Weekly Challenges

一拖再拖，忍无可忍，自己懒的没救了。

一周前就结束的比赛，到现在才想起来补。

最后一题貌似又遇到了splay。。。还是不会！！！shit

题目链接

Ａ题－－快速幂

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

const int MOD = 1e9 + 7;

int mod_pow(int a, int b) {
    int res = 1;
    while (b) {
        if (b & 1) res = (res * (long long)a) % MOD;
        a = (a * (long long)a) % MOD;
        b >>= 1;
    }
    return res;
}

int main() {
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */   
    int T;
    ios::sync_with_stdio(false);
    cin >> T;
    while (T--) {
        int t;
        cin >> t;
        cout << (2 + mod_pow(2, t + 1)) % MOD << endl;
    }
    return 0;
}

Ｂ题

如果说这题的困难之处，那就是要看出来其实这种数的个数很少。所以先把所有的范围之内的数打表出来。

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

typedef long long LL;
#define rep(i, n) for (int i = (1); i <= (n); i++)
LL pw10[20];
const LL MAX_N = (LL)1e18;

int main() {
    pw10[0] = 1LL;
    rep (i, 19) pw10[i] = pw10[i - 1] * 10;
    vector<LL> ans;
    rep (i, 9) ans.push_back(i);
    int cur = 0, cur_len = 2, next = -1;
    while (true) {
        if (next == - 1 && ans[cur] * (cur_len + 1) >= pw10[cur_len]) {
            next = cur;
        }
        LL x = ans[cur] * cur_len;
        if (x >= MAX_N) break;
        if (pw10[cur_len - 1] <= x && x < pw10[cur_len]) {
            ans.push_back(x);
        } else if (x >= pw10[cur_len]) {
            cur = next;
            next = -1;
            cur_len++;
            continue;
        }
        cur++;
    }
    //rep (i, 25) cerr << ans[i] << endl;
    int T;
    cin >> T;
    while (T--) {
        LL A, B;
        cin >> A >> B;
        LL rgt = upper_bound(ans.begin(), ans.end(), B) - ans.begin();
        LL lft = lower_bound(ans.begin(), ans.end(), A) - ans.begin();
        LL res = rgt - lft;
        if (A == 0) res++;
        cout << res << endl;
    }
    
    return 0;
}

Ｃ题

题意：给出Ｎ个结点的树，每个结点有一个权值，在树上有一种操作，可以任选一个结点，然后删除以该结点为根的子树。最多执行Ｋ次这种操作

　　　使得树上的权值总和最大。

首先，如果按照前序遍历访问每个结点的顺序给结点做标记，那么以某个结点为根的子树中所有结点的标记一定该结点的后面。

dfs时顺便维护一下以每个结点为根的子树中结点的个数tot[i]。

dp[i][j]表示到dfs中第ｉ个遍历的结点为止操作了ｊ次可以获得的最大权值。

这样当到达第ｉ个结点时，dp[i + 1][j] = max(dp[i + 1][j], dp[i][j] + value[i])   //不删除

　　　　　　　　　　　　dp[i + tot[i]][j + 1] = max(dp[i + tot[i]][j + 1], dp[i][j])　／／删除

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

const int MAX_N = 100050;
typedef long long LL;
const LL INF = (LL)1e19;
vector<int> G[MAX_N];
int w[MAX_N], vs[MAX_N], tot[MAX_N], cnt;
LL dp[MAX_N][205];

void dfs(int u, int fa) {
     int sz = (int)G[u].size();
     vs[++cnt] = u;
     tot[u] = 1;
     if (sz == 1) {
         return ;
     }
     for (int i = 0; i < sz; i++) {
        int v = G[u][i];
        if (v == fa) continue;
        dfs(v, u);
        tot[u] += tot[v];
    }
}

void read(int &res) {
    res = 0;
    int sign = 1;
    char c = ' ';
    while (c < '0' || c > '9') {
        if (c == '-') sign = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9') res = (res<<3) + (res<<1) + c - '0', c = getchar();
    if (sign == -1) res = -res;
}

int main() {
    ios::sync_with_stdio(false);
    int N, K;
    cin >> N >> K;
    //read(N); read(K);
    for (int i = 1; i <= N; i++) cin >> w[i];//read(w[i]);
    for (int i = 1; i <  N; i++) {
        int u, v;
        cin >> u >> v;
        //read(u), read(v);
        G[u].push_back(v);
        G[v].push_back(u);
    }
    cnt = 0;
    dfs(1, -1);
    for (int i = 0; i <= K; i++) for (int j = 1; j <= N; j++) dp[j][i] = -INF;
    //for (int i = 0; i <= K; i++) dp[0][i] = 0;
    dp[1][0] = 0;
    for (int i = 1; i <= N; i++) {
        int u = vs[i];
        for (int j = 0; j <= K; j++) if (dp[i][j] != -INF) {
            dp[i + 1][j] = max(dp[i + 1][j], dp[i][j] + w[u]);
            if (j < K)
                dp[i + tot[u]][j + 1] = max(dp[i + tot[u]][j + 1], dp[i][j]);
        }
    }
    LL res = -INF;
    for (int i = 0; i <= K; i++) res = max(res, dp[N + 1][i]);
    cout << res << endl;
    
    return 0;
}

Ｄ题

题意：Ｎ种技能，第ｉ种技能有ci个人学会，有Ｔ个巫师，每个巫师有两个集合Ａ和Ｂ，他可以把一个拥有Ａ集合中的技能的人转移到集合Ｂ中的某个技能中去。

问最多可以使多少个技能至少有一个人学会。

增加一个源点和汇点，源点到Ｎ种技能连一条容量为ci的边，从Ｎ种技能到汇点连一条容量为１的边。

对于每个巫师，从Ａ集合中的技能连一条容量为１的边到该巫师，从该巫师连一条容量为１的边到Ｂ集合中的每种技能。

最大流即为答案。。。

#include <queue>
#include <cmath>
#include <cstdio>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std; 

struct edge {
    int to, cap, rev;
    edge() {}
    edge(int _to, int _cap, int _rev):to(_to), cap(_cap), rev(_rev) {}
};
const int INF = 0x3f3f3f3f;
const int MAX_V = 250; // ????`
vector<edge> G[MAX_V]; // ?????
int level[MAX_V]; // ??????????
int iter[MAX_V];  // ????????????????
int Q[MAX_V]; // BFS??

