寒假：Day24

Day24

继续图论。

346. 走廊泼水节 - AcWing题库

Kruskal模板应用，先把每个点看作一个集合，然后从小到大枚举边，每次把两个集合合并时，两个集合互相连边直至成局部完全图即可

#include
using namespace std;
const int N = 6010;

struct Edge
{
    int a, b, w;
    bool operator< (const Edge &t) const
    {
        return w < t.w;
    }
}e[N];
int p[N], cnt[N];

int find(int x)
{
    if(x != p[x]) p[x] = find(p[x]);
    return p[x];
}

int main(void)
{
    int t;
    cin >> t;
    while (t--)
    {
        int n;
        cin >> n;
        for (int i = 0; i < n - 1; i++)
        {
            int a, b, w;
            cin >> a >> b >> w;
            e[i] = {a, b, w};
        }
        sort(e, e + n - 1);
        for (int i = 1; i <= n; i++) p[i] = i, cnt[i] = 1;
        int res = 0;
        for (int i = 0; i < n - 1; i++)
        {
            int a = find(e[i].a), b = find(e[i].b), w = e[i].w;
            if (a != b)
            {
                res += (cnt[a] * cnt[b] - 1) * (w + 1);
                cnt[b] += cnt[a];
                p[a] = b;
            }
        }
        cout << res << endl ;
    }
    return 0;
}

1148. 秘密的牛奶运输 - AcWing题库

#include
using namespace std;
typedef long long LL;
const int N = 510, M = 10010;

int n, m;
struct Edge
{
    int a, b, w;
    bool f;
    bool operator< (const Edge &t) const
    {
        return w < t.w;
    }
}edge[M];
int p[N];
int dist1[N][N], dist2[N][N];
int h[N], e[N * 2], w[N * 2], ne[N * 2], idx;

void add(int a, int b, int c)
{
    e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx ++ ;
}

int find(int x)
{
    if (x != p[x]) p[x] = find(p[x]);
    return p[x];
}

void dfs(int u, int fa, int maxd1, int maxd2, int d1[], int d2[])
{
    d1[u] = maxd1, d2[u] = maxd2;
    for (int i = h[u]; i != -1; i = ne[i])
    {
        int j = e[i];
        if (j != fa)
        {
            int td1 = maxd1, td2 = maxd2;
            if (w[i] > td1) td2 = td1, td1 = w[i];
            else if (w[i] < td1 && w[i] > td2) td2 = w[i];
            dfs(j, u, td1, td2, d1, d2);
        }
    }
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> n >> m;
    memset(h, -1, sizeof h);
    for (int i = 0; i < m; i ++ )
    {
        int a, b, w;
        cin >> a >> b >> w;
        edge[i] = {a, b, w};
    }

    sort(edge, edge + m);
    for (int i = 1; i <= n; i ++ ) p[i] = i;

    LL sum = 0;
    for (int i = 0; i < m; i ++ )
    {
        int a = edge[i].a, b = edge[i].b, w = edge[i].w;
        int pa = find(a), pb = find(b);
        if (pa != pb)
        {
            p[pa] = pb;
            sum += w;
            add(a, b, w), add(b, a, w);
            edge[i].f = true;
        }
    }
    
    for (int i = 1; i <= n; i ++ ) dfs(i, -1, -1e9, -1e9, dist1[i], dist2[i]);

    LL res = 1e18;
    for (int i = 0; i < m; i ++ )
        if (!edge[i].f)
        {
            int a = edge[i].a, b = edge[i].b, w = edge[i].w;
            LL t;
            if (w > dist1[a][b])
                t = sum + w - dist1[a][b];
            else if (w > dist2[a][b])
                t = sum + w - dist2[a][b];
            res = min(res, t);
        }
    cout << res << endl ;
    return 0;
}

904. 虫洞 - AcWing题库

spfa判断负环

#include
using namespace std;
const int N = 510, M = 5210;

int n, m1, m2;
int h[N], e[M], ne[M], w[M], idx;
int dist[N];
int q[N], cnt[N];
bool st[N];

void add(int a, int b, int c)
{
    e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx++;   
}

bool spfa()
{
    int hh = 0, tt = 0;
    memset(dist, 0, sizeof dist);
    memset(st, 0, sizeof st);
    memset(cnt, 0, sizeof cnt);
    
    for (int i = 1; i <= n; i++)
    {
        q[tt++] = i;
        st[i] = true;
    }
    
    while (hh != tt)
    {
        int t = q[hh++];
        if (hh == N) hh = 0;
        st[t] = false;
        
        for (int i = h[t]; ~i; 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])
                {
                    q[tt++] = j;
                    if (tt == N) tt = 0;
                    st[j] = true;
                }
            }
        }
    }
    return false;
}
int main(void)
{
    int T;
    cin >> T;
    while (T--)
    {
        cin >> n >> m1 >> m2;
        memset(h, -1, sizeof h);
        idx = 0;
        
        while (m1--)
        {
            int a, b, c;
            cin >> a >> b >> c;
            add(a, b, c), add(b, a, c);
        }
        
        while (m2--)
        {
            int a, b, c;
            cin >> a >> b >> c;
            add(a, b, -c);
        }
        if (spfa()) puts("YES");
        else puts("NO");
    }
    return 0;
}

361. 观光奶牛 - AcWing题库

01分数规划问题，先二分一个可能的答案，然后通过整理不等式，最后得出是否存在满足条件的正环

#include
using namespace std;
const int N = 1010, M = 5010;

int n, m;
int wf[N];
int h[N], e[M], wt[M], ne[M], idx;
double dist[N];
int q[N], cnt[N];
bool st[N];

void add(int a, int b, int c)
{
    e[idx] = b, wt[idx] = c, ne[idx] = h[a], h[a] = idx ++ ;
}

bool check(double mid)
{
    memset(dist, 0, sizeof dist);
    memset(st, 0, sizeof st);
    memset(cnt, 0, sizeof cnt);

    int hh = 0, tt = 0;
    for (int i = 1; i <= n; i ++ )
    {
        q[tt ++ ] = i;
        st[i] = true;
    }

    while (hh != tt)
    {
        int t = q[hh ++ ];
        if (hh == N) hh = 0;
        st[t] = false;

        for (int i = h[t]; ~i; i = ne[i])
        {
            int j = e[i];
            if (dist[j] < dist[t] + wf[t] - mid * wt[i])
            {
                dist[j] = dist[t] + wf[t] - mid * wt[i];
                cnt[j] = cnt[t] + 1;
                if (cnt[j] >= n) return true;
                if (!st[j])
                {
                    q[tt ++ ] = j;
                    if (tt == N) tt = 0;
                    st[j] = true;
                }
            }
        }
    }
    return false;
}

int main()
{
    cin >> n >> m;
    for (int i = 1; i <= n; i ++ ) cin >> wf[i];

    memset(h, -1, sizeof h);
    for (int j = 0; j < m; j ++ )
    {
        int a, b, c;
        cin >> a >> b >> c;
        add(a, b, c);
    }

    double l = 0, r = 1e6; // 浮点数二分
    while (r - l > 1e-4)
    {
        double mid = (l + r) / 2;
        if (check(mid)) l = mid;
        else r = mid;
    }
    printf("%.2lf\n", l);
    return 0;
}