;
    bool operator < (const node &n) const {
        return d > n.d;
    }
};

long long int read(void) {
    char ch;
    do {
        ch = getchar();
    } while (!isdigit(ch));
    long long int ret = 0;
    while (isdigit(ch)) {
        ret *= 10;
        ret += ch - '0';
        ch = getchar();
    }
    return ret;
}

void dijkstra(vector <vector <int>> &graph, vector <edge> &edges, vector <long long int> &dis, int s) {
    priority_queue <node> pq;
    pq.push({s, dis[s] = 0});
    vector <int> vis(graph.size());
    while (!pq.empty()) {
        auto tmp = pq.top();
        pq.pop();
        if (vis[tmp.u]) {
            continue;
        }
        vis[tmp.u] = true;
        for (unsigned int i = 0; i < graph[tmp.u].size(); i++) {
            edge &e = edges[graph[tmp.u][i]];
            if (dis[e.v] > dis[e.u] + e.w) {
                pq.push((node) {e.v, dis[e.v] = dis[e.u] + e.w});
            }
        }
    }
}

int main(void) {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int T;
    T = read();
    while (T--) {
        int n = read(), m = read(), s = read(), k = read(), c = read();
        int ptr = 0;
        vector <edge> edges(2 * m + k);
        vector <vector <int>> graph(n + 1);
        auto add_edge = [&] (int u, int v, long long int w) {
            graph[u].emplace_back(ptr);
            edges[ptr++] = ((edge) {u, v, w});
        };
        for (int i = 0; i < k; i++) {
            int v = read();
            add_edge(0, v, 0);
        }
        for (int i = 0; i < m; i++) {
            int u = read(), v = read();
            long long int w = read();
            add_edge(u, v, w);
            add_edge(v, u, w);
        }

        vector <long long int> dis(n + 1, INF);
        dijkstra(graph, edges, dis, s);
        auto ans_s = *max_element(dis.begin() + 1, dis.end());
        for (int i = 0; i <= n; i++) {
            dis[i] = INF;
        }
        dijkstra(graph, edges, dis, 0);
        auto ans_k = *max_element(dis.begin() + 1, dis.end());

        if (ans_k * c < ans_s) {
            cout << ans_k << endl;
        } else {
            cout << ans_s << endl;
        }
    }
    return 0;
}

C：

一开始以为是莫队后来发现不对，就以为是字符串乱搞，结果是线段树维护dp矩阵区间合并。原题是Codeforces goodbye 2016 E。(tourist都没做出来的题这么多人做出来，有点假啊)

E：

一个n有4e7的约瑟夫问题居然是模拟，死都不信，垃圾数据。

G：

签到。

/* basic header */
#include <bits/stdc++.h>
/* define */
#define ll long long
#define dou double
#define pb emplace_back
#define mp make_pair
#define sot(a,b) sort(a+1,a+1+b)
#define rep1(i,a,b) for(int i=a;i<=b;++i)
#define rep0(i,a,b) for(int i=a;i<b;++i)
#define eps 1e-8
#define int_inf 0x3f3f3f3f
#define ll_inf 0x7f7f7f7f7f7f7f7f
#define lson (curpos<<1)
#define rson (curpos<<1|1)
/* namespace */
using namespace std;
/* header end */

int t;

int main() {
    scanf("%d", &t);
    while (t--) {
        ll n; scanf("%lld", &n);
        if (n == 1) puts("18000");
        else puts("0");
    }
    return 0;
}

H：

二分矩阵快速幂+记忆化。

#include<bits/stdc++.h>
typedef long long ull;

using namespace std;

const int maxn = 1e6 + 10;
const ull mod = 998244353;

struct Matrix {
    ull mem[2][2];
    void init() {
        memset(mem, 0, sizeof(mem));
    }
    Matrix operator*(const Matrix &rhs)const {
        Matrix ret; ret.init();
        for (int i = 0; i < 2; i++) {
            for (int j = 0; j < 2; j++) {
                for (int k = 0; k < 2; k++)
                    ret.mem[i][j] += mem[i][k] * rhs.mem[k][j];
                ret.mem[i][j] %= mod;
            }
        }
        return ret;
    }
} mAns, mBase;
unordered_map<ull, Matrix> M;

int tot = 0;
Matrix solve(ull n) {
    if (M.count(n)) return M[n];
    if (n == 1) {
        M[n] = mBase;
        return mBase;
    }
    Matrix ret = solve(n / 2ull);
    ret = ret * ret;
    if (n % 2ull == 1ull) ret = ret * mBase;
    return M[n] = ret;
}

int main() {
    ull q, n;
    cin >> q >> n;
    ull lastans = 0ull;
    ull ans = 0;
    for (int i = 1ull; i <= q; i++) {
        mAns.mem[0][0] = 1, mAns.mem[0][1] = 0;
        mBase.mem[0][0] = 3, mBase.mem[0][1] = 1, mBase.mem[1][0] = 2;
        if (i >= 2ull) n = (n ^ (ans * ans));
        Matrix val = solve(n % (mod - 1));
        ans = val.mem[0][1];
        lastans = (lastans ^ ans);
    }
    cout << lastans << endl;
    return 0;
}

L：

三位偏序，cdq分治。待补。