2020杭电第九场

1001. tree

1001. tree

题意

一棵根为1的树，要求选择两个节点加一条单向边，使得的二元组数量最多

思路

从一个点出发可以到达他的所有子节点和他自己，添加一条边使得(x,y)（从x到y）最多，添加的这条边的终点最好的情况就是连一个子节点最多的点，也就是直接连到根节点1是最优的情况

第一次搜索，用来记录未加边时

用dept记录深度，dp记录元组数，sum记录元组和

第二次搜索，用来记录加边后重复元组的数量

n * dept - tt 为增量

代码

/*************************************************************************
 > FileName:
 > Author:      Lance
 > Mail:        lancelot_hcs@qq.com
 > Date:        9102.1.8
 > Description:
 ************************************************************************/
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
const double pi = acos(-1.0);
const double eps = 1e-6;
const int mod = 1e9 + 7;
#define debug(a) cout << "*" << a << "*" << endl
const int INF = 0x3f3f3f3f;//int2147483647//ll9e18//unsigned ll 1e19
const int maxn = 500005;
//sacnf("%lf") printf("%f")
ll read()
{
    ll x = 0,f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9')
    {
        if (ch == '-')
        f = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9')
    {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return x * f;
}
ll t, n;
vector<ll> G[maxn];

ll dp[maxn];
ll dept[maxn];
ll tt[maxn], sum = 0;
void dfs(int x, int fa) {
    if (x == 1) dept[x] = 1;
    else dept[x] = dept[fa] + 1;
    dp[x] = 1;
    for (int i = 0; i < G[x].size(); i++) {
        int tem = G[x][i];
        if (tem != fa) {
            dfs(tem, x);
            dp[x] += dp[tem];
        }
    }
    sum += dp[x]; 
}

void dfs1(int x, int fa) {
    tt[x] = tt[fa] + dp[x];
    for (int i = 0; i < G[x].size(); i++) {
        int tem = G[x][i];
        if (tem != fa) {
            dfs1(tem, x);
        }
    }
}


void solve()
{
    t = read();
    while (t--) {
        sum = 0;
        n = read();
        int a;
        for (int i = 2; i <= n; i++) {
            a = read();
            G[a].push_back(i);
        }
        dfs(1, 0);
        dfs1(1, 0);
        ll max_ = -INF, index = 0;
        for (int i = 1; i <= n; i++) {
            max_ = max(max_, n * dept[i] - tt[i]);
        }
        printf("%lld
", sum + max_);
        for (int i = 1; i <= n; i++) {
            G[i].clear();
            dp[i] = 0;
            dept[i] = 0;
            tt[i] = 0;
        }
    }
}

int main()
{

//    freopen("F:/Overflow/in.txt","r",stdin);
//    ios::sync_with_stdio(false);
    solve();
    return 0;
}

1003. Slime and Stones

1003.Slime and Stones

题意

威尔逊博弈的拓展, 威佐夫博弈是本题的k = 0 的情况

思路

待续

代码

///*************************************************************************
// > FileName:
// > Author:      Lance
// > Mail:        lancelot_hcs@qq.com
// > Date:        9102.1.8
// > Description:
// ************************************************************************/
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
const double pi = acos(-1.0);
const double eps = 1e-6;
const int mod = 1e9 + 7;
#define debug(a) cout << "*" << a << "*" << endl
const int INF = 0x3f3f3f3f;//int2147483647//ll9e18//unsigned ll 1e19
const int maxn = 1000005;
//sacnf("%lf") printf("%f")
ll read()
{
   ll x = 0,f = 1;
   char ch = getchar();
   while (ch < '0' || ch > '9')
    {
        if (ch == '-')
        f = -1;
        ch = getchar();
    }
   while (ch >= '0' && ch <= '9')
    {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
   return x * f;
}
ll t, n, a, b, k;

void solve()
{
    t =read();
    while (t--) {
        a = read(), b = read(), k = read();
        if (a > b) {
            swap(a, b);
        }
        k++;
        n = b - a;
        if (n % k) {
            puts("1");
            continue;
        } else {
            if (int(n * ((2 - k) * 1ll + sqrt(k * k * 1.0 + 4.0)) / 2 / k) == a) {
                puts("0");
            } else {
                puts("1");
            }
        }
        
    }
}

int main()
{

//    freopen("F:/Overflow/in.txt","r",stdin);
//    ios::sync_with_stdio(false);
   solve();
   return 0;
}

参考博客

https://www.cnblogs.com/stelayuri/p/13525300.html