ICPC Latin American Regional Contests 2019 EGIKLM

E (尺取法)

#include <bits/stdc++.h>
#define ll long long
using namespace std;

const int maxn = 2e5 + 5;

int n;
char s[maxn];

int main()
{
    cin >> s + 1 >> n;
    int len = strlen(s + 1);
    for (int i = 1; i <= len; ++i) s[i + len] = s[i];
    ll ans = 0;
    for (int i = 1, x = 0, j = 0; i <= len; ++i)
    {
        while (!x && j + 1 <= (len << 1))
            if (s[++j] == 'E') ++x;
        if (x) ans += max(0, n - j + i);
        if (s[i] == 'E') --x;
    }
    cout << ans;
    return 0;
}

　　

G(后缀自动机)

#include <bits/stdc++.h>
using namespace std;

const int maxn = 4e5 + 10;

struct node
{
    int fa, len;
    map<char, int> nxt;
}tr[maxn];

int sz, las;

void init()
{
    sz = las = 1;
    tr[1].len = tr[1].fa = 0;
    map<char, int>().swap(tr[1].nxt);
}

void add(char c)
{
    int p = las; int cur = las = ++sz;
    tr[cur].len = tr[p].len + 1;
    for (; p && !tr[p].nxt.count(c); p = tr[p].fa)
        tr[p].nxt[c] = cur;
    if (p == 0) { tr[cur].fa = 1; return; }
    int q = tr[p].nxt[c];
    if (tr[q].len == tr[p].len + 1) { tr[cur].fa = q; return; }
    int nq = ++sz; tr[nq] = tr[q];
    tr[nq].len = tr[p].len + 1;
    for (; p && tr[p].nxt[c] == q; p = tr[p].fa)
        tr[p].nxt[c] = nq;
    tr[q].fa = tr[cur].fa = nq;
}

int dp[maxn], n;
char s[maxn >> 1];

void solve()
{
    int lenx = 1, p = 1, tmp =  0;
    for(int& i = lenx; s[i]; ++i)
    {
        if (tr[p].nxt.count(s[i])) p = tr[p].nxt[s[i]], ++tmp; 
        else 
        {
            while (!tr[p].nxt.count(s[i]) && p) p = tr[p].fa;
            if(p == 0) p = 1, tmp = 0;
            else tmp = tr[p].len + 1, p = tr[p].nxt[s[i]];
        }
        if(!tmp) { puts("-1"); return; }
        dp[i] = dp[i - tmp] + 1;
    }--lenx;
    printf("%d
", dp[lenx]);
}

int main()
{
    scanf("%s%d", s + 1, &n); init();
    for (int i = 1; s[i]; ++i) add(s[i]);
    for (int i = 1; i <= n; ++i)
        scanf("%s", s + 1), solve();
    return 0;
}

　　

I(记忆化搜索)

#include <bits/stdc++.h>
#define ll long long
using namespace std;

int n, m, a[1005][2005];
int v[2005], mod = 1e9 + 7;
ll ans2, ans[1005];

ll work(int k)
{
    if (v[k]) { return ans[k];}
    v[k] = 1;
    for (int i = 1; i <= a[k][0]; ++i)
    {
        if (a[k][i] <= m) ans[k] = (ans[k] + work(a[k][i])) % mod;
        else 
        {
            if (!v[a[k][i]]) ++ans2, v[a[k][i]] = 1;
            ans[k] = (ans[k] + 1) % mod;
        }
    }
    return ans[k];
}

int main()
{
    cin >> n >> m;
    for (int i = 1; i <= m; ++i)
    {
        cin >> a[i][0];
        for (int j = 1; j <= a[i][0]; ++j) cin >> a[i][j];
    }
    work(1);
    cout << ans[1] << ' ' << ans2;
    return 0;
}

　　

K(给定方程的零点公式，拆括号号，复杂度够，直接暴力拆)

#include <bits/stdc++.h>
#define ll long long
using namespace std;

const int maxn = 1e4 + 5;

char s[maxn];
int tot;
ll a[maxn], b[maxn];

int main()
{
    cin >> s + 1; b[0] = 1;
    for (int i = 2; s[i]; ++i)
        if (s[i] != s[i - 1]) a[++tot] -= (i - 1 << 1) + 1;
    for (int i = 1; i <= tot; ++i)
    {
        b[i] = b[i - 1];
        for (int j = i - 1; j; --j)
            b[j] =  b[j - 1] + b[j] * a[i];  
        b[0] *= a[i];
    }
    int flag = 1;
    if (s[1] == 'H' && tot & 1) flag = -1;
    else if (s[1] == 'A' && tot % 2 == 0) flag = -1;
    printf("%d
%lld", tot, flag * b[tot]);
    for (int i = tot - 1; i >= 0; --i) printf(" %lld", flag * b[i]);
    return 0;
}

　　

L

#include <bits/stdc++.h>
#define ll long long
#define P pair<int, int>
using namespace std;

int n, m, a, b[1005][1005], ans;

inline int read()
{
    char c = getchar();
    while (c != 'G' && c != 'B') c = getchar();
    return c == 'G';
}

void init()
{
    for (int i = 1; i <= n; ++i)
    {
        a = read(); b[i][1] = 1;
        for (int j = 2, c; j <= m; ++j, a = c)
        {
            c = read();
            if (a == c) b[i][j] = b[i][j - 1] + 1;
            else b[i][j] = 1;
        }
    }
}

void work(stack<P> &st, int k, int j)
{
    int w = 1;
    for (P p = st.top(); b[k][j] < b[p.second][j]; st.pop(), p = st.top())
    {
        int c = min(p.first + k - p.second - 1, b[p.second][j]);
        ans = max(ans, c * c); w += p.first;
    }
    st.push({ w, k });
}

int main()
{
    cin >> n >> m; init();
    for (int i = m; i; --i)
    {
        stack<P> st; st.push({0, 0});
        for (int j = 1; j <= n + 1; ++j) work(st, j, i);
    }
    cout << ans;
    return 0;
}

　　

M（签到题）

#include <bits/stdc++.h>
using namespace std;

const int maxn = 1005;

int n, a[maxn], x, d[maxn], mx;

int main()
{
    cin >> n >> x;
    for (int i = 1; i <= n; ++i) cin >> a[i];
    for (int i = 1; i <= n; ++i)
    {
        d[i] = 1; int ls = a[i];
        for (int j = i + 1; j <= n; ++j)
            if (a[j] - ls <= x) ls = a[j], ++d[i];
        mx = max(mx, d[i]);
    }
    cout << mx;
    return 0;
}