2018-2019 ACM-ICPC Brazil Subregional Programming Contest

B. Marbles

image-20230916201212126

题解

  • 显然如果存在棋子位于\((x,x)\),那么一定先手必胜
  • 容易发现必败态位于\((1, 2)\)\((2,1)\),那么我们可以通过\(sg\)函数暴力打表得到
  • 并且玩家一定不会将棋子移动至\((0,i),(i,0),(i,i)\)这三种情况上,因为谁移动到这些位置,对手一定处于必胜态
int n, f[N][N];
pair<int, int> p[M];

int sg(int x, int y)
{
    if (f[x][y] != -1)
        return f[x][y];
    unordered_map<int, int> mp;
    for (int i = x - 1; i >= 1; --i)
    {
        if (i != y)
            mp[sg(i, y)]++;
    }
    for (int j = y - 1; j >= 1; --j)
        if (x != j)
            mp[sg(x, j)]++;
    for (int i = 1; i <= min(x - 1, y - 1); ++i)
        mp[sg(x - i, y - i)]++;
    for (int i = 0;; ++i)
        if (!mp.count(i))
            return f[x][y] = i;
}

void solve()
{
    cin >> n;
    for (int i = 1; i <= n; ++i)
        cin >> p[i].first >> p[i].second;
    memset(f, -1, sizeof f);
    f[1][2] = f[2][1] = 0;
    int ans = 0;
    for (int i = 1; i <= n; ++i)
    {
        auto [x, y] = p[i];
        ans ^= sg(x, y);
        if (x == y)
        {
            cout << "Y" << endl;
            return;
        }
    }
    if (ans > 0)
        cout << "Y" << endl;
    else
        cout << "N" << endl;
}

F. Music Festival

\(n\)个舞台,每个舞台上都有节目,所有节目个数之和不超过\(1000\),每个节目都有开始时间和结束时间以及喜爱程度,要求你挑选一些节目观看,使得节目之间时间不冲突,并且在所有舞台上都看了节目的条件下,能够得到的最多喜爱程度

\(1 \leq n \leq 10\)

题解:状压\(DP\)

  • 我们设计状态\(dp[i][j]\)代表舞台状态为\(i\)时,在时间点的\(j\)的时刻下,能够得到的最多喜爱程度

  • 这是一个经典的转移问题,一个节目的右端点肯定是通过其左端点转移过来的,所以我们考虑将左端点挂在右端点的\(vector\)

  • 考虑状态转移:

\[dp[i][r] = dp[i \oplus (1 \ll id) ][l] + v,id是节目[l,r]的舞台编号\\ dp[i][r] = dp[i][l] + v \]

  • 初始状态:\(dp[i][j] = -INF,dp[0][0] = 0\)
int n;
int dp[(1ll << 11)][N];
vector<array<int, 3>> seg[N];

void solve()
{
    cin >> n;
    int mx = 0;
    for (int i = 0; i < n; ++i)
    {
        int k;
        cin >> k;
        for (int j = 1; j <= k; ++j)
        {
            int l, r, v;
            cin >> l >> r >> v;
            mx = max(mx, r);
            seg[r].push_back({l, v, i});
        }
    }
    for (int i = 0; i < (1ll << n); ++i)
        for (int j = 0; j <= mx; ++j)
            dp[i][j] = -inf;
    for (int j = 0; j <= mx; ++j)
        dp[0][j] = 0;
    for (int i = 0; i < (1ll << n); ++i)
    {
        for (int j = 1; j <= mx; ++j)
        {
            dp[i][j] = dp[i][j - 1];
            for (auto [l, v, id] : seg[j])
            {
                if (i >> id & 1)
                {
                    dp[i][j] = max(dp[i][j], dp[i][l] + v);
                    dp[i][j] = max(dp[i][j], dp[i ^ (1ll << id)][l] + v);
                }
            }
        }
    }
    int ans = -inf;
    for (int j = 0; j <= mx; ++j)
        ans = max(ans, dp[(1ll << n) - 1][j]);
    if (ans <= 0)
        cout << -1 << endl;
    else
        cout << ans << endl;
}

H. Police Hypothesis

给定一颗树,树上每个节点都有字母,给定模式串\(P\),存在两个操作:

  • \(1 , u, v\):询问从\(u\)\(v\)的路径中\(P\)出现的次数
  • \(2, u, ch\):将\(u\)上的字母改成\(ch\)

\(1 \leq |P| \leq 100\)

题解:树链剖分 + 线段树维护字符串哈希

  • 由于\(P\)的长度比较小,我们可以在线段树上维护答案\(ans\),区间长度\(len\),以及长度为\(100\)的字符串哈希前后缀数组\(pre[],suf[]\)

  • 考虑区间合并:

\(ans:=lson.ans+rson.ans + (lson.suf[]和rson.pre[]之间产生的贡献)\)

\(pre[] := lson.pre[] + rson.pre[]\),通过字符串哈希合并

\(suf[]:=rson.suf[] + lson.suf[]\),通过字符串哈希合并,注意应该倒着合并

\(len:=lson.len + rson.len\)

