//目录

ACM Arabella Collegiate Programming Contest 2015 H. Capital City 边连通分量

题目链接:http://codeforces.com/gym/100676/attachments

题意:

有 n 个点,m 条边,图中,边强连通分量之间可以直达,即距离为 0 ,找一个点当做首都,其他点到首都的最大距离最小。

参考:http://www.cnblogs.com/ost-xg/p/6395100.html

仔细研究了大佬的思路,很牛逼,但是也有一点细节遗漏了,但是还是能AC,interesting;

 

分析:

边双连通:条件比点双连通松一点,体现在程序中,就是存在一个子节点能到达的最早祖先在父节点之后(没有等于);这样桥就找到了,然后再去找边双连通分量——dfs,从一个点出发,是桥就不是同一个边双连通分量;

这样就把题目中的边双连通分量找出来了;

然后缩点重新建图ed的边集!!!(遍历每一条边,是桥就加上来,大佬的这里多加了哦)

 

此时,怎么选首都呢? 

其实,现在图已经是一棵树了,遍历每一个点跑最短路是傻逼行为;

先找到树的直径,首都一定在这条直径上,枚举直径上的点,最远的,不过是直径上的两个部分;

  1 #include <bits/stdc++.h>
  2 
  3 using namespace std;
  4 
  5 const int maxn = 100005;
  6 const long long inf = 1e15 + 5;
  7 
  8 struct Edge
  9 {
 10     long long u, v, c;
 11 } e[maxn<<2];
 12 struct Ed
 13 {
 14     long long v, c;
 15 };
 16 
 17 int n, m, low[maxn], pre[maxn], tim, ebcc_cnt, du[maxn];
 18 long long k, len, dis[maxn][2];
 19 vector<int> G[maxn];
 20 vector<Ed> ed[maxn];
 21 int isbri[maxn<<4];
 22 bool vis[maxn];
 23 
 24 void init()
 25 {
 26     ebcc_cnt = tim = 0;
 27     for (int i = 1; i <= n; i++) G[i].clear();
 28     memset(isbri, 0, sizeof(isbri));
 29     memset(pre, 0, sizeof(pre));
 30     memset(du, 0, sizeof(du));
 31 }
 32 
 33 void tarjan(int u, int fa)
 34 {
 35     low[u] = pre[u] = ++tim;
 36     for (int i = 0; i < G[u].size(); i++)
 37     {
 38         int tmp = G[u][i];
 39         int v = e[tmp].v;
 40         if (!pre[v])
 41         {
 42             tarjan(v, u);
 43             low[u] = min(low[u], low[v]);
 44             if (low[v] > pre[u])
 45                 isbri[tmp] = isbri[tmp^1] = true; //标记为桥
 46         }
 47         else if (fa != v)
 48             low[u] = min(low[u], pre[v]);
 49     }
 50 }
 51 
 52 void dfs(int u)
 53 {
 54     pre[u] = ebcc_cnt;
 55     for (int i = 0; i < G[u].size(); i++)
 56     {
 57         int tmp = G[u][i];
 58         if (isbri[tmp]) continue;
 59         int v = e[tmp].v;
 60         if (pre[v]) continue;
 61         dfs(v);
 62     }
 63 }
 64 
 65 void find_ebcc()
 66 {
 67     tarjan(1, -1);
 68     memset(pre, 0, sizeof(pre));
 69     for (int i = 1; i <= n; i++)
 70     {
 71         if (!pre[i])
 72         {
 73             ebcc_cnt++;
 74             dfs(i);
 75         }
 76     }
 77 }
 78 
 79 void BFS(int s, int ca)
 80 {
 81     memset(vis, 0, sizeof(vis));
 82     queue<Ed> q;
 83     q.push((Ed)
 84     {
 85         s, 0
 86     });
 87     vis[s] = 1;
 88     while (q.size())
 89     {
 90         Ed tmp = q.front();
 91         q.pop();
 92         dis[tmp.v][ca] = tmp.c;
 93         for (int i = 0; i < ed[tmp.v].size(); i++)
 94         {
 95             Ed xx = ed[tmp.v][i];
 96             if (!vis[xx.v])
 97             {
 98                 vis[xx.v] = 1;
 99                 q.push((Ed)
100                 {
101                     xx.v, xx.c + tmp.c
102                 });
103             }
104         }
105     }
106 }
107 
108 void dfs_len(int x, int fa, long long dep)
109 {
110     if (dep > len)
111     {
112         k = x;
113         len = dep;
114     }
115     for (int i = 0; i < ed[x].size(); i++)
116     {
117         Ed tmp = ed[x][i];
118         if (tmp.v == fa) continue;
119         dfs_len(tmp.v, x, dep + tmp.c);
120     }
121 }
122 
123 int main()
124 {
125     int t = 1;
126     cin >> t;
127 
128     while (t--)
129     {
130         cin >> n >> m;
131         init();
132         for (int i = 1; i <= m; i++)
133         {
134             int u, v, c;
135             scanf("%d%d%d", &u, &v, &c);
136             e[i<<1|1].u = u, e[i<<1|1].v = v, e[i<<1|1].c = c;
137             e[i<<1].u = v, e[i<<1].v = u, e[i<<1].c = c;
138             G[u].push_back(i<<1|1);
139             G[v].push_back(i<<1);
140         }
141         find_ebcc();
142         int tot = m<<1|1;
143         for (int i = 1; i <= ebcc_cnt; i++) ed[i].clear();
144         for (int i = 3; i <= tot; i += 2)
145         {
146             if (isbri[i])
147             {
148                 int u = e[i].v, v = e[i].u;
149                 ed[pre[u]].push_back((Ed)
150                 {
151                     pre[v], e[i].c
152                 });
153                 ed[pre[v]].push_back((Ed)
154                 {
155                     pre[u], e[i].c
156                 });
157             }
158         }
159         len = -1;
160         dfs_len(1, -1, 0);
161         int st = k;
162         len = -1;
163         dfs_len(st, -1, 0);
164         BFS(st, 0);
165         BFS(k, 1);
166         long long inx = n + 1, dd = inf;
167         for (int i = 1; i <= n; i++)
168         {
169             int pr = pre[i];
170             if (dis[pr][0] + dis[pr][1] != len) continue;
171             long long tmp = max(dis[pr][0], dis[pr][1]);
172             if (tmp < dd)
173             {
174                 inx = i;
175                 dd = tmp;
176             }
177         }
178         printf("%lld %lld\n",inx,dd);
179     }
180     return 0;
181 }
View Code

 

posted @ 2017-04-21 16:41  小草的大树梦  阅读(235)  评论(0编辑  收藏  举报