Codeforces Beta Round #95 (Div. 2) D.Subway

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题目链接：http://codeforces.com/problemset/problem/131/D

思路：题目的意思是说给定一个无向图，求图中的顶点到环上顶点的最短距离（有且仅有一个环，并且环上顶点的距离不计）。

一开始我是直接用Tarjan求的无向图的双连通分量，然后标记连通分量上的点（如果某一个连通分量上的顶点的个数大于1，那么就是环了，其余的都只有一个点），然后即使重新建图，spfa求最短路径。

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <vector>#include <stack>#include <queue>#define REP(i, a, b) for (int i = (a); i < (b); ++i)#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)using namespace std;const int MAX_N = (3000 + 300);int dfn[MAX_N], low[MAX_N], cnt, N, _count, color[MAX_N];int st, dist[MAX_N];bool mark[MAX_N];vector<int > g[MAX_N], reg[MAX_N];stack<int > S;void Tarjan(int u, int father){    low[u] = dfn[u] = ++cnt;    S.push(u);    mark[u] = true;    REP(i, 0, (int)g[u].size()) {        int v = g[u][i];        if (father == v) continue;        if (dfn[v] == 0) {            Tarjan(v, u);            low[u] = min(low[u], low[v]);        } else if (mark[v]) {            low[u] = min(low[u], dfn[v]);        }    }    if (low[u] == dfn[u]) {        int x, num = 0;        ++_count;        do {            x = S.top();            S.pop();            mark[x] = false;            color[x] = _count;            ++num;        } while (x != u);        if (num > 1) st = _count;    }}void spfa(int st){    queue<int > que;    memset(dist, 0x3f, sizeof(dist));    memset(mark, false, sizeof(mark));    dist[st] = 0;    que.push(st);    while (!que.empty()) {        int u = que.front();        que.pop();        REP(i, 0, (int)reg[u].size()) {            int v = reg[u][i];            if (dist[u] + 1 < dist[v]) {                dist[v] = dist[u] + 1;                if (!mark[v]) {                    mark[v] = true; que.push(v);                }            }        }    }}int main(){    while (cin >> N) {        FOR(i, 1, N) g[i].clear(), reg[i].clear();        FOR(i, 1, N) {            int u, v; cin >> u >> v;            g[u].push_back(v);            g[v].push_back(u);        }        cnt = _count = 0;        memset(dfn, 0, sizeof(dfn));        memset(mark, false, sizeof(mark));        FOR(i, 1, N) if (!dfn[i]) Tarjan(i, -1);        FOR(u, 1, N) {            REP(i, 0, (int)g[u].size()) {                int v = g[u][i];                if (color[u] != color[v]) reg[color[u]].push_back(color[v]), reg[color[v]].push_back(color[u]);            }        }        spfa(st);        FOR(i, 1, N) {            cout << dist[color[i]];            if (i == N) cout << endl;            else cout << " ";        }    }    return 0;}

后来我发现自己想的太复杂了，其实只要一遍dfs就能求出这个环上的点了，具体的做法是从某一点开始深搜，然后如果遇上之前搜过的点，那么说明形成一个环，用一个变量记录这个点，然后回退的时候判断是否遇到过这个点，如果没有遇到过，就把回退路径上的点都标记为环上的点，否则，继续回退。最后即使一遍bfs就可以求出最短路径。

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <vector>#include <queue>#define REP(i, a, b) for (int i = (a); i < (b); ++i)#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)using namespace std;const int MAX_N = (3000 + 300);int N, flag[MAX_N], dist[MAX_N], mark[MAX_N], found, st, Ok;vector<int > g[MAX_N];void dfs(int u, int father){    mark[u] = true;    REP(i, 0, (int)g[u].size()) {        int v = g[u][i];        if (v == father) continue;        if (!mark[v]) dfs(v, u);        else { found = 1; st = v; flag[u] = 1; return; }        if (found) {            if (Ok) return;            if (st == u) Ok = 1;            flag[u] = 1;            return;        }    }}void bfs(int st){    queue<int > que;    memset(mark, false, sizeof(mark));    mark[st] = true;    dist[st] = 0;    que.push(st);    while (!que.empty()) {        int u = que.front();        que.pop();        REP(i, 0, (int)g[u].size()) {            int v = g[u][i];            if (mark[v]) continue;            mark[v] = true;            if (flag[v]) dist[v] = 0;            else dist[v] = dist[u] + 1;            que.push(v);        }    }}int main(){    while (cin >> N) {        FOR(i, 1, N) g[i].clear(), flag[i] = mark[i] = 0;        FOR(i, 1, N) {            int u, v; cin >> u >> v;            g[u].push_back(v);            g[v].push_back(u);        }        found = Ok = 0;        dfs(1, 1);        bfs(st);        FOR(i, 1, N) {            cout << dist[i];            if (i == N) cout << endl;            else cout << " ";        }    }    return 0;}

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