9.18 城市道路设计

题意

给定\(n\)个点\(n\)条边的连通图，要求删去一条边，使得剩下的图仍是一个连通图，并且图中距离最远的两个点的距离最小

解法

\(n\)个点\(n\)条边的连通图即为基环树

要删去一条边使得剩下的图仍是连通图，我们删除的一定是环上的边，剩下的图一定是一颗树

那么树上距离最远的两个点的距离就是树的直径

暴力枚举环上断边求直径是\(O(N^2)\)的，考虑优化

单独考虑环上的点，把环上的点编号为\(1\)至\(m\)

断边后的直径有两种情况：

• 直径不经过环\(\rightarrow\) 答案即为所有基环外向树中的直径

• 直径经过环 \(\rightarrow\) 此时又要分两种情况进行考虑

  • 使用了\(1 \to m\)这条边

    预处理出倒\(L\)形路径的前缀后缀最大值

  • 未使用\(1 \to m\)这条边

    直径一定是从某个子树的最深处出发，在环上遍历了某些点，再回到某个子树中去的

    预处理出\(\Pi\)形路径的前缀后缀最大值

有很多细节，巨难打。。

代码

#include <bits/stdc++.h> 

using namespace std;

const int N = 1e6 + 10;

int read();

int n;

int cap;
int head[N], to[N], nxt[N], val[N];

int deg[N];

long long ans;

inline void add(int x, int y, int z) {
    to[++cap] = y, nxt[cap] = head[x], head[x] = cap, val[cap] = z; 
}

int p;
int cir[N], mrk[N], book[N];

long long DIS, len[N];

void Dfs_cir(int x) {
    book[x] = mrk[x] = 1, cir[++p] = x;
    for (int i = head[x]; i; i = nxt[i]) {
        if (!mrk[to[i]]) {
            Dfs_cir(to[i]);
            break;
        }
    }
}

long long f[N][5];

void Dfs_diametre(int x, int fa) {
    for (int i = head[x]; i; i = nxt[i]) {
        if (to[i] == fa || book[to[i]]) continue;
        Dfs_diametre(to[i], x);
        if (f[to[i]][1] + val[i] >= f[x][1])
            f[x][2] = f[x][1], f[x][1] = f[to[i]][1] + val[i];
        else
            f[x][2] = max(f[x][2], f[to[i]][1] + val[i]);
    }
}

int top;
int stk[N];

void Get_cir() {
    for (int i = 1; i <= n; ++i)
        if (deg[i] == 1)
            stk[++top] = i;
    while (top) {
        int u = stk[top--];
        mrk[u] = 1;
        for (int i = head[u]; i; i = nxt[i]) {
            deg[u]--, deg[to[i]]--;
            if (deg[to[i]] == 1)
                stk[++top] = to[i];
        }
    }
    for (int i = 1; i <= n; ++i)
        if (!mrk[i])    Dfs_cir(i);
}

void Get_diametre() {
    for (int i = 1; i <= p; ++i) {
        Dfs_diametre(cir[i], 0);
        ans = max(ans, f[cir[i]][1] + f[cir[i]][2]);
    }
}

long long pre[N], suf[N];
long long f1[N], f2[N], g1[N], g2[N];

void prepare_for_ans() {
    for (int i = 1; i < p; ++i)
        for (int j = head[cir[i]]; j; j = nxt[j]) 
            if (to[j] == cir[i + 1])  
                len[i] = val[j];
    
    for (int i = head[cir[p]]; i; i = nxt[i])
        if (to[i] == cir[1]) {
            DIS = val[i];
            break;  
        }
    
    for (int i = 1; i < p; ++i)         pre[i] = pre[i - 1] + len[i];
    for (int i = p - 1; i >= 1; --i)    suf[i] = suf[i + 1] + len[i];
    
    for (int i = 1; i <= p; ++i)   f1[i] = max(f1[i - 1], pre[i - 1] + f[cir[i]][1]);
    for (int i = p; i >= 1; --i)   f2[i] = max(f2[i + 1], suf[i] + f[cir[i]][1]);
    
    long long now = f[cir[1]][1];
    for (int i = 2; i <= p; ++i) {
        now += len[i - 1];
        g1[i] = max(g1[i - 1], now + f[cir[i]][1]);
        now = max(now, f[cir[i]][1]);
    }
    
    now = f[cir[p]][1]; 
    for (int i = p - 1; i >= 1; --i) {
        now += len[i];
        g2[i] = max(g2[i + 1], now + f[cir[i]][1]);
        now = max(now, f[cir[i]][1]);
    }
    
}

void Get_ans() {
    prepare_for_ans();
    
    long long res = g2[1];
    
    for (int i = 1; i < p; ++i) 
        res = min(res, max(max(g1[i], g2[i + 1]), f1[i] + f2[i + 1] + DIS));
    ans = max(ans, res);
}

int main() {
    
    n = read();
    
    for (int i = 1; i <= n; ++i) {
        int u = read(), v = read(), w = read();
        add(u, v, w), add(v, u, w); 
        deg[u]++, deg[v]++;
    }
    
    Get_cir();
    Get_diametre();
    Get_ans();
    
    printf("%lld\n", ans);
    
    return 0;
}

#define gc getchar
int read() {
    int x = 0, c = gc();
    while (c < '0' || c > '9')    c = gc();
    while (c >= '0' && c <= '9')  x = x * 10 + c - 48, c = gc();
    return x;   
}