BZOJ 2125: 最短路

2125: 最短路
思路：构建圆方树，然后如果两个点的lca是圆点，直接算，否则跳到环上相应的位置，再求环上两个点的最短距离。
代码1(在重链上跳)：

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#include<bits/stdc++.h>
using namespace std;
#define y1 y11
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
//#define mp make_pair
#define pb push_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pli pair<LL, int>
#define pii pair<int, int>
#define piii pair<pii, int>
#define pdd pair<double, double>
#define mem(a, b) memset(a, b, sizeof(a))
#define debug(x) cerr << #x << " = " << x << "\n";
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
//head

struct Circle_Square_Tree {
    const static int N = 2e4 + 10;
    vector<pair<int, LL> > g[N];
    int fa[N], dp[N], sz[N], son[N], top[N], dfn[N], to[N], cnt, n;
    LL dis[N], cir[N]; //环的大小
    bool vis[N];//记录到方点的最短距离是否经过回边

    inline void dfs1(int u, int o) {
        fa[u] = o;
        sz[u] = 1;
        dp[u] = dp[o] + 1;
        for (int i = 0; i < g[u].size(); ++i) {
            int v = g[u][i].fi;
            LL w = g[u][i].se;
            if(v != o) {
                dis[v] = dis[u] + w;
                dfs1(v, u);
                sz[u] += sz[v];
                if(sz[v] > sz[son[u]]) son[u] = v;
            }
        }
    }
    inline void dfs2(int u, int t) {
        top[u] = t;
        dfn[u] = ++cnt;
        to[cnt] = u;
        if(!son[u]) return ;
        dfs2(son[u], t);
        for (int i = 0; i < g[u].size(); ++i) {
            int v = g[u][i].fi;
            if(v != fa[u] && v != son[u]) dfs2(v, v);
        }
    }
    inline void init(int _n) {
        cnt = 0;
        n = _n;
        dp[0] = 0;
        dfs1(1, 0);
        dfs2(1, 1);
    }
    inline int lca(int u, int v) {
        int fu = top[u], fv = top[v];
        while(fu != fv) {
            if(dp[fu] >= dp[fv]) u = fa[fu], fu = top[u];
            else v = fa[fv], fv = top[v];
        }
        if(dp[u] <= dp[v]) return u;
        else return v;
    }
    inline int jump(int u, int lca) {
        int res;
        while(top[u] != top[lca]) res = top[u], u = fa[top[u]];
        if(u == lca) return res;
        else return to[dfn[lca]+1];
    }
    inline LL query(int u, int v) {
        int l = lca(u, v);
        if(l <= n) return dis[u]+dis[v]-2*dis[l];
        int uu = jump(u, l), vv = jump(v, l);
        LL d1 = dis[uu]-dis[l], d2 = dis[vv]-dis[l];
        if(!vis[uu]) d1 = cir[l]-d1;
        if(!vis[vv]) d2 = cir[l]-d2;
        return dis[u]-dis[uu]+dis[v]-dis[vv]+min(abs(d1-d2), cir[l]-abs(d1-d2));
    }
}CST;
const int N = 1e4 + 10;
int n, m, q, u, v, w;
vector<pii> g[N];
int dfn[N], low[N], fa[N], cnt = 0, tot = 0;
LL dep[N];
int a[N];
inline void solve(int u, int v, int d) {
    int cnt = 0;
    LL sum = d;
    for (int i = v; i != u; i = fa[i]) sum += dep[i] - dep[fa[i]], a[++cnt] = i;
    a[++cnt] = u;
    LL dis = 0;
    ++tot;
    for (int i = cnt; i >= 1; --i) {
        LL D = min(dis, sum-dis);
        if(D == dis) CST.vis[a[i]] = true;
        else CST.vis[a[i]] = false;
        CST.g[a[i]].pb({tot, D});
        CST.g[tot].pb({a[i], D});
        dis += dep[a[i-1]]-dep[a[i]];
    }
    CST.cir[tot] = sum;
}
inline void tarjan(int u, int o) {
    fa[u] = o;
    dfn[u] = low[u] = ++cnt;
    for (int i = 0; i < g[u].size(); ++i) {
        int v = g[u][i].fi;
        int w = g[u][i].se;
        if(v == o) continue;
        if(!dfn[v]) {
            dep[v] = dep[u] + w;
            tarjan(v, u);
            low[u] = min(low[u], low[v]);
        }
        else low[u] = min(low[u], dfn[v]);
        if(low[v] > dfn[u]) {
            CST.g[u].pb({v, w});
            CST.g[v].pb({u, w});
        }
    }
    for (int i = 0; i < g[u].size(); ++i) {
        int v = g[u][i].fi;
        int w = g[u][i].se;
        if(v == o) continue;
        if(fa[v] != u && dfn[v] > dfn[u]) {
            solve(u, v, w);
        }
    }
}
int main() {
    scanf("%d %d %d", &n, &m, &q);
    for (int i = 1; i <= m; ++i) scanf("%d %d %d", &u, &v, &w), g[u].pb({v, w}), g[v].pb({u, w});
    tot = n;
    tarjan(1, 0);
    CST.init(n);
    while(q--) {
        scanf("%d %d", &u, &v);
        printf("%lld\n", CST.query(u, v));
    }
    return 0;
}

