【LG4103】[HEOI2014]大工程

題面

洛谷

題解

先建虛樹，下面所有討論均是在虛樹上的。

對於第一問：直接統計所有樹邊對答案的貢獻即可。

對於第\(2,3\)問：記\(f[x]\)表示在\(x\)的子樹內離\(x\)距離最遠的關鍵點的距離，\(g[x]\)表示在\(x\)的子樹內離\(x\)距離最近的關鍵點的距離。

具體更新以\(f[x]\)為例：

訪問到\(v\in son_x\)，

如果以前訪問過的點中有關鍵點，則有\(f[x]=max(f[x],f[v]+dis(u,v)+f[x])\)，

每次還要向上傳遞，即\(f[x]=max(f[x],f[v]+dis(u,v))\)。

代碼

#include <iostream> 
#include <cstdio> 
#include <cstdlib> 
#include <cstring> 
#include <cmath> 
#include <algorithm> 
using namespace std; 
inline int gi() { 
    register int data = 0, w = 1; 
    register char ch = 0; 
    while (!isdigit(ch) && ch != '-') ch = getchar(); 
    if (ch == '-') w = -1, ch = getchar(); 
    while (isdigit(ch)) data = 10 * data + ch - '0', ch = getchar(); 
    return w * data; 
} 
const int MAX_N = 1e6 + 5; 
struct Graph { int to, next; } e[MAX_N << 2];
int fir1[MAX_N], fir2[MAX_N], e_cnt;
void clearGraph() {
    memset(fir1, -1, sizeof(fir1)); 
    memset(fir2, -1, sizeof(fir2)); 
} 
void Add_Edge(int *fir, int u, int v) { 
    e[e_cnt] = (Graph){v, fir[u]}; 
    fir[u] = e_cnt++; 
}
namespace Tree { 
    int fa[MAX_N], dep[MAX_N], size[MAX_N], top[MAX_N], son[MAX_N], dfn[MAX_N], tim; 
    void dfs1(int x) {
        dfn[x] = ++tim; 
        size[x] = 1, dep[x] = dep[fa[x]] + 1; 
        for (int i = fir1[x]; ~i; i = e[i].next) {
            int v = e[i].to; if (v == fa[x]) continue; 
            fa[v] = x; dfs1(v); size[x] += size[v]; 
            if (size[v] > size[son[x]]) son[x] = v; 
        } 
    } 
    void dfs2(int x, int tp) {
        top[x] = tp; 
        if (son[x]) dfs2(son[x], tp); 
        for (int i = fir1[x]; ~i; i = e[i].next) {
            int v = e[i].to; if (v == fa[x] || v == son[x]) continue; 
            dfs2(v, v); 
        } 
    } 
    int LCA(int x, int y) { 
        while (top[x] != top[y]) { 
            if (dep[top[x]] < dep[top[y]]) swap(x, y); 
            x = fa[top[x]]; 
        } 
        return dep[x] < dep[y] ? x : y; 
    } 
} 
using Tree::LCA; using Tree::dfn; using Tree::dep; 
int N, M, K, a[MAX_N];
bool key[MAX_N];
int f[MAX_N], g[MAX_N], s[MAX_N]; 
bool cmp(int i, int j) { return dfn[i] < dfn[j]; } 
void build() { 
    static int stk[MAX_N], top; 
    sort(&a[1], &a[K + 1], cmp); 
    stk[top = 1] = 1; fir2[1] = -1;
    e_cnt = 0; 
    for (int i = 1; i <= K; i++) {
        key[a[i]] = 1; 
        if (a[i] == 1) continue; 
        int lca = LCA(stk[top], a[i]); 
        if (lca != stk[top]) { 
            while (dfn[lca] < dfn[stk[top - 1]]) { 
                int u = stk[top], v = stk[top - 1]; 
                Add_Edge(fir2, u, v), Add_Edge(fir2, v, u); 
                --top; 
            } 
            if (dfn[lca] > dfn[stk[top - 1]]) { 
                fir2[lca] = -1; int u = stk[top], v = lca; 
                Add_Edge(fir2, u, v), Add_Edge(fir2, v, u); 
                stk[top] = lca; 
            }
            else { 
                int u = lca, v = stk[top--]; 
                Add_Edge(fir2, u, v), Add_Edge(fir2, v, u); 
            } 
        }
        fir2[a[i]] = -1, stk[++top] = a[i]; 
    } 
    for (int i = 1; i < top; i++) {
        int u = stk[i], v = stk[i + 1]; 
        Add_Edge(fir2, u, v), Add_Edge(fir2, v, u); 
    } 
} 
long long ans1;
int ans2, ans3; 
void Dp(int x, int fa) { 
    s[x] = key[x], f[x] = 0, g[x] = (key[x] ? 0 : 1e9); 
    for (int i = fir2[x]; ~i; i = e[i].next) { 
        int v = e[i].to; if (v == fa) continue; 
        Dp(v, x); 
    } 
    for (int i = fir2[x]; ~i; i = e[i].next) { 
        int v = e[i].to, w = dep[v] - dep[x]; 
        if (v == fa) continue; 
        ans1 += 1ll * (K - s[v]) * s[v] * w; 
        if (s[x] > 0) { 
            ans2 = min(ans2, g[x] + w + g[v]); 
            ans3 = max(ans3, f[x] + w + f[v]); 
        } 
        g[x] = min(g[x], g[v] + w); 
        f[x] = max(f[x], f[v] + w);
        s[x] += s[v]; 
    } 
    key[x] = 0; 
} 
int main () {
#ifndef ONLINE_JUDGE 
    freopen("cpp.in", "r", stdin); 
#endif
    clearGraph(); 
    N = gi(); 
    for (int i = 1; i < N; i++) { 
        int u = gi(), v = gi(); 
        Add_Edge(fir1, u, v), Add_Edge(fir1, v, u); 
    }
    Tree::dfs1(1), Tree::dfs2(1, 1); 
    M = gi(); 
    while (M--) { 
        ans1 = 0, ans2 = 1e9, ans3 = 0; 
        K = gi(); for (int i = 1; i <= K; i++) a[i] = gi(); 
        build(); 
        Dp(1, 0); 
        printf("%lld %d %d\n", ans1, ans2, ans3); 
    } 
    return 0; 
} 
 

【LG4103】[HEOI2014]大工程