Codeforces 178B1-B3 Greedy Merchants

题意：给出一张n个点m条边的无向图，然后又q组查询，询问从s节点走到m节点至少要经过多少条桥

思路：先dfs找桥，并给所有边-双连通分量蓝色。根据找到的桥以及桥的两端的节点的颜色，重新建图（每个节点代表一个双连通分量，其实就是形成一棵树），dfs一遍求各个点的深度，然后就是离线的求LCA，距离就是dis[u]+dis[v]-2*dis[LCA(u,v)]

#include <cstdio>
#include <cstring>
#include <stack>
#include <vector>
#include <algorithm>
#define maxn 100010
using namespace std;

int pre[maxn],bccno[maxn],dfs_clock,bcc_cnt,bg_cnt;
int depth,dep[maxn],ancestor[maxn],f[maxn],vis[maxn],ans[maxn];
pair<int,int> bridge[maxn];
vector<int> G[maxn],g[maxn],qes[maxn],rank[maxn];
stack<int> S;

int dfs(int u,int fa){
    //printf("dfs(%d,%d)
",u,fa);
    int lowu = pre[u] = ++dfs_clock;
    S.push(u);
    for(int i = 0;i < G[u].size();i++){
        int v = G[u][i];
        if(!pre[v]){
            int lowv = dfs(v,u);
            lowu = min(lowu,lowv);
            if(lowv > pre[u]){
                bridge[bg_cnt++] = make_pair(u,v);
            }
        }else if(pre[v] < pre[u] && v != fa){
            lowu = min(lowu,pre[v]);
        }
    }
    if(lowu == pre[u]){
        bcc_cnt++;
        while(1){
            int x = S.top();S.pop();
            bccno[x] = bcc_cnt;
            if(x == u)  break;
        }
    }
    return lowu;
}

void find_bcc(int n){
    memset(pre,0,sizeof(pre));
    dfs_clock = bg_cnt = bcc_cnt = 0;
    dfs(1,-1);
}

void dfs2(int u,int fa,int d){
    //printf("dfs2(%d,%d,%d)
",u,fa,d);
    dep[u] = d;
    for(int i = 0;i < g[u].size();i++){
        int v = g[u][i];
        if(v == fa) continue;
        dfs2(v,u,d+1);
    }
}

int find(int x){
    return x == f[x] ? x : f[x] = find(f[x]);
}

void LCA(int u,int fa){
    //printf("LCA(%d,%d)
",u,fa);
    ancestor[u] = u;
    for(int i = 0;i < g[u].size();i++){
        int v = g[u][i];
        if(v == fa) continue;
        LCA(v,u);
        int fx = find(u),fy = find(v);
        if(fx != fy)  f[fx] = fy;
        ancestor[find(u)] = u;
    }
    vis[u] = 1;
    for(int i = 0;i < qes[u].size();i++){
        int v = qes[u][i];
        if(vis[v]){
            //printf("both ancestor of (%d,%d) is %d
",u,v,ancestor[find(v)]);
            int tmp = dep[u] + dep[v] - 2 * dep[ancestor[find(v)]];
            ans[rank[u][i]] = tmp;
        }
    }
}

int main(){
    int n,m,q;
    while(scanf("%d%d",&n,&m) == 2){
        while(!S.empty())   S.pop();
        for(int i = 1;i <= n;i++)   G[i].clear();
        for(int i = 0;i < m;i++){
            int a,b;
            scanf("%d%d",&a,&b);
            G[a].push_back(b);
            G[b].push_back(a);
        }
        find_bcc(n);
        //for(int i = 1;i <= n;i++){
        //    printf("bccno[%d] = %d
",i,bccno[i]);
        //}
        //printf("bg:
");
        //for(int i = 0;i < bg_cnt;i++){
        //    printf("%d %d bccno: %d %d
",bridge[i].first,bridge[i].second,bccno[bridge[i].first],bccno[bridge[i].second]);
        //}
        for(int i = 1;i <= bcc_cnt;i++){
            g[i].clear();
            f[i] = i;
            vis[i] = 0;
        }
        for(int i = 0;i < bg_cnt;i++){
            int a = bridge[i].first;
            int b = bridge[i].second;
            g[bccno[a]].push_back(bccno[b]);
            g[bccno[b]].push_back(bccno[a]);
        }
        //for(int i = 1;i <= bcc_cnt;i++){
        //    printf("g %d:
",i);
        //    for(int j = 0;j < g[i].size();j++)
        //        printf("%d ",g[i][j]);
        //    printf("
");
        //}
        depth = 0;
        dfs2(1,-1,1);
        //for(int i = 1;i <= bcc_cnt;i++){
        //    printf("dep[%d] = %d
",i,dep[i]);
        //}
        scanf("%d",&q);
        for(int i = 0;i < q;i++){
            int a,b,x,y;
            scanf("%d%d",&x,&y);
            a = bccno[x];
            b = bccno[y];
            qes[a].push_back(b);
            qes[b].push_back(a);
            rank[a].push_back(i);
            rank[b].push_back(i);
        }
        LCA(1,-1);
        for(int i = 0;i < q;i++){
            printf("%d
",ans[i]);
        }
    }
    return 0;
}