Codeforces Round #343 (Div. 2)

 居然补完了

组合 A - Far Relative’s Birthday Cake

import java.util.*;
import java.io.*;

public class Main   {
    public static void main(String[] args)  {
        Scanner cin = new Scanner (new BufferedInputStream (System.in));
        int n = cin.nextInt ();
        int[] col = new int[105];
        String str;
        long ans = 0;
        for (int i=0; i<n; ++i) {
            str = cin.next ();
            int num = 0;
            for (int j=0; j<str.length (); ++j)  {
                if (str.charAt (j) == 'C')  {
                    col[j]++;   num++;
                }
            }
            ans += num * (num - 1) / 2;
        }
        for (int i=0; i<n; ++i) ans += col[i] * (col[i] - 1) / 2;
        System.out.println (ans);
    }
}

枚举 B - Far Relative’s Problem

import java.util.*;
import java.io.*;

public class Main   {
    public static void main(String[] args)  {
        Scanner cin = new Scanner (new BufferedInputStream (System.in));
        int n = cin.nextInt ();
        String[] sex = new String[5005];
        int[] btime = new int[5005];
        int[] etime = new int[5005];
        for (int i=0; i<n; ++i) {
            sex[i] = cin.next ();
            btime[i] = cin.nextInt ();
            etime[i] = cin.nextInt ();            
        }
        int ans = 0;
        for (int i=1; i<=366; ++i)  {
            int male = 0, female = 0;
            for (int j=0; j<n; ++j) {
                if (i >= btime[j] && i <= etime[j]) {
                    if (sex[j].charAt (0) == 'M')   male++;
                    else    female++;
                }
            }
            if (male < female)  {
                if (male * 2 > ans) ans = male * 2;
            }
            else    {
                if (female * 2 > ans)   ans = female * 2;
            }
        }
        System.out.println (ans);
    }
}

DP C - Famil Door and Brackets

题意：n长的字符串有‘（’和‘）’组成，现在已知其中m长的字串，问满足任意前缀字串‘（’数量不小于‘）’数量且最后两者数量相等的原字符串的方案数。

分析：‘（’看作+1，‘（’看作-1。dp[i][j]表示前i个字符，和为j(j >= 0)的方案数，然后枚举m字符串前的字符个数p，那么m后q的个数也知道，根据总和为0可以得到前后组合，这里dp[n-m-i][j+now]用到对称思想。

import java.util.*;
import java.io.*;

public class Main   {
    public static final int MOD = 1000000007;
    public static void main(String[] args)  {
        Scanner cin = new Scanner (new BufferedInputStream (System.in));
        Main ma = new Main ();
        int n = cin.nextInt ();
        int m = cin.nextInt ();
        char[] str = cin.next ().toCharArray ();
        long[][] dp = new long[2005][2005];
        dp[0][0] = 1;
        for (int i=1; i<=n-m; ++i)  {
            for (int j=0; j<=i; ++j)    {
                if (j > 0)  {
                    dp[i][j] = ma.add (dp[i][j], dp[i-1][j-1]);
                }
                dp[i][j] = ma.add (dp[i][j], dp[i-1][j+1]);
            }
        }
        int mn = 10000000;
        int now = 0;
        for (int i=0; i<m; ++i)    {
            if (str[i] == '(')  now++;
            else    now--;
            if (now < mn)   mn = now;
        }
        long ans = 0;
        for (int i=0; i<=n-m; ++i)  {
            for (int j=0; j<=i; ++j)    {
                if (j + mn >= 0 && j + now <= n - m - i)    {
                    ans = ma.add (ans, dp[i][j] * dp[n-m-i][j+now] % MOD);
                }
            }
        }
        System.out.println (ans);
    }
    public long add(long a, long b)  {
        a += b;
        if (a >= MOD)   a -= MOD;
        return a;
    }
}

线段树+DP D - Babaei and Birthday Cake

题意：求最大上升序列和

分析：dp[i] 表示前i得到的最大上升序列和，dp[i] = dp[j] + vol[i] (vol[i] > vol[j])。用线段树优化动态统计前rk[i] - 1的最大值即dp[j]，rk[i]是vol[i]离散化后的排名。

#include <bits/stdc++.h>

#define lson l, mid, o << 1
#define rson mid + 1, r, o << 1 | 1
const double PI = acos (-1.0);
const int N = 1e5 + 5;
struct Segment_Tree {
    double v[N<<2], mx[N<<2];
    void push_up(int o) {
        mx[o] = std::max (mx[o<<1], mx[o<<1|1]);
    }
    void build(int l, int r, int o) {
        if (l == r) {
            v[o] = mx[o] = 0;
            return ;
        }
        int mid = l + r >> 1;
        build (lson);   build (rson);
        push_up (o);
    }
    void updata(int p, double x, int l, int r, int o)   {
        if (l == r && l == p)   {
            v[o] = mx[o] = x;
            return ;
        }
        int mid = l + r >> 1;
        if (p <= mid)   updata (p, x, lson);
        else    updata (p, x, rson);
        push_up (o);
    }
    double query(int ql, int qr, int l, int r, int o)   {
        if (ql <= l && r <= qr) {
            return mx[o];
        }
        int mid = l + r >> 1;   double ret = 0;
        if (ql <= mid)  ret = std::max (ret, query (ql, qr, lson));
        if (qr > mid)   ret = std::max (ret, query (ql, qr, rson));
        return ret;
    }
};
double dp[N];
int r[N], h[N];
double vol[N], V[N];

