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yolandaw




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活动打卡代码 AcWing 3827. 最小正整数

yolandaw
15天前
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>

using namespace std;

int n, k, t;
long long gcd(long long a,long long b)
{
    return b == 0 ? a:gcd(b, a % b);
}

int main()
{

    cin>>t;

    while (t --)
    {
        cin >> n >> k;
        int m = pow(10, k);
        cout<< (long long) n * m / gcd(n, m) << endl;
    }

    return 0;
}


活动打卡代码 AcWing 3826. 青蛙跳

yolandaw
15天前
#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long LL;

int t;

int main()
{
    cin >> t;
    while (t -- )
    {
        LL a, b, k;
        cin >> a >> b >> k;
        LL c = a - b;
        LL e = k / 2;
        k %= 2;
        if (k)
            cout << c * e + a << endl;
        else
            cout << c * e << endl;
    }
    return 0;
}



yolandaw
1个月前
#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 510, INF = 0x3f3f3f3f;

int n, m;
int g[N][N];
int dist[N];
bool st[N];

int prim()
{
    memset(dist, 0x3f, sizeof dist);

    int res = 0;
    for (int i = 0; i < n; i ++ )
    {
        int t = -1;
        for (int j = 1; j <= n; j ++ )
            if (!st[j] && (t == -1 || dist[t] > dist[j]))
                t = j;

        if (i && dist[t] == INF) return INF;
        if (i) res += dist[t];

        for (int j = 1; j <= n; j ++ ) dist[j] = min(dist[j], g[t][j]);

        st[t] = true;
    }

    return res;
}

int main()
{
    cin >> n >> m;

    memset(g, 0x3f, sizeof g);

    while (m -- )
    {
        int a, b, c;
        cin >> a >> b >> c;
        g[a][b] = g[b][a] = min(g[a][b], c);
    }

    int t = prim();

    if (t == INF) puts("impossible");
    else cout << t;

    return 0;
}



yolandaw
1个月前
#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 510, INF = 0x3f3f3f3f;

int n, m;
int g[N][N];
int dist[N];
bool st[N];

int prim()
{
    memset(dist, 0x3f, sizeof dist);

    int res = 0;
    for (int i = 0; i < n; i ++ )
    {
        int t = -1;
        for (int j = 1; j <= n; j ++ )
            if (!st[j] && (t == -1 || dist[t] > dist[j]))
                t = j;

        if (i && dist[t] == INF) return INF;
        if (i) res += dist[t];

        for (int j = 1; j <= n; j ++ ) dist[j] = min(dist[j], g[t][j]);

        st[t] = true;
    }

    return res;
}

int main()
{
    cin >> n >> m;

    memset(g, 0x3f, sizeof g);

    while (m -- )
    {
        int a, b, c;
        cin >> a >> b >> c;
        g[a][b] = g[b][a] = min(g[a][b], c);
    }

    int t = prim();

    if (t == INF) puts("impossible");
    else cout << t;

    return 0;
}


活动打卡代码 AcWing 854. Floyd求最短路

yolandaw
1个月前
#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 210, INF = 1e9;

int n, m, Q;
int d[N][N];

void floyd()
{
    for (int k = 1; k <= n; k ++ )
        for (int i = 1; i <= n; i ++ )
            for (int j = 1; j <= n; j ++ )
                d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
}

int main()
{
    cin >> n >> m >> Q;

    for (int i = 1; i <= n; i ++ )
        for (int j = 1; j <= n; j ++ )
            if (i == j) d[i][j] = 0;
            else d[i][j] = INF;

    while (m -- )
    {
        int a, b, w;
        cin >> a >> b >> w;

        d[a][b] = min(d[a][b], w);
    }

    floyd();

    while (Q -- )
    {
        int a, b;
        cin >> a >> b;

        if (d[a][b] > INF / 2) puts("impossible");
        else cout << d[a][b] << endl;
    }

    return 0;
}



yolandaw
1个月前
模板级补全收录了floyd算法吗


活动打卡代码 AcWing 852. spfa判断负环

yolandaw
1个月前
#include <iostream>
#include <cstring>
#include <algorithm>
#include <queue>

using namespace std;

typedef pair<int, int> PII;

const int N = 150010;

int n, m;
int h[N], e[N], w[N], ne[N], idx;
int dist[N], cnt[N];
bool st[N];

void add(int a, int b, int c)  // 添加一条边a->b,边权为c
{
    e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx ++ ;
}

int spfa()  // 求1号点到n号点的最短路距离,如果从1号点无法走到n号点则返回-1
{
    queue<int> q;
    for (int i = 1; i <= n; i ++ )
    {
        st[i] = true;
        q.push(i);
    }

    while (q.size())
    {
        int t = q.front();
        q.pop();
        st[t] = false;

