HDU 4005 The war（边双连通）

题意

​ 给定一张 \(n\) 个点 \(m\) 条边的无向连通图，加入一条边，使得图中权值最小的桥权值最大，如果能使图中没有桥则输出 \(-1\)。

思路

​ 先对原图边双缩点，然后变成了一棵树。在树上加一条边等价于使一条路径上的边都不是桥，那么原题转化为在树上删一条路径，使得最小的边最大。固定一条最小的边之后模拟即可。

代码

#include<bits/stdc++.h>
#define FOR(i, x, y) for(int i = (x), i##END = (y); i <= i##END; ++i)
#define DOR(i, x, y) for(int i = (x), i##END = (y); i >= i##END; --i)
template<typename T, typename _T> inline bool chk_min(T &x, const _T &y) {return y < x ? x = y, 1 : 0;}
template<typename T, typename _T> inline bool chk_max(T &x, const _T &y) {return x < y ? x = y, 1 : 0;}
typedef long long ll;
const int N = 10005;
const int M = 100005;

template<const int N, const int M, typename T> struct Linked_List
{
    int head[N], nxt[M], tot; T to[M];
    Linked_List() {clear();}
    T &operator [](const int x) {return to[x];}
    void clear() {memset(head, -1, sizeof(head)), tot = 0;}
    void add(int u, T v) {to[tot] = v, nxt[tot] = head[u], head[u] = tot++;}
    #define EOR(i, G, u) for(int i = G.head[u]; ~i; i = G.nxt[i])
};

struct edge {int to, cost;};
Linked_List<N, M << 1, edge> G;
Linked_List<N, N << 1, edge> T;

int dfn[N], low[N], stk[N], bel[N], dfn_idx, tp, bcc;
int miner[N], son[N];
int n, m;

void tarjan(int u, int fa_e)
{
    dfn[u] = low[u] = ++dfn_idx, stk[++tp] = u;
    EOR(i, G, u)
    {
        if(i == (fa_e ^ 1)) continue;
        int v = G[i].to;
        if(!dfn[v]) tarjan(v, i), chk_min(low[u], low[v]);
        else if(dfn[v] < dfn[u]) chk_min(low[u], dfn[v]);
    }
    if(dfn[u] == low[u])
    {
        bcc++;
        do bel[stk[tp]] = bcc; while(stk[tp--] != u);
    }
}

void dfs(int u, int f)
{
    miner[u] = 2e9, son[u] = 0;
    EOR(i, T, u)
    {
        int v = T[i].to, w = T[i].cost;
        if(v == f) continue;
        dfs(v, u);
        if(chk_min(miner[u], miner[v])) son[u] = v;
        if(chk_min(miner[u], w)) son[u] = v;
    }
}

int redfs(int u, int f)
{
    if(!son[u]) return 2e9;
    int res = redfs(son[u], u);
    EOR(i, T, u)
    {
        int v = T[i].to, w = T[i].cost;
        if(v == f || v == son[u]) continue;
        chk_min(res, miner[v]);
        chk_min(res, w);
    }
    return res;
}

int main()
{
    while(~scanf("%d%d", &n, &m))
    {
        G.tot = T.tot = 0;
        FOR(i, 1, n) G.head[i] = T.head[i] = -1;
        FOR(i, 1, n) dfn[i] = 0;
        bcc = dfn_idx = 0;

        FOR(i, 1, m)
        {
            int u, v, w;
            scanf("%d%d%d", &u, &v, &w);
            G.add(u, (edge){v, w}), G.add(v, (edge){u, w});
        }

        tarjan(1, -1);

        int s, t, mi = 2e9;
        FOR(u, 1, n) EOR(i, G, u)
        {
            int v = G[i].to, w = G[i].cost;
            if(bel[u] < bel[v])
            {
                if(chk_min(mi, w)) s = bel[u], t = bel[v];
                T.add(bel[u], (edge){bel[v], w}), T.add(bel[v], (edge){bel[u], w});
            }
        }

        dfs(s, t), dfs(t, s);
        int res = std::min(redfs(s, t), redfs(t, s));
        printf("%d\n", (res > 1e9 ? -1 : res));
    }
    return 0;
}

标签：HDU,边双,bel,min,int,chk,4005,dfn,head
来源： https://www.cnblogs.com/Paulliant/p/12048559.html