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AtCoder Beginner Contest 375（A，B，C，D，E，F）（大模拟，前缀和，dp，离线处理，Floyd）

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AtCoder Beginner Contest 375

A题

代码
#pragma GCC optimize("O2")
#pragma GCC optimize("O3")
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N = 2e5 + 5, M = 1e6 + 5;
const int inf = 0x3f3f3f3f3f3f3f3f;
int n;
char s[N];
void solve()
{
    cin >> n;
    for (int i = 1; i <= n; i++)
    {
        cin >> s[i];
    }
    int ans = 0;
    for (int i = 1; i <= n - 2; i++)
    {
        if (s[i] == '#' && s[i + 1] == '.' && s[i + 2] == '#')
            ans++;
    }
    cout << ans << endl;
}

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    int test = 1;
    // cin >> test;
    for (int i = 1; i <= test; i++)
    {
        solve();
    }
    return 0;
}

B题

代码
#pragma GCC optimize("O2")
#pragma GCC optimize("O3")
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N = 2e5 + 5, M = 1e6 + 5;
const int inf = 0x3f3f3f3f3f3f3f3f;
int n;
struct Point
{ // 点
    double x, y;
    Point() {}
    Point(double x, double y) : x(x), y(y) {}

    Point operator+(Point B) { return Point(x + B.x, y + B.y); }
    Point operator-(Point B) { return Point(x - B.x, y - B.y); }
} p[N];
double Distance(Point A, Point B)
{ // 两点之间的欧氏距离
    return sqrtl((A.x - B.x) * (A.x - B.x) + (A.y - B.y) * (A.y - B.y));
}
void solve()
{
    cin >> n;
    p[0] = {0, 0};
    for (int i = 1; i <= n; i++)
    {
        cin >> p[i].x >> p[i].y;
    }
    double ans = 0;
    n++;
    for (int i = 0; i < n; i++)
    {
        int j = (i + 1) % n;
        ans += Distance(p[i], p[j]);
    }
    cout << fixed << setprecision(8) << ans << endl;
}

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    int test = 1;
    // cin >> test;
    for (int i = 1; i <= test; i++)
    {
        solve();
    }
    return 0;
}

C题

思路

其实就是从外圈到内圈的局部旋转，注意取模，每个圈最多旋转3次。

代码
#pragma GCC optimize("O2")
#pragma GCC optimize("O3")
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N = 3e3 + 5, M = 1e6 + 5;
const int inf = 0x3f3f3f3f3f3f3f3f;
int n;
char a[N][N], b[N][N];
void solve()
{
    cin >> n;
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= n; j++)
        {
            cin >> a[i][j];
        }
    }
    for (int k = 1; k <= n / 2; k++)
    {
        int num = k % 4;
        vector<char> v[4];
        for (int j = k + 1; j <= n - k; j++)
        {
            v[0].push_back(a[k][j]);
        }
        for (int i = k + 1; i <= n - k; i++)
        {
            v[1].push_back(a[i][n - k + 1]);
        }
        for (int j = n - k; j >= k + 1; j--)
        {
            v[2].push_back(a[n - k + 1][j]);
        }
        for (int i = n - k; i >= k + 1; i--)
        {
            v[3].push_back(a[i][k]);
        }
        vector<char> op[4];
        for (int i = 0; i < 4; i++)
        {
            op[(i + num) % 4] = v[i];
        }
        int idx = 0;
        for (int j = k + 1; j <= n - k; j++)
        {
            b[k][j] = op[0][idx++];
        }
        idx = 0;
        for (int i = k + 1; i <= n - k; i++)
        {
            b[i][n - k + 1] = op[1][idx++];
        }
        idx = 0;
        for (int j = n - k; j >= k + 1; j--)
        {
            b[n - k + 1][j] = op[2][idx++];
        }
        idx = 0;
        for (int i = n - k; i >= k + 1; i--)
        {
            b[i][k] = op[3][idx++];
        }

        vector<int> ji;
        ji.push_back(a[k][k]);
        ji.push_back(a[k][n - k + 1]);
        ji.push_back(a[n - k + 1][n - k + 1]);
        ji.push_back(a[n - k + 1][k]);

        vector<int> kk(4);
        for (int i = 0; i < 4; i++)
        {
            kk[(i + num) % 4] = ji[i];
        }

        b[k][k] = kk[0];
        b[k][n - k + 1] = kk[1];
        b[n - k + 1][n - k + 1] = kk[2];
        b[n - k + 1][k] = kk[3];
    }
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= n; j++)
        {
            cout << b[i][j];
        }
        cout << endl;
    }
}