代码2（倍增跳）：

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#include<bits/stdc++.h>
using namespace std;
#define y1 y11
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
//#define mp make_pair
#define pb push_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pli pair<LL, int>
#define pii pair<int, int>
#define piii pair<pii, int>
#define pdd pair<double, double>
#define mem(a, b) memset(a, b, sizeof(a))
#define debug(x) cerr << #x << " = " << x << "\n";
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
//head

struct Circle_Square_Tree {
    const static int N = 2e4 + 10;
    vector<pair<int, LL> > G[N];
    int dp[N], anc[N][18], n;
    LL dis[N], cir[N]; //环的大小
    bool vis[N];//记录到方点的最短距离是否经过回边

    vector<pii> g[N];
    int dfn[N], low[N], fa[N], cnt, tot;
    int a[N];
    inline void solve(int u, int v, int d) {
        int cnt = 0;
        LL sum = d;
        for (int i = v; i != u; i = fa[i]) sum += dis[i] - dis[fa[i]], a[++cnt] = i;
        a[++cnt] = u;
        LL DIS = 0;
        ++tot;
        for (int i = cnt; i >= 1; --i) {
            LL D = min(DIS, sum-DIS);
            if(D == DIS) vis[a[i]] = true;
            else vis[a[i]] = false;
            G[a[i]].pb({tot, D});
            G[tot].pb({a[i], D});
            DIS += dis[a[i-1]]-dis[a[i]];
        }
        cir[tot] = sum;
    }
    inline void tarjan(int u, int o) {
        fa[u] = o;
        dfn[u] = low[u] = ++cnt;
        for (int i = 0; i < g[u].size(); ++i) {
            int v = g[u][i].fi;
            int w = g[u][i].se;
            if(v == o) continue;
            if(!dfn[v]) {
                dis[v] = dis[u] + w;
                tarjan(v, u);
                low[u] = min(low[u], low[v]);
            }
            else low[u] = min(low[u], dfn[v]);
            if(low[v] > dfn[u]) {
                G[u].pb({v, w});
                G[v].pb({u, w});
            }
        }
        for (int i = 0; i < g[u].size(); ++i) {
            int v = g[u][i].fi;
            int w = g[u][i].se;
            if(v == o) continue;
            if(fa[v] != u && dfn[v] > dfn[u]) {
                solve(u, v, w);
            }
        }
    }
    inline void dfs(int u, int o) {
        anc[u][0] = o;
        dp[u] = dp[o] + 1;
        for (int i = 1; i < 18; ++i) anc[u][i] = anc[anc[u][i-1]][i-1];
        for (int i = 0; i < G[u].size(); ++i) {
            int v = G[u][i].fi;
            LL w = G[u][i].se;
            if(v == o) continue;
            dis[v] = dis[u] + w;
            dfs(v, u);
        }
    }
    inline void init(int _n) {
        tot = n = _n;
        cnt = 0;
        dis[0] = dp[1] = 0;
        tarjan(1, 0);
        dfs(1, 1);
    }
    inline int lca(int u, int v) {
        if(dp[u] < dp[v]) swap(u, v);
        for (int i = 17; i >= 0; --i) if(dp[anc[u][i]] >= dp[v]) u = anc[u][i];
        if(u == v) return u;
        for (int i = 17; i >= 0; --i) if(anc[u][i] != anc[v][i]) u = anc[u][i], v = anc[v][i];
        return anc[u][0];
    }
    inline int jump(int u, int lca) {
        for (int i = 17; i >= 0; --i) if(dp[anc[u][i]] > dp[lca]) u = anc[u][i];
        return u;
    }
    inline LL query(int u, int v) {
        int l = lca(u, v);
        if(l <= n) return dis[u]+dis[v]-2*dis[l];
        int uu = jump(u, l), vv = jump(v, l);
        LL d1 = dis[uu]-dis[l], d2 = dis[vv]-dis[l];
        if(!vis[uu]) d1 = cir[l]-d1;
        if(!vis[vv]) d2 = cir[l]-d2;
        return dis[u]-dis[uu]+dis[v]-dis[vv]+min(abs(d1-d2), cir[l]-abs(d1-d2));
    }
}CST;
int n, m, q, u, v, w;
int main() {
    scanf("%d %d %d", &n, &m, &q);
    for (int i = 1; i <= m; ++i) scanf("%d %d %d", &u, &v, &w), CST.g[u].pb({v, w}), CST.g[v].pb({u, w});
    CST.init(n);
    while(q--) {
        scanf("%d %d", &u, &v);
        printf("%lld\n", CST.query(u, v));
    }
    return 0;
}