void add_edge(int from, int to, int cap) {
    G[from].push_back(edge(to, cap, G[to].size()));
    G[to].push_back(edge(from, 0, G[from].size() - 1));
}

// ?????????????????
void bfs(int s) {
    memset(level, -1, sizeof(level));
    int ini = 0, end = 0;
    Q[end++] = s;
    level[s] = 0; 
    while (end > ini) {
        int v = Q[ini++];
        for (int i = 0; i < G[v].size(); i++) {
            edge &e = G[v][i];
            if (e.cap > 0 && level[e.to] < 0) {
                level[e.to] = level[v] + 1;
                Q[end++] = e.to;
            }
        }
    }
}

// ??????????
int dfs(int v, int t, int f) {
    if (v == t) return f;
    for (int &i = iter[v]; i < G[v].size(); i++) {
        edge &e = G[v][i];
        if (e.cap > 0 && level[v] < level[e.to]) {
            int d = dfs(e.to, t, min(f, e.cap));
            if (d > 0) {
                e.cap -= d;
                G[e.to][e.rev].cap += d;
                return d;
            }
        }
    }
    return 0;
}

// ??????????
int Dinic(int s, int t) {
    int flow = 0;
    for (;;) {
        bfs(s);
        if (level[t] < 0) return flow;
        memset(iter, 0, sizeof(iter));
        int f;
        while ((f = dfs(s, t, INF)) > 0) {
            flow += f;
        }
    }
}

int main() {
    ios_base::sync_with_stdio(false);
    
    int N, T;
    cin >> N >> T;
    int s = 0, t = N + T + 1;
    
    for (int i = 1; i <= N; i++) {
        int C;
        cin >> C;
        add_edge(s, i, C);
        add_edge(i, t, 1);
    }
    for (int i = 1; i <= T; i++) {
        int A, B;
        cin >> A;
        for (int j = 1; j <= A; j++) {
            int x;
            cin >> x;
            add_edge(x, N + i, 1);
        }
        cin >> B;
        for (int j = 1; j <= B; j++) {
            int x;
            cin >> x;
            add_edge(N + i, x, 1);
        }
    }
    
    cout << Dinic(s, t) << endl;
    return 0;
}

#include <queue>
#include <cmath>
#include <cstdio>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std; 

struct edge {
    int to, cap, rev;
    edge() {}
    edge(int _to, int _cap, int _rev):to(_to), cap(_cap), rev(_rev) {}
};
const int INF = 0x3f3f3f3f;
const int MAX_V = 250; // ????`
vector<edge> G[MAX_V]; // ?????
int level[MAX_V]; // ??????????
int iter[MAX_V];  // ????????????????
int Q[MAX_V]; // BFS??

void add_edge(int from, int to, int cap) {
    G[from].push_back(edge(to, cap, G[to].size()));
    G[to].push_back(edge(from, 0, G[from].size() - 1));
}

// ?????????????????
void bfs(int s) {
    memset(level, -1, sizeof(level));
    int ini = 0, end = 0;
    Q[end++] = s;
    level[s] = 0; 
    while (end > ini) {
        int v = Q[ini++];
        for (int i = 0; i < G[v].size(); i++) {
            edge &e = G[v][i];
            if (e.cap > 0 && level[e.to] < 0) {
                level[e.to] = level[v] + 1;
                Q[end++] = e.to;
            }
        }
    }
}

// ??????????
int dfs(int v, int t, int f) {
    if (v == t) return f;
    for (int &i = iter[v]; i < G[v].size(); i++) {
        edge &e = G[v][i];
        if (e.cap > 0 && level[v] < level[e.to]) {
            int d = dfs(e.to, t, min(f, e.cap));
            if (d > 0) {
                e.cap -= d;
                G[e.to][e.rev].cap += d;
                return d;
            }
        }
    }
    return 0;
}

// ??????????
int Dinic(int s, int t) {
    int flow = 0;
    for (;;) {
        bfs(s);
        if (level[t] < 0) return flow;
        memset(iter, 0, sizeof(iter));
        int f;
        while ((f = dfs(s, t, INF)) > 0) {
            flow += f;
        }
    }
}

int main() {
    ios_base::sync_with_stdio(false);
    
    int N, T;
    cin >> N >> T;
    int s = 0, t = N + T + 1;
    
    for (int i = 1; i <= N; i++) {
        int C;
        cin >> C;
        add_edge(s, i, C);
        add_edge(i, t, 1);
    }
    for (int i = 1; i <= T; i++) {
        int A, B;
        cin >> A;
        for (int j = 1; j <= A; j++) {
            int x;
            cin >> x;
            add_edge(x, N + i, 1);
        }
        cin >> B;
        for (int j = 1; j <= B; j++) {
            int x;
            cin >> x;
            add_edge(N + i, x, 1);
        }
    }
    
    cout << Dinic(s, t) << endl;
    return 0;
}

Ｅ题