Codeforces Round #328 (Div. 2)

这场CF，准备充足，回寝室洗了澡，睡了一觉，可结果。。。

 

水 A - PawnChess

第一次忘记判断相等时A先走算A赢，hack掉。后来才知道自己的代码写错了(摔

for (int i=1; i<=8; ++i)	{
	scanf ("%s", s[i]);    //！！！
}

数学(找规律) B - The Monster and the Squirrel

题意：多边形每个顶点向其它的点引射线，如果碰到其他射线则停止，问最后多边形被分成多少个区域

分析：搬题解：

Problem B. The monster and the squirrel

After drawing the rays from the first vertex (n - 2) triangles are formed. The subsequent rays will generate independently sub-regions in these triangles. Let's analyse the triangle determined by vertices 1, i, i + 1, after drawing the rays from vertex i and (i + 1) the triangle will be divided into (n - i) + (i - 2) = n - 2 regions. Therefore the total number of convex regions is (n - 2)2

If the squirrel starts from the region that have 1 as a vertex, then she can go through each region of triangle (1, i, i + 1) once. That implies that the squirrel can collect all the walnuts in (n - 2)2 jumps.

 

我一直在猜结论，试了一些公式，但和3，4的情况对不上。(卒

注意爆int

 

数学 C - The Big Race

题意：两个在比赛，每个人的速度固定，终点<=t，问两人不分胜负的可能情况

分析：分类讨论，如果LCM (b, w) <= t, 那么每个LCM的倍数的点以及之后跟着的min (b, w) - 1都是不分胜负的，其他情况不详细分析。。。

注意的是LCM可能爆long long，可以先除和t比较或者转换成double log函数比较

/************************************************
* Author        :Running_Time
* Created Time  :2015/10/31 星期六 22:41:05
* File Name     :C.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = acos (-1.0);

ll GCD(ll a, ll b)	{
	return b ? GCD (b, a % b) : a;
}

int main(void)    {
	ll t, w, b;	scanf ("%I64d%I64d%I64d", &t, &w, &b);
	if (w > b)	swap (w, b);
	if (w > t && b > t)	{
		printf ("1/1\n");
	}
	else if (b > t)	{
		ll x = GCD (w - 1, t);
		printf ("%I64d/%I64d\n", (w - 1) / x, t / x);
	}
	else if (log ((double) w) + log ((double) b) - log ((double) GCD (w, b)) > log ((double) t))	{
		ll x = GCD (w - 1, t);
		printf ("%I64d/%I64d\n", (w - 1) / x, t / x);
	}
	else	{
		ll lcm = b / GCD (w, b) * w;
		if (t % lcm == 0)	{
			ll y = (w - 1) + (t / lcm - 1) * w + 1;
			ll x = GCD (y, t);
			printf ("%I64d/%I64d\n", y / x, t / x);
		}
		else	{
			ll y = (w - 1) + (t / lcm - 1) * w + (1 + min (t % lcm, w - 1));
			ll x = GCD (y, t);
			printf ("%I64d/%I64d\n", y / x, t / x);
		}
	}

   //cout << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";

    return 0;
}

　　

虚树的直径 D - Super M

题意：一棵树，有若干个点要走，问从哪个点出发可以使得路径最短，走完不用走回起点

分析：如果回到起点的话就是路径上的边的两倍，不回起点就减去一条最长的回路(直径)。首先找出包含所有要走的点的子树，从任意一点DFS找深度最大的点再DFS一遍，这样就可以找到直径，最后贪心找字典序小的直径的端点

/************************************************
* Author        :Running_Time
* Created Time  :2015/10/31 星期六 22:41:05
* File Name     :D.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 123456 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = acos (-1.0);

bool mark[N];
int d[N];
int cnt[N];
vector<int> G[N];

void DFS(int u, int fa) {
    cnt[u] = 0;
    if (mark[u])    cnt[u] = 1;
    for (int v, i=0; i<G[u].size (); ++i)   {
        v = G[u][i];
        if (v == fa)    continue;
        d[v] = d[u] + 1;
        DFS (v, u);
        cnt[u] += cnt[v];
    }
}

int main(void)    {
    int n, m;   scanf ("%d%d", &n, &m);
    for (int u, v, i=1; i<n; ++i)   {
        scanf ("%d%d", &u, &v);
        G[u].push_back (v);
        G[v].push_back (u);
    }
    int id = -1;
    for (int u, i=1; i<=m; ++i) {
        scanf ("%d", &u);
        mark[u] = true;
        if (id == -1 || id > u) {
            id = u;
        }
    }
    if (m == 1) {
        printf ("%d\n0\n", id); return 0;
    }
    DFS (1, 0);
    id = -1;
    for (int i=1; i<=n; ++i)    {
        if (mark[i] && (id == -1 || d[id] < d[i]))  {
            id = i;
        }
    }
    int es = 0, len = 0;
    memset (d, 0, sizeof (d));
    DFS (id, 0);                         //从深度最深的点开始搜
    for (int i=1; i<=n; ++i)    {
        if (cnt[i] > 0 && m - cnt[i] > 0)   {       //这个cnt数组记录了以i为根的子树中被攻击的城市的个数
            es += 2;
        }
        if (mark[i])    len = max (len, d[i]);
    }
    for (int i=1; i<=n; ++i)    {       //找到直径后取id小的端点
        if (mark[i] && i < id && d[i] == len)    {
            id = i; break;
        }
    }
    printf ("%d\n%d\n", id, es - len);

   //cout << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";

    return 0;
}