Codeforces数据结构刷题

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最新推荐文章于 2021-02-03 22:31:12 发布

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分类专栏： 1900分数据结构

本文链接：https://blog.youkuaiyun.com/Zenith_Habitant/article/details/111143391

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本文解析四道Codeforces算法题，涉及线段树、RMQ查询、单调栈等数据结构及技巧，通过具体实现展示了不同问题的高效解决方案。

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题目链接：https://codeforces.com/problemset/problem/1439/C
解题思路：本题通过简单分析，发现一操作可以分解为线段树上二分和区间修改，而操作只需要暴力处理一下即可

const int N = 2e5 + 5;
int n , m;
ll a[N];
struct node
{
    int l , r;
    ll tag, maxn, minn;
    ll sum;
}tr[N << 2];
void push_up (int i)
{
    tr[i].minn = min (tr[i<<1].minn, tr[i<<1|1].minn);
    tr[i].maxn = max (tr[i<<1].maxn, tr[i<<1|1].maxn);
    tr[i].sum = tr[i<<1].sum + tr[i<<1|1].sum;
}
void build (int i , int l , int r)
{
    tr[i].l = l , tr[i].r = r;
    if (l == r)
    {
        tr[i].sum = tr[i].maxn = tr[i].minn = a[l];
        return ;
    }
    int mid = (l + r) >> 1;
    build (i<<1, l, mid);
    build (i<<1|1, mid + 1 , r);
    push_up (i);
}
void push_down (int i)
{
    if (tr[i].tag)
    {
        tr[i<<1].maxn = tr[i].tag;
        tr[i<<1].minn = tr[i].tag;
        tr[i<<1|1].maxn = tr[i].tag;
        tr[i<<1|1].minn = tr[i].tag;
        tr[i<<1].sum = 1ll * tr[i].tag * (tr[i<<1].r - tr[i<<1].l + 1);
        tr[i<<1|1].sum = 1ll * tr[i].tag * (tr[i<<1|1].r - tr[i<<1|1].l + 1);
        tr[i<<1].tag = tr[i<<1|1].tag = tr[i].tag;
        tr[i].tag = 0;
    }
}
int querypos (int i , int l ,int r, ll y)
{
    //cout << i << ' ' << l << ' ' << r << ' ' << tr[i].minn << endl;
    if (l == r)
        return l;
    int mid = (l + r) >> 1;
    push_down (i);
    if (tr[i<<1].minn <= y)
        return querypos (i<<1,l,mid,y);
    return querypos (i<<1|1,mid+1,r,y);

}
void update (int i , int l , int r, int L , int R, ll y)
{
    if (l >= L && r <= R)
    {
        tr[i].minn = tr[i].maxn = y;
        tr[i].tag = y;
        tr[i].sum = (r - l + 1) * y;
        return ;
    }
    push_down(i);
    int mid = (l + r) >> 1;
    if (L <= mid) update (i<<1,l,mid,L,R,y);
    if (R > mid) update (i<<1|1,mid+1,r,L,R,y);
    push_up(i);
}
ll getsum (int i , int l , int r, int L, int R)
{
    ll sum = 0;
    if (L > R)
        return 0;
    if (l >= L && r <= R)
        return tr[i].sum;
    push_down(i);
    int mid = (l + r) >> 1;
    if (L <= mid)
        sum += getsum(i<<1,l,mid,L,R);
    if (R > mid)
        sum += getsum(i<<1|1,mid+1,r,L,R);
    return sum;
}
ll cost;
int query (int i , int l, int r)
{
    //cout << i << ' ' << l << ' ' << r << ' ' << tr[i].sum << ' ' << cost << endl;
    int ans = 0;
    if (cost >= tr[i].sum)
    {
        cost -= tr[i].sum;
        return (tr[i].r - tr[i].l + 1);
    }
    push_down(i);
    int mid = (l + r) >> 1;
    if (tr[i<<1].minn <= cost)
        ans += query (i<<1, l, mid);
    if (tr[i<<1|1].minn <= cost)
        ans += query (i<<1|1,mid+1,r);
    return ans;
}
int main ()
{
    CLOSE;
    cin >> n >> m;
    for (int i = 1 ; i <= n ; i ++)
        cin >> a[i];
    build (1, 1, n);
    while (m --)
    {
        int op, x;
        ll y;
        cin >> op >> x >> y;
        if (op == 1) {
            int pos = querypos (1, 1, n, y);
            if (pos > x) continue;
            //cout << pos << endl;
            //cout << pos << endl;
            update(1, 1, n, pos, x, y);
        }
        else
        {
            cost = y;
            ll p = getsum(1, 1, n, 1, x - 1);
            cost += p;
            //cout << cost << endl;
            int ans = query (1, 1, n);
            ans -= (x-1);
            cout << ans << endl;
        }
    }
    return 0;
}

