AtCoder Beginner Contest 199 游记（AB水题，C字符串操作，D搜索，E状压）

A - Square Inequality

水题

B - Intersection

水题，就是找公共区间，维护一下 Lmax，Rmin即可

void solve() {
    int n, a, b;
    int maxa = -1, minb = 0x3f3f3f3f;
    cin >> n;
    for (int i = 0; i < n; ++i) {
        cin >> a;
        maxa = max(maxa, a);
    }
    for (int i = 0; i < n; ++i) {
        cin >> b;
        minb = min(minb, b);
    }
    cout << (minb - maxa + 1 > 0 ? minb - maxa + 1 : 0);
}

C - IPFL

交换两个或者把前一半和后一半交换，求 次变换后的结果。

模拟会T，把字符串前后两个分开存，然后进行操作

void solve() {
    string s, x, y;
    int n, q, a, b, op;
    cin >> n >> s >> q;
    x = s.substr(0, n), y = s.substr(n, n);
    while (q--) {
        cin >> op >> a >> b;
        if (op == 2) swap(x, y);
        else {
            a--, b--;
            if (a > b) swap(a, b);
            if (b < n) swap(x[a], x[b]);
            else if (a >= n)
                swap(y[a - n], y[b - n]);
            else
                swap(x[a], y[b - n]);
        }
    }
    cout << x << y;
}

D - RGB Coloring 2

个点 条边的无向图，求用三种颜色染色后每条边相连的点不同的图有几种。

一看数据范围显然是一个搜索题，不过直接搜索会有一些问题，因为先搜到哪个的不同可能图染色后是相同的，会造成重复，既然如此，那我们随便指定一个顺序就可以了。

using ll    = long long;
const int N = 30;
vector<int> e[N], a;
int col[N], v[N][4];
ll ans = 1, cur;
bool vis[N];
void dfs0(int x) {
    a.push_back(x);
    vis[x] = true;
    for (int &y : e[x])
        if (!vis[y]) dfs0(y);
}
void dfs(int x) {
    if (x == a.size()) {
        cur++;
        return;
    }
    for (int i = 1; i <= 3; ++i) {
        if (!v[a[x]][i]) {
            for (int &y : e[a[x]])
                if (!v[y][i]) v[y][i] = a[x];
            dfs(x + 1);
            for (int &y : e[a[x]])
                if (v[y][i] == a[x]) v[y][i] = 0;
        }
    }
}
void solve() {
    int n, m;
    cin >> n >> m;
    for (int i = 0, u, v; i < m; ++i) {
        cin >> u >> v;
        e[u].push_back(v);
        e[v].push_back(u);
    }
    for (int i = 1; i <= n; ++i) {
        if (!vis[i]) {
            cur = 0;
            a.resize(0);
            dfs0(i);
            for (int &y : e[i])
                if (!v[y][1]) v[y][1] = i;
            dfs(1);
            ans = ans * cur * 3;
        }
    }
    cout << ans << "\n";
}

E - Permutation 状压DP

借用一下 Acfboy 的思路

// Murabito-B 21/04/26
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N = 20;
struct node {
    int x, y, z;
    bool operator<(node b) const { return x < b.x; }
} a[N];
int n, m, L[N], R[N], g[N], f[(1 << 19) + 5];
bool flag[(1 << 19) + 5];
int popcount(int x) { return x == 0 ? 0 : popcount(x & (x - 1)) + 1; }
void solve() {
    cin >> n >> m;
    for (int i = 1; i <= m; ++i) cin >> a[i].x >> a[i].y >> a[i].z;
    sort(a + 1, a + 1 + m);
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= m; ++j) {
            if (a[j].x == i) {
                L[i] = j;
                break;
            }
        }
        for (int j = m; j > 0; --j) {
            if (a[j].x == i) {
                R[i] = j;
                break;
            }
        }
    }
    for (int S = 1; S < (1 << n); ++S) {
        flag[S] = true;
        int k   = popcount(S);
        if (L[k] == 0) continue;
        for (int i = 1; i <= n; ++i) g[i] = 0;
        for (int i = 1, j = 1; i <= k; ++i) {
            while ((S & (1 << j)) == 0) j++;
            for (int l = j; l <= n; ++l) g[l]++;
            j++;
        }
        for (int i = L[k]; i <= R[k]; ++i)
            flag[S] &= (g[a[i].y] <= a[i].z);
    }
    f[0] = 1;
    for (int S = 0; S < (1 << n); ++S)
        for (int i = 0; i <= 18; ++i)
            if (((S & (1 << i)) == 0) && flag[S | (1 << i)]) f[S | (1 << i)] += f[S];
    cout << f[(1 << n) - 1];
}
signed main() {
    ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    solve();
}