Codeforces Round #627 (Div. 3)

题库链接

A. Yet Another Tetris Problem

俄罗斯方块，\(2 \times 1\)的方块，显然只有高度差是偶数时才能消除全部

#pragma GCC optimize(2)
#include <bits/stdc++.h>
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
#define pii pair<int,int>
#define gcd __gcd
const int inf = 0x3f3f3f3f;
const int maxn = 201110;
const int M = 1e9+7;
int n,m,k,ok;

int a[maxn];

signed main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("data.in", "r", stdin);
#endif
    int T;
    cin>>T;
    while(T--)
    {
        cin>>n;
        for(int i = 1; i <= n; i++) 
        {
            cin>>a[i];
        }
        sort(a+1,a+1+n);
        ok = 1;
        for(int i = 2; i <= n; i++) 
        {
            if((a[i]-a[i-1])%2) ok = 0;
        }
        if(ok) puts("YES");
        else puts("NO");
    }
    return 0;
}

B. Yet Another Palindrome Problem

显然两个相等的数中间还有数，那么一定符合条件，暴力枚举就行了

#pragma GCC optimize(2)
#include <bits/stdc++.h>
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
#define pii pair<int,int>
#define int long long
#define gcd __gcd
const int inf = 0x3f3f3f3f;
const int maxn = 201110;
const int M = 1e9+7;
int n,m,k,ok;

int a[maxn];

signed main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("data.in", "r", stdin);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    int T;
    cin>>T;
    while(T--)
    {
        cin>>n;
        for(int i = 1; i <= n; i++) 
        {
            cin>>a[i];
        }
        ok = 0;
        for(int i = 1; i <= n && !ok; i++) 
        {
            for(int j = i+2; j <= n; j++)
            {
                if(a[i] == a[j]) ok = 1;
            }
        }
        if(ok) puts("YES");
        else puts("NO");
    }
    return 0;
}

C. Frog Jumps

有一段Ｌ，显然跳跃长度必须大于这一段的长度才能跳过去

#pragma GCC optimize(2)
#include <bits/stdc++.h>
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
#define pii pair<int,int>
#define int long long
#define gcd __gcd
const int inf = 0x3f3f3f3f;
const int maxn = 201110;
const int M = 1e9+7;
int n,m,k,ok;

signed main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("data.in", "r", stdin);
#endif
    int T;
    cin>>T;
    while(T--)
    {
        string s;
        cin>>s;
        n = s.size();
        int ans = 0, i = 0,res = 0;
        while(i < n)
        {
            if(s[i] == 'L')
            {
                res++;
            }
            else
            {
                ans = max(ans,res);
                res = 0;
            }
            i++;
        }
        ans = max(ans,res);
        cout<<ans+1<<endl;
    }
    return 0;
}

D. Pair of Topics

\(a_i + a_j > b_i + b_j\)，移项，\((a_i - b_i) + (a_j - b_j) > 0\)
那么对于每个\(i\)，我们只要找大于\((-(a_i-b_i))\)的\(j\)的个数就好了
排序之后二分，复杂度\(O(nlogn)\)

#pragma GCC optimize(2)
#include <bits/stdc++.h>
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
#define pii pair<int,int>
#define int long long
#define gcd __gcd
const int inf = 0x3f3f3f3f;
const int maxn = 201110;
const int M = 1e9+7;
int n,m,k,ok;

int a[maxn];

signed main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("data.in", "r", stdin);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin>>n;
    for(int i = 1; i <= n; i++) 
    {
        cin>>a[i];
    }
    for(int i = 1,x; i <= n; i++) 
    {
        cin>>x;
        a[i] -= x;
    }
    sort(a+1,a+1+n);
    int ans = 0;
    for(int i = 1; i <= n; i++) 
    {
        int pos = upper_bound(a+1+i,a+1+n,-a[i])-a;
        ans += (n+1-pos);
    }
    cout<<ans<<endl;
    return 0;
}

E. Sleeping Schedule

定义\(dp[i][j]\)表示第\(i\)次在第\(j\)小时睡觉，所达到的最大睡眠次数
复杂度\(O(n^2)\)

#pragma GCC optimize(2)
#include <bits/stdc++.h>
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
#define pii pair<int,int>
#define gcd __gcd
const int inf = 0x3f3f3f3f;
const int maxn = 2110;
const int M = 1e9+7;
int n,h,l,r;

int dp[maxn][maxn];

int a[maxn];

signed main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("data.in", "r", stdin);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin>>n>>h>>l>>r;
    for(int i = 1; i <= n; i++) 
    {
        cin>>a[i];
    }
    mem(dp,-inf);dp[0][0] = 0;
    for(int i = 1; i <= n; i++) 
    {
        for(int j = 0; j < h; j++) 
        {
            int k = (j+a[i])%h;
            dp[i][k] = max(dp[i][k],dp[i-1][j]+(k >= l && k <= r));
            k = (j+a[i]-1+h)%h;
            dp[i][k] = max(dp[i][k],dp[i-1][j]+(k >= l && k <= r));
        }
    }
    int ans = 0;
    for(int i = 0; i <= h; i++) 
    {
        ans = max(ans,dp[n][i]);
    }
    cout<<ans<<endl;
    return 0;
}

F. Maximum White Subtree

先把无根树看成有根树，对于每个点，我们把答案分成两部分，以\(i\)为根的子树部分，和除去\(i\)的子树的剩余部分
以\(i\)为根的子树部分可以利用树型dp求解

#pragma GCC optimize(2)
#include <bits/stdc++.h>
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
#define pii pair<int,int>
#define gcd __gcd
const int inf = 0x3f3f3f3f;
const int maxn = 401110;
const int M = 1e9+7;
int n,m,k,ok;


int head[maxn],to[maxn],Next[maxn],cnt = 2;

void add(int u,int v)
{
    to[cnt] = v;Next[cnt] = head[u];head[u] = cnt;cnt++;
}

int a[maxn];
int f[maxn];
int h[maxn];

void dfs(int u,int pre)
{   
    if(a[u] == 1) f[u] = 1;
    else f[u] = -1;
    for(int i = head[u]; i ; i = Next[i])
    {
        int v = to[i];
        if(v == pre) continue;
        dfs(v,u);
        f[u] += max(f[v],0);        //选与不选
    }
}

void dfs2(int u,int pre)
{
    int sum = max(0,h[u]);
    if(a[u]) sum++;
    else sum--;
    for(int i = head[u]; i ; i = Next[i])
    {
        int v = to[i];
        if(v == pre) continue;
        sum += max(f[v],0);
    }
    for(int i = head[u]; i ; i = Next[i])
    {
        int v = to[i];
        if(v == pre) continue;
        h[v] = max(0,sum-max(f[v],0));
    }
    for(int i = head[u]; i ; i = Next[i])
    {
        int v = to[i];
        if(v == pre) continue;
        dfs2(v,u);
    }
}


signed main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("data.in", "r", stdin);
#endif
    cin>>n;
    for(int i = 1; i <= n; i++) 
    {
        cin>>a[i];
    }
    for(int i = 1,x,y; i < n; i++) 
    {
        cin>>x>>y;
        add(x,y);add(y,x);
    }
    dfs(1,0);
    dfs2(1,0);
    for(int i = 1; i <= n; i++) 
    {
        cout<<f[i]+h[i]<<' ';
    }
    return 0;
}

posted @ 2020-03-13 10:45 hezongdnf 阅读(74) 评论(0) 收藏举报

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