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    int test = 1;
    // cin >> test;
    for (int i = 1; i <= test; i++)
    {
        solve();
    }
    return 0;
}

D题

思路

预处理出每一个字符的前缀和与后缀和，最后遍历一遍直接计算即可。

代码
#pragma GCC optimize("O2")
#pragma GCC optimize("O3")
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N = 2e5 + 5, M = 1e6 + 5;
const int inf = 0x3f3f3f3f3f3f3f3f;
int n;
int pre[N][26], nxt[N][26];
string s;
void solve()
{
    cin >> s;
    n = s.size();
    s = "#" + s;
    for (int i = 1, j = n; i <= n; i++, j--)
    {
        int op1 = s[i] - 'A';
        int op2 = s[j] - 'A';
        for (int k = 0; k < 26; k++)
        {
            pre[i][k] = pre[i - 1][k];
            nxt[j][k] = nxt[j + 1][k];
        }
        pre[i][op1]++, nxt[j][op2]++;
    }
    int ans = 0;
    for (int i = 2; i < n; i++)
    {
        for (int k = 0; k < 26; k++)
        {
            int low = pre[i - 1][k];
            int high = nxt[i + 1][k];
            ans += (low * high);
        }
    }
    cout << ans << endl;
}

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    int test = 1;
    // cin >> test;
    for (int i = 1; i <= test; i++)
    {
        solve();
    }
    return 0;
}

E题

思路

注意到， n n n最大为100，强度的和最大值为1500。因为有三组，所以每一组强度的和最大值为500。

我们令 d p [ i ] [ j ] [ k ] dp[i][j][k] dp[i][j][k]表示前 i i i个已经分配完成，第 1 1 1组的强度为 j j j，第 2 2 2组的强度为 k k k时需要执行换队操作的最小数量。

状态转移方程为：

d p [ i + 1 ] [ j + b [ i + 1 ] ] [ k ] = m i n ( d p [ i + 1 ] [ j + b [ i + 1 ] ] [ k ] , d p [ i ] [ j ] [ k ] + ( a [ i + 1 ] ! = 1 ) ) dp[i + 1][j + b[i + 1]][k] = min(dp[i + 1][j + b[i + 1]][k], dp[i][j][k] + (a[i + 1] != 1)) dp[i+1][j+b[i+1]][k]=min(dp[i+1][j+b[i+1]][k],dp[i][j][k]+(a[i+1]!=1))

d p [ i + 1 ] [ j ] [ k + b [ i + 1 ] ] = m i n ( d p [ i + 1 ] [ j ] [ k + b [ i + 1 ] ] , d p [ i ] [ j ] [ k ] + ( a [ i + 1 ] ! = 2 ) ) dp[i + 1][j][k + b[i + 1]] = min(dp[i + 1][j][k + b[i + 1]], dp[i][j][k] + (a[i + 1] != 2)) dp[i+1][j][k+b[i+1]]=min(dp[i+1][j][k+b[i+1]],dp[i][j][k]+(a[i+1]!=2))

d p [ i + 1 ] [ j ] [ k ] = m i n ( d p [ i + 1 ] [ j ] [ k ] , d p [ i ] [ j ] [ k ] + ( a [ i + 1 ] ! = 3 ) ) dp[i + 1][j][k] = min(dp[i + 1][j][k], dp[i][j][k] + (a[i + 1] != 3)) dp[i+1][j][k]=min(dp[i+1][j][k],dp[i][j][k]+(a[i+1]!=3))

