LCA题目选讲4

两道紫题，哭了

T1 [BZOJ 1977]次小生成树

我们可以使用 (kruskal) 来求出最小生成树，然后通过最小生成树来生成严格次小生成树。

可以证明存在一个严格次小生成树与最小生成树之有一条边的区别，我们枚举非树边加入到树中形成一个奇环树，我们再枚举环上的边断掉一条，我们为了让树边权和尽可能小，我们选择断掉最长的那条边，为了求严格次小生成树，假如最长边与非树边长度相同，则断掉严格次大边。

求环上的边权我们可以使用倍增来维护。

#include <cstdio>
#include <cstring>
#include <algorithm>
#define ll long long
using namespace std;

int read() {
    int a = 0, x = 1;
    char ch = getchar();
    while (ch > '9' || ch < '0') {
        if (ch == '-')
            x = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
        a = a * 10 + ch - '0';
        ch = getchar();
    }
    return a * x;
}
const int N = 1e5 + 7, M = 1e6 + 7;
const ll inf = 1e18 + 7;

int n, m;

int f[N];
int find(int s) {
    if (f[s] == s)
        return s;
    return f[s] = find(f[s]);
}

int head[N], go[M], nxt[M], val[M], cnt;
void add(int u, int v, int w) {
    go[++cnt] = v;
    nxt[cnt] = head[u];
    head[u] = cnt;
    val[cnt] = w;
}

int fa[N][31], g[N][31], h[N][31];

void upd(int &a, int &b, int c, int d) {
    int ret1 = max(a, c), ret2 = max(b, d);
    if (ret1 != a)
        ret2 = max(ret2, a);
    if (ret1 != c)
        ret2 = max(ret2, c);
    a = ret1, b = ret2;
}
int dis[N];
void LCA(int a, int b, int &mx1, int &mx2) {
    if (dis[a] < dis[b])
        swap(a, b);
    for (int i = 30; i >= 0; i--) {
        if (dis[fa[a][i]] >= dis[b]) {
            upd(mx1, mx2, h[a][i], g[a][i]);
            a = fa[a][i];
        }
    }
    if (a == b)
        return;
    for (int i = 30; i >= 0; i--) {
        if (fa[a][i] != fa[b][i]) {
            upd(mx1, mx2, h[a][i], g[a][i]);
            upd(mx1, mx2, h[b][i], g[b][i]);
            a = fa[a][i], b = fa[b][i];
        }
    }
    upd(mx1, mx2, h[a][0], g[a][0]);
    upd(mx1, mx2, h[b][0], g[b][0]);
}

void dfs(int u) {
    for (int i = 1; i <= 30; i++) {
        fa[u][i] = fa[fa[u][i - 1]][i - 1];
        h[u][i] = h[u][i - 1], g[u][i] = g[u][i - 1];
        upd(h[u][i], g[u][i], h[fa[u][i - 1]][i - 1], g[fa[u][i - 1]][i - 1]);
    }
    for (int e = head[u]; e; e = nxt[e]) {
        int v = go[e];
        if (v == fa[u][0])
            continue;
        fa[v][0] = u, dis[v] = dis[u] + 1;
        h[v][0] = val[e];
        dfs(v);
    }
}

struct node {
    int u, v, w;
    friend bool operator<(node a, node b) { return a.w < b.w; }
} arr[M];
ll sum = 0;
bool vis[M];
int main() {
    // freopen("in.in","r",stdin);
    // freopen("out.out","w",stdout);
    n = read(), m = read();
    for (int i = 1; i <= m; i++) {
        arr[i].u = read(), arr[i].v = read(), arr[i].w = read();
    }
    for (int i = 1; i <= n; i++) f[i] = i;
    sort(arr + 1, arr + 1 + m);
    for (int i = 1; i <= m; i++) {
        if (find(arr[i].u) != find(arr[i].v)) {
            sum += arr[i].w;
            f[find(arr[i].u)] = find(arr[i].v);
            vis[i] = 1;
            add(arr[i].u, arr[i].v, arr[i].w);
            add(arr[i].v, arr[i].u, arr[i].w);
        }
    }
    dfs(1);
    ll ans = inf;
    for (int i = 1; i <= m; i++) {
        if (vis[i])
            continue;
        int mx1(0), mx2(0);
        LCA(arr[i].u, arr[i].v, mx1, mx2);
        if (!mx1)
            continue;
        //   printf("--%d %d--
",mx1,mx2);
        if (mx1 != arr[i].w) {
            ans = min(ans, (ll)arr[i].w - mx1);
        } else if (mx2)
            ans = min(ans, (ll)arr[i].w - mx2);
    }
    printf("%lld", sum + ans);
    return 0;
}

T2 [BZOJ 2306][CTSC 2011]幸福路径

正解是倍增Flody求解。 这和LCA什么关系啊？

#include <cstdio>
#include <cstring>
#include <algorithm>
#define ll long long
#define double long double
using namespace std;

int read() {
    int a = 0, x = 1;
    char ch = getchar();
    while (ch > '9' || ch < '0') {
        if (ch == '-')
            x = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
        a = a * 10 + ch - '0';
        ch = getchar();
    }
    return a * x;
}
const int N = 1e2 + 7, M = 1e6 + 7, inf = 2e9 + 7;
const double eps = 1e-12;
int n, m;
double w[N];
int s;
double p;
double dp[N][N], f[N][N];
int main() {
    // freopen("in.in","r",stdin);
    // freopen("out.out","w",stdout);
    n = read(), m = read();
    for (int i = 1; i <= n; i++) {
        scanf("%Lf", &w[i]);
    }
    s = read();
    scanf("%Lf", &p);
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            if (i != j)
                f[i][j] = -inf;
        }
    }
    for (int i = 1; i <= m; i++) {
        int a = read(), b = read();
        f[a][b] = w[b] * p;
    }
    double tmp = p;
    for (int h = 1; h <= 60; h++) {
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= n; j++) {
                dp[i][j] = -inf;
            }
        }

        for (int k = 1; k <= n; k++) {
            for (int i = 1; i <= n; i++) {
                for (int j = 1; j <= n; j++) {
                    dp[i][j] = max(dp[i][j], f[i][k] + p * f[k][j]);
                    //			printf(" %d %d %d %Lf %Lf %Lf
                    //%Lf
",i,k,j,dp[i][j],f[i][k],f[k][j],f[i][k]+p*f[k][j]);
                }
            }
        }
        p *= p;
        // for(int i = 1;i <= n;i ++,putchar('
')) {
        // 	for(int j = 1;j <= n;j ++) {
        // 		printf("dp[%d][%d] = %13.1Lf ",i,j,dp[i][j]);
        // 	}
        // }putchar('
');
        memcpy(f, dp, sizeof(dp));
        if (p < eps)
            break;
    }
    double ans = 0;
    for (int i = 1; i <= n; i++) {
        ans = max(ans, f[s][i]);
    }
    printf("%.1Lf", ans + w[s]);
    return 0;
}