Codeforces 965 枚举轮数贪心分糖果 青蛙跳石头最大流=最小割思想 trie启发式合并

A

/*#include<cstring>#include<algorithm>#include<queue>#include<vector>#include<cstdio>#include<cmath>#include<iostream>*/
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 300005;
int main()
{
        int k, n, s, p;
        cin >> k >> n >> s >> p;
        int ans = n / s + (1 - (n % s == 0));
        ans *= k;
        cout << ans / p + (1 - (ans % p == 0)) << endl;
        return 0;
}

B

/*#include<cstring>#include<algorithm>#include<queue>#include<vector>#include<cstdio>#include<cmath>#include<iostream>*/
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 300005;
char f[105][105];
int ans;
int aimc = 1;
int aimr = 1;
int flag;
int now;
int main()
{
        int n, k;
        cin >> n >> k;
        for (int i = 1; i <= n; i++)
        {
                scanf("%s", f[i] + 1);
        }
        for (int i = 1; i <= n; i++)
        {
                for (int j = 1; j <= n; j++)
                {
                        if (f[i][j] == '#')
                        {
                                continue;
                        }
                        now = 0;
                        for (int dx = j - k + 1; dx <= j; dx++)
                        {
                                if (dx + k > n + 1)
                                {
                                        break;
                                }
                                if (dx < 1)
                                {
                                        continue;
                                }
                                flag = 1;
                                for (int w = dx; w <= dx + k - 1; w++)
                                {
                                        if (f[i][w] == '#')
                                        {
                                                flag = 0;
                                                break;
                                        }
                                }
                                if (flag)
                                {
                                        now++;
                                }
                        }
                        for (int dy = i - k + 1; dy <= i; dy++)
                        {
                                if (dy + k > n + 1)
                                {
                                        break;
                                }
                                if (dy < 1)
                                {
                                        continue;
                                }
                                flag = 1;
                                for (int w = dy; w <= dy + k - 1; w++)
                                {
                                        if (f[w][j] == '#')
                                        {
                                                flag = 0;
                                                break;
                                        }
                                }
                                if (flag)
                                {
                                        now++;
                                }
                        }
                        if (now > ans)
                        {
                                ans = now;
                                aimc = i, aimr = j;
                        }
                }
        }
        cout << aimc << " " << aimr << endl;
        return 0;
}

C

首先用贪心的思想可以知道 如果是一整轮一整轮地分 肯定是X越大越好

当加上题目剩下的不小于X的也要分的时候 最佳肯定是当X尽量大且最后多分给A1一次的时候最佳

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int main()
{
        ll n,k,m,d;
        cin >> n >> k >> m >> d;
        ll anser=0;
        for(ll i=1;i<=d;i++)
        {
                if(i!=1)
                {
                        ll now=n/(i-1);
                        if(now<k)
                        continue;
                }
                ll sum=n/(k*i-k+1);
                if(sum>m)
                {
                        if(m*i>=(n-n%m)/k)
                        sum=m;
                        else
                        continue;
                }
                anser=max(anser,sum*i);
        }
        cout<<anser<<endl;
        return 0;
}

D

这题的建模是一个网络流  第i个石头对每个[i+1,i+l]都有一条容量为a[i]的边 源点与左岸相连 汇点与右岸相连 算最大流

但其实可以用最大流最小割思想简化 因为你跳的次序并不会影响最后的答案 所以我们可以认定每次全部青蛙都在一个长度为L的窗口内

所以答案就是min(sum(ai~ai+l)) 即视连续L个石头为一个节点 前一个节点有指向后一个节点sum(ai~ai+l)的边 所以最大流是最小的那条边

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int a[100005];
int main()
{
        int n, l;
        cin >> n >> l;
        n--;
        int minn = INT_MAX;
        int sum = 0;
        for (int i = 0; i < n; i++)
        {
                cin >> a[i];
        }
        for (int i = 0; i < l; i++)
        {
                sum += a[i];
        }
        minn = sum;
        for (int i = l; i <= n - 1; i++)
        {
                sum -= a[i - l];
                sum += a[i];
                minn = min(minn, sum);
        }
        cout << minn << endl;
        return 0;
}

 E