  • 但是询问存在方向性,所以我们需要再开一颗线段树维护反串,合并时右儿子合并左儿子即可

  • 对于一次询问来说,因为存在方向性,所以我们可以维护\(u\)\(v\)\(lca\)的左侧和右侧两部分的串,最后将这两部分合并

  • 复杂度:\(O(100nlog^2n)\)

int n, m, q, sz[N], hson[N], dep[N], fa[N], l[N], mp[N], idx, top[N], pw[N], base = 131, hash_P;
string P;
char a[N];
vector<int> g[N];

void dfs1(int u, int par)
{
    sz[u] = 1;
    hson[u] = -1;
    fa[u] = par;
    dep[u] = dep[par] + 1;
    for (auto v : g[u])
    {
        if (v == par)
            continue;
        dfs1(v, u);
        sz[u] += sz[v];
        if (hson[u] == -1 || sz[v] > sz[hson[u]])
            hson[u] = v;
    }
}
void dfs2(int u, int head)
{
    top[u] = head;
    l[u] = ++idx;
    mp[idx] = u;
    if (hson[u] != -1) 
        dfs2(hson[u], head);
    for (auto v : g[u])
    {
        if (v != fa[u] && v != hson[u])
            dfs2(v, v);
    }
}

struct info
{
    int ans, len, pre[101], suf[101];
    friend info operator+(const info &a, const info &b)
    {
        if (a.len == 0)
            return b;
        if (b.len == 0)
            return a;
        info c;
        c.len = a.len + b.len;
        c.ans = a.ans + b.ans;
        c.pre[0] = 0;
        for (int i = 1; i <= min(100ll, c.len); ++i)
        {
            if (i <= a.len)
                c.pre[i] = a.pre[i];
            else
                c.pre[i] = (a.pre[a.len] * pw[i - a.len] % mod + b.pre[i - a.len]) % mod;
            if (i <= b.len)
                c.suf[i] = b.suf[i];
            else
                c.suf[i] = (a.suf[i - b.len] * pw[b.len] % mod + b.suf[b.len]) % mod;
        }
        for (int i = max(1ll, m - b.len); i <= a.len && m - i >= 1; ++i)
            if ((a.suf[i] * pw[m - i] % mod + b.pre[m - i]) % mod == hash_P)
                ++c.ans;
        return c;
    }
};
struct SEG
{
    info val;
} seg[2][N << 2];
void up(int id)
{
    seg[0][id].val = seg[0][lson].val + seg[0][rson].val;
    seg[1][id].val = seg[1][rson].val + seg[1][lson].val;
}
void build(int id, int l, int r)
{
    if (l == r)
    {
        if (m == 1 && a[mp[l]] == P[0])
        {
            seg[0][id].val.ans = 1;
            seg[1][id].val.ans = 1;
        }
        else
        {
            seg[0][id].val.ans = 0;
            seg[1][id].val.ans = 0;
        }
        seg[0][id].val.len = seg[1][id].val.len = 1;
        seg[0][id].val.pre[1] = seg[1][id].val.pre[1] = a[mp[l]];
        seg[0][id].val.suf[1] = seg[1][id].val.suf[1] = a[mp[l]];
        return;
    }
    int mid = l + r >> 1;
    build(lson, l, mid);
    build(rson, mid + 1, r);
    up(id);
}

void change(int id, int l, int r, int x, char val)
{
    if (l == r)
    {
        a[mp[l]] = val;
        if (m == 1 && a[mp[l]] == P[0])
        {
            seg[0][id].val.ans = 1;
            seg[1][id].val.ans = 1;
        }
        else
        {
            seg[0][id].val.ans = 0;
            seg[1][id].val.ans = 0;
        }
        seg[0][id].val.pre[1] = seg[1][id].val.pre[1] = a[mp[l]];
        seg[0][id].val.suf[1] = seg[1][id].val.suf[1] = a[mp[l]];
        return;
    }
    int mid = l + r >> 1;
    if (x <= mid)
        change(lson, l, mid, x, val);
    else
        change(rson, mid + 1, r, x, val);
    up(id);
}

info query(int id, int l, int r, int ql, int qr, int op)
{
    if (ql <= l && r <= qr)
        return seg[op][id].val;
    int mid = l + r >> 1;
    if (qr <= mid)
        return query(lson, l, mid, ql, qr, op);
    else if (ql > mid)
        return query(rson, mid + 1, r, ql, qr, op);
    else if (op == 0)
        return query(lson, l, mid, ql, qr, op) + query(rson, mid + 1, r, ql, qr, op);
    else
        return query(rson, mid + 1, r, ql, qr, op) + query(lson, l, mid, ql, qr, op);
}

// u --> v
int query(int u, int v)
{
    info L, R, M;
    L.ans = L.len = R.ans = R.len = M.ans = M.len = 0;
    while (top[u] != top[v])
    {
        if (dep[top[u]] > dep[top[v]])
        {
            L = L + query(1, 1, n, l[top[u]], l[u], 1);
            u = fa[top[u]];
        }
        else
        {
            R = query(1, 1, n, l[top[v]], l[v], 0) + R;
            v = fa[top[v]];
        }
    }
    if (dep[u] < dep[v])
        M = query(1, 1, n, l[u], l[v], 0);
    else
        M = query(1, 1, n, l[v], l[u], 1);
    return (L + M + R).ans;
}

void solve()
{
    cin >> n >> q >> P;
    m = (int)P.size();
    for (auto ch : P)
        hash_P = (hash_P * base % mod + ch) % mod;
    pw[0] = 1ll;
    for (int i = 1; i <= n; ++i)
    {
        cin >> a[i];
        pw[i] = pw[i - 1] * base % mod;
    }
    for (int i = 1, u, v; i < n; ++i)
    {
        cin >> u >> v;
        g[u].push_back(v);
        g[v].push_back(u);
    }
    dfs1(1, 0);
    dfs2(1, 1);
    build(1, 1, n);
    while (q--)
    {
        int op, u, v;
        char c;
        cin >> op;
        if (op == 1)
        {
            cin >> u >> v;
            cout << query(u, v) << endl;
        }
        else
        {
            cin >> u >> c;
            change(1, 1, n, l[u], c);
        }
    }
}
posted @ 2023-09-16 20:50  Zeoy_kkk  阅读(26)  评论(0编辑  收藏  举报