int main(void)  {
    int n;  scanf ("%d", &n);
    for (int i=0; i<n; ++i) {
        scanf ("%d%d", r + i, h + i);
        vol[i] = PI * r[i] * r[i] * h[i];
        V[i] = vol[i];
    }
    std::sort (V, V+n);
    Segment_Tree st;
    st.build (1, n, 1);
    for (int i=0; i<n; ++i) {
        int pos = std::lower_bound (V, V+n, vol[i]) - V + 1;
        if (pos == 1)   dp[i] = vol[i];
        else    dp[i] = st.query (1, pos - 1, 1, n, 1) + vol[i];
        st.updata (pos, dp[i], 1, n, 1);
    }
    double ans = 0;
    for (int i=0; i<n; ++i) {
        if (ans < dp[i])    ans = dp[i];
    }
    printf ("%.10f
", ans);

    return 0;
}

LCA + DP + DFS E - Famil Door and Roads

题意：加一条边，使形成简单环(无重边)，u和v在其中的方案数

分析：加一条边一定是从u或v引出一条边到w而且w能到另一个点。无重边就是w不能选择u到v路径上的点。

　　dep[u]：根节点1到u的距离　　　sz[u]：u的子树包括u的结点数　　sdown[u]：在u的子树下到u的距离和，树形dp

　　sall[u]：所有点到u的距离和，由sdown[u]得到，也是树形DP

　　一共有3种情况：1.LCA (u, v) == v，除v子树外所有点到v的距离和/点数 + u子树所有点到u的距离 / 点数

　　　　　　　　　　2.LCA (v, u) == u，同1

　　　　　　　　　　3.除1，2的情况，只能在u或v的子树选择一节点才能构成环，u子树所有点到u的距离 / 点数 + v子树所有点到v的距离 / 点数

　　最后还要加上不变的距离 dis (u, v) + 1。学习大牛的代码，获益匪浅

#include <bits/stdc++.h>

const int N = 1e5 + 5;
const int D = 20;
std::vector<int> G[N];
int dep[N], sz[N];
long long sdown[N], sall[N];
int rt[N][D];
int n, m;

void DFS(int u, int fa) {       //get rt[v][0], sdown[u] and dep[v]
    sdown[u] = 0;   sz[u] = 1;
    for (int i=0; i<G[u].size (); ++i)  {
        int v = G[u][i];
        if (v == fa || dep[v] != 0)    continue;
        rt[v][0] = u;
        dep[v] = dep[u] + 1;
        DFS (v, u);
        sdown[u] += sdown[v] + sz[v];
        sz[u] += sz[v];
    }
}

void DFS2(int u, int fa)    {       //get sall[v]
    for (int i=0; i<G[u].size (); ++i)  {
        int v = G[u][i];
        if (v == fa)    continue;
        sall[v] = sall[u] + n - 2 * sz[v];
        DFS2 (v, u);
    }
}

void init_LCA(void) {
    for (int i=1; i<D; ++i) {
        for (int j=1; j<=n; ++j)    {
            rt[j][i] = rt[j][i-1] == 0 ? 0 : rt[rt[j][i-1]][i-1];
        }
    }
}

int up(int u, int d)    {
    for (int i=D-1; i>=0; --i)  {
        if (d < (1 << i))   continue;
        u = rt[u][i];   d -= (1 << i);
    }
    return u;
}

int LCA(int u, int v)   {
    if (dep[u] < dep[v])    std::swap (u, v);
    for (int i=0; i<D; ++i) {
        if ((dep[u] - dep[v]) >> i & 1) {
            u = rt[u][i];
        }
    }
    if (u == v) return u;
    for (int i=D-1; i>=0; --i)  {
        if (rt[u][i] != rt[v][i])   {
            u = rt[u][i];
            v = rt[v][i];
        }
    }
    return rt[u][0];
}

int main(void)  {
    scanf ("%d%d", &n, &m);
    for (int u, v, i=0; i<n-1; ++i) {
        scanf ("%d%d", &u, &v);
        G[u].push_back (v);
        G[v].push_back (u);
    }
    dep[1] = 0;
    DFS (1, 0);
    sall[1] = sdown[1];
    DFS2 (1, 0);
    init_LCA ();
    while (m--) {
        int u, v;   scanf ("%d%d", &u, &v);
        int lca = LCA (u, v);
        double ans = dep[u] + dep[v] - 2 * dep[lca] + 1;
        if (lca == v || lca == u)   {
            if (lca == u)   std::swap (u, v);
            int v2 = up (u, dep[u] - dep[v] - 1);
            long long supv = sall[v] - sdown[v2] - sz[v2];
            ans += 1.0 * supv / (n - sz[v2]) + 1.0 * sdown[u] / sz[u];
        }
        else    {
            ans += 1.0 * sdown[u] / sz[u] + 1.0 * sdown[v] / sz[v];
        }
        printf ("%.12f
", ans);
    }

    return 0;
}