题目链接：https://codeforces.com/contest/1454/problem/F
解题思路：本题通过简单分析，通过RMQ查询区间最值，通过枚举x之后，发现区间的单调性从而使用两次二分求解

const int N = 2e5 + 5;
int n, T, a[N];
int maxn[N][30], minn[N][30];
void getST()
{
    for(int i = 1; i <= n; ++i)
        maxn[i][0] = minn[i][0] = a[i];
    for(int j = 1; (1 << j) <= n; ++j)
        for(int i = 1; i + (1 << j) - 1 <= n; ++i)
        {
            minn[i][j] = min(minn[i][j - 1], minn[i + (1 << (j - 1))][j - 1]);
            maxn[i][j] = max(maxn[i][j - 1], maxn[i + (1 << (j - 1))][j - 1]);
        }
}
int RMQ(int l, int r, int op)
{
    int k = log2(r - l + 1);
    if (op)
        return max(maxn[l][k], maxn[r-(1<<k)+1][k]);
    else
        return min(minn[l][k], minn[r-(1<<k)+1][k]);
}
int main ()
{
    read (T);
    while (T --)
    {
        cin >> n;
        for (int i = 1 ; i <= n ; i ++)
            scanf ("%d",&a[i]);
        getST();
        int mid1 = -1 , mid2 = -1;
        for(int i = 1 ; i <= n - 2 ; i ++)
        {
            int x = RMQ(1,i,1);
            int l = i + 1 , r = n - 1 , MID;
            while( r>=l )
            {
                MID = l+r>>1;
                int two = RMQ(i+1,MID,0);
                int three = RMQ(MID+1,n,1);
                if(two < x)	r = MID - 1;
                else if(two > x) l = MID + 1;
                else
                {
                    if (three < x) r = MID - 1;
                    else if (three > x)	l = MID + 1;
                    else { mid1 = i , mid2 = MID ; break; }
                }
            }
            if(mid1 != -1)
                break;
        }
        if(mid1 != -1)
            printf("YES\n%d %d %d\n",mid1,mid2-mid1,n-mid2);
        else
            printf("NO\n");
    }
}

题目链接：https://codeforces.com/problemset/problem/1446/D1
解题思路：本题有一个性质：答案一定包含众数，可以感性理解一下(扩张思维)，然后就可以枚举搞了，这里一定要用unordered_map不然会TLE

const int N = 2e5 + 5;
int n;
int a[N];
int buck[N], cnt, val;
unordered_map <int,int> ma;
int main ()
{
    CLOSE;
    cin >> n;
    int maxn = 0;
    for (int i = 1 ; i <= n ; i ++)
    {
        cin >> a[i];
        buck[a[i]] ++;
        if (buck[a[i]] > maxn)
            maxn = buck[a[i]], cnt = 1, val = a[i];
        else if (buck[a[i]] == maxn)
            cnt ++;
    }
    if (cnt > 1)
        cout << n << endl;
    else
    {
        int ans = 0;
        for (int i = 1 ; i <= 100 ; i ++)
        {
            if (i == val) continue;
            int v = 0;
            ma.clear();
            ma[0] = 0;
            for (int j = 1 ; j <= n ; j ++)
            {
                if (a[j] == i) v ++;
                if (a[j] == val) v --;
                if (!ma.count(v))
                    ma[v] = j;
                else
                    ans = max(ans, j - ma[v]);
            }
        }
        cout << ans << endl;
    }
}

题目链接：https://codeforces.com/contest/1442/problem/D
解题思路：正常来说，是很容易想到 $O(n^2k)$ 的分组背包，可惜时间不够 $A C$ 不得，于是我们观察题目，发现有一个很特殊的点，每个栈是单调栈。那么从网上的结论可知，选择的序列，仅存在一个选不满，然后，用线段树分治的策略搞一搞就好了

const int N = 3005;
int n , k;
int L[N];
vector <ll> sum[N];
ll dp[N], ans = 0;
void solve (int l , int r)
{
    if (l == r)
    {
        for (int i = 0 ; i <= min(L[l],k) ; i ++)
            ans = max (ans , dp[k - i] + sum[l][i]);
        return ;
    }
    int mid = (l + r) >> 1;
    ll cpy[N];
    for (int i = 0 ; i <= k ; i ++)
        cpy[i] = dp[i];
    for (int i = mid + 1 ; i <= r ; i ++)
    {
        for (int j = k ; j >= L[i] ; j --)
            dp[j] = max (dp[j], dp[j-L[i]] + sum[i][L[i]]);
    }
    solve (l , mid);
    for (int i = 0 ; i <= k ; i ++)
        dp[i] = cpy[i];
    for (int i = l ; i <= mid ; i ++)
    {
        for (int j = k ; j >= L[i] ; j --)
            dp[j] = max (dp[j], dp[j-L[i]] + sum[i][L[i]]);
    }
    solve (mid + 1, r);
}
int main ()
{
    CLOSE;
    cin >> n >> k;
    for (int i = 1 ; i <= n ; i ++)
    {
        cin >> L[i];
        sum[i].pb(0);
        for (int j = 1 ; j <= L[i] ; j ++)
        {
            int x;
            cin >> x;
            ll p = sum[i][j-1] + x;
            sum[i].pb(p);
        }
    }
    solve (1, n);
    cout << ans << endl;
}