代码
#pragma GCC optimize("O2")
#pragma GCC optimize("O3")
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N = 1e2 + 5, M = 5e2 + 5;
const int inf = 0x3f3f3f3f3f3f3f3f;
int n;
int a[N], b[N], dp[N][M][M];
void solve()
{
    cin >> n;
    int sum = 0;
    for (int i = 1; i <= n; i++)
    {
        cin >> a[i] >> b[i];
        sum += b[i];
    }
    if (sum % 3 != 0)
    {
        cout << -1 << endl;
        return;
    }
    memset(dp, inf, sizeof dp);
    dp[0][0][0] = 0, sum /= 3;
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j <= sum; j++)
        {
            for (int k = 0; k <= sum; k++)
            {
                if (dp[i][j][k] == inf)
                    continue;
                if (j + b[i + 1] <= sum)
                    dp[i + 1][j + b[i + 1]][k] = min(dp[i + 1][j + b[i + 1]][k], dp[i][j][k] + (a[i + 1] != 1));
                if (k + b[i + 1] <= sum)
                    dp[i + 1][j][k + b[i + 1]] = min(dp[i + 1][j][k + b[i + 1]], dp[i][j][k] + (a[i + 1] != 2));
                dp[i + 1][j][k] = min(dp[i + 1][j][k], dp[i][j][k] + (a[i + 1] != 3));
            }
        }
    }
    if (dp[n][sum][sum] == inf)
    {
        cout << -1 << endl;
    }
    else
        cout << dp[n][sum][sum] << endl;
}

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    int test = 1;
    // cin >> test;
    for (int i = 1; i <= test; i++)
    {
        solve();
    }
    return 0;
}

F题

思路

对于所有的查询，我们考虑倒着进行离线操作。先用 F l o y d Floyd Floyd进行一次预处理求出所有的 d i s t [ x ] [ y ] dist[x][y] dist[x][y]，即点到点 y y y的最短距离。

之后根据查询的条件，不断将新的边加到图中，并根据新边用 F l o y d Floyd Floyd进行 n 2 n^2 n2的更新操作。

代码
#pragma GCC optimize("O2")
#pragma GCC optimize("O3")
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N = 3e2 + 5, M = 1e6 + 5;
const int inf = 1e14;
int n, m, q;
int g[N][N];
struct edge
{
    int a, b, c;
};
vector<edge> v;
struct query
{
    int id, x, y;
};
vector<query> op;
void solve()
{
    cin >> n >> m >> q;
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= n; j++)
        {
            g[i][j] = inf;
        }
    }
    for (int i = 1, a, b, c; i <= m; i++)
    {
        cin >> a >> b >> c;
        g[a][b] = c;
        g[b][a] = c;
        v.push_back({a, b, c});
    }
    for (int i = 1, id, x, y; i <= q; i++)
    {
        cin >> id >> x;
        if (id == 1)
        {
            int a = v[x - 1].a;
            int b = v[x - 1].b;
            int c = v[x - 1].c;
            g[a][b] = inf;
            g[b][a] = inf;
        }
        y = 0;
        if (id == 2)
            cin >> y;
        op.push_back({id, x, y});
    }
    for (int k = 1; k <= n; k++)
    {
        for (int i = 1; i <= n; i++)
        {
            for (int j = 1; j <= n; j++)
            {
                g[i][j] = min(g[i][j], g[i][k] + g[k][j]);
            }
        }
    }
    vector<int> ans;
    for (int num = q - 1; num >= 0; num--)
    {
        if (op[num].id == 1)
        {
            int x = op[num].x;
            int a = v[x - 1].a;
            int b = v[x - 1].b;
            int c = v[x - 1].c;
            g[a][b] = min(g[a][b], c);
            g[b][a] = min(g[b][a], c);
            for (int i = 1; i <= n; i++)
            {
                for (int j = 1; j <= n; j++)
                {
                    g[i][j] = min(g[i][j], g[i][a] + g[a][j]);
                }
            }
            for (int i = 1; i <= n; i++)
            {
                for (int j = 1; j <= n; j++)
                {
                    g[i][j] = min(g[i][j], g[i][b] + g[b][j]);
                }
            }
        }
        else
        {
            int x = op[num].x;
            int y = op[num].y;
            if (g[x][y] >= inf / 2)
                ans.push_back(-1);
            else
                ans.push_back(g[x][y]);
        }
    }
    reverse(ans.begin(), ans.end());
    for (int val : ans)
    {
        cout << val << endl;
    }
}

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    int test = 1;
    // cin >> test;
    for (int i = 1; i <= test; i++)
    {
        solve();
    }
    return 0;
}

悦读

道可道，非常道；名可名，非常名。 无名，天地之始，有名，万物之母。 故常无欲，以观其妙，常有欲，以观其徼。 此两者，同出而异名，同谓之玄，玄之又玄，众妙之门。

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