        for (int i = h[t]; i != -1; i = ne[i])
        {
            int j = e[i];
            if (dist[j] > dist[t] + w[i])
            {
                dist[j] = dist[t] + w[i];
                cnt[j] = cnt[t] + 1;

                if (cnt[j] >= n) return true;
                if (!st[j])     // 如果队列中已存在j,则不需要将j重复插入
                {
                    q.push(j);
                    st[j] = true;
                }
            }
        }
    }

    return false;
}


int main()
{
    cin >> n >> m;

    memset(h, -1, sizeof h);

    while (m -- )
    {
        int a, b, c;
        cin >> a >> b >> c;
        add(a, b, c);
    }

    if (spfa()) puts("Yes");
    else puts("No");

    return 0;
}  


活动打卡代码 AcWing 851. spfa求最短路

yolandaw
1个月前
#include <iostream>
#include <cstring>
#include <algorithm>
#include <queue>

using namespace std;

typedef pair<int, int> PII;

const int N = 150010;

int n, m;
int h[N], e[N], w[N], ne[N], idx;
int dist[N];
bool st[N];

void add(int a, int b, int c)  // 添加一条边a->b,边权为c
{
    e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx ++ ;
}

int spfa()  // 求1号点到n号点的最短路距离,如果从1号点无法走到n号点则返回-1
{
    queue<int> q;
    memset(dist, 0x3f, sizeof dist);
    dist[1] = 0;
    q.push(1);
    st[1] = true;

    while (q.size())
    {
        int t = q.front();
        q.pop();
        st[t] = false;

        for (int i = h[t]; i != -1; i = ne[i])
        {
            int j = e[i];
            if (dist[j] > dist[t] + w[i])
            {
                dist[j] = dist[t] + w[i];
                if (!st[j])     // 如果队列中已存在j,则不需要将j重复插入
                {
                    q.push(j);
                    st[j] = true;
                }
            }
        }
    }

    if (dist[n] == 0x3f3f3f3f) return -1;
    return dist[n];
}


int main()
{
    cin >> n >> m;

    memset(h, -1, sizeof h);

    while (m -- )
    {
        int a, b, c;
        cin >> a >> b >> c;
        add(a, b, c);
    }

    int t = spfa();

    if (t == -1) cout << "impossible";
    else cout << t;

    return 0;
}



yolandaw
1个月前
#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 510, M = 10010;

int n, m, k;
int dist[N], bakcup[N];

struct Edge
{
    int a, b, w;
}edges[M];

int bellman_ford()
{
    memset(dist, 0x3f, sizeof dist);
    dist[1] = 0;

    for (int i = 0; i < k; i ++ )
    {
        memcpy(bakcup, dist, sizeof dist);
        for (int j = 0; j < m; j ++ )
        {
            int a = edges[j].a, b = edges[j].b, w = edges[j].w;
            dist[b] = min(dist[b], bakcup[a] + w);
        }
    }

    if (dist[n] > 0x3f3f3f3f / 2) return -1;
    return dist[n];
}

int main()
{
    cin >> n >> m >> k;

    for (int i = 0; i < m; i ++ )
    {
        int a, b, w;
        cin >> a >> b >> w;
        edges[i] = {a, b, w};
    }

    int t = bellman_ford();

    if (t == -1) puts("impossible");
    else cout << t;

    return 0;
}



yolandaw
1个月前
#include <iostream>
#include <cstring>
#include <algorithm>
#include <queue>

using namespace std;

typedef pair<int, int> PII;

const int N = 150010;

int n, m;
int h[N], e[N], w[N], ne[N], idx;
int dist[N];
bool st[N];

void add(int a, int b, int c)  // 添加一条边a->b,边权为c
{
    e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx ++ ;
}

int dijkstra()  // 求1号点到n号点的最短路距离,如果从1号点无法走到n号点则返回-1
{
    memset(dist, 0x3f, sizeof dist);
    dist[1] = 0;
    priority_queue<PII, vector<PII>, greater<PII>> heap;
    heap.push({0, 1});

    while (heap.size())
    {
        auto t = heap.top();
        heap.pop();

        int ver = t.second, distance = t.first;

        if (st[ver]) continue;
        st[ver] = true;

        for (int i = h[ver]; i != -1; i = ne[i])
        {
            int j = e[i];
            if (dist[j] > distance + w[i])
            {
                dist[j] = distance + w[i];
                heap.push({dist[j], j});
            }
        }
    }

    if (dist[n] == 0x3f3f3f3f) return -1;
    return dist[n];
}

int main()
{
    cin >> n >> m;

    memset(h, -1, sizeof h);

    while (m -- )
    {
        int a, b, c;
        cin >> a >> b >> c;
        add(a, b, c);
    }

    int t = dijkstra();

    cout << t;

    return 0;
}