• poj 1986(LCA tarjan算法)


    题目:给一棵带权树,然后让你求任意两点之间的距离。

    思路:tarjan一边dfs求出dis[i]每个点到根的距离,ans[i]第i组查询的lca。ans = dis[u]+dis[v]-2*dis[lca]。

    代码如下:

     1 /**************************************************
     2  * Author     : xiaohao Z
     3  * Blog     : http://www.cnblogs.com/shu-xiaohao/
     4  * Last modified : 2014-02-09 09:18
     5  * Filename     : hdu_2586.cpp
     6  * Description     : 
     7  * ************************************************/
     8 
     9 #include <iostream>
    10 #include <cstdio>
    11 #include <cstring>
    12 #include <cstdlib>
    13 #include <cmath>
    14 #include <algorithm>
    15 #include <queue>
    16 #include <stack>
    17 #include <vector>
    18 #include <set>
    19 #include <map>
    20 #define MP(a, b) make_pair(a, b)
    21 #define PB(a) push_back(a)
    22 
    23 using namespace std;
    24 typedef long long ll;
    25 typedef pair<int, int> pii;
    26 typedef pair<unsigned int,unsigned int> puu;
    27 typedef pair<int, double> pid;
    28 typedef pair<ll, int> pli;
    29 typedef pair<int, ll> pil;
    30 
    31 const int INF = 0x3f3f3f3f;
    32 const double eps = 1E-6;
    33 const int LEN = 40000+10;
    34 vector<pii> Map[LEN], que[LEN];
    35 pii qv[LEN];
    36 int n, m, q, vis[LEN], fa[LEN], ancestor[LEN], ans[LEN], dis[LEN];
    37 
    38 void init(){
    39     for(int i=0; i<LEN; i++) {
    40         Map[i].clear();
    41         que[i].clear();
    42         fa[i] = i;     
    43     }
    44     memset(vis, 0, sizeof vis);
    45     memset(ancestor, 0, sizeof ancestor);
    46 }
    47 //UFSet
    48 int Find(int x){return fa[x] == x ? x : fa[x] = Find(fa[x]);}
    49 void Union(int a, int b){int pa = Find(a), pb = Find(b);if(pa != pb) fa[pa] = pb;}
    50 
    51 void tarjan(int u){
    52     ancestor[u] = u;
    53     vis[u] = 1;
    54     for(int i=0; i<Map[u].size(); i++){
    55         int v = Map[u][i].first;
    56         if(vis[v]) continue;
    57         dis[v] = dis[u] + Map[u][i].second;
    58         tarjan(v);
    59         Union(u, v);
    60         ancestor[Find(u)] = u;
    61     }
    62     for(int i=0; i<que[u].size(); i++){
    63         int v = que[u][i].first, idx = que[u][i].second;
    64         if(vis[v])ans[idx] = ancestor[Find(v)];
    65     }
    66 }
    67 
    68 int main()
    69 {
    70 //    freopen("in.txt", "r", stdin);
    71 
    72     int a, b, val;
    73     char tc[10];
    74     while(scanf("%d%d", &n, &m)!=EOF){
    75         init();
    76         for(int i=0; i<m; i++) {
    77             scanf("%d%d%d", &a, &b, &val);
    78             gets(tc);
    79             Map[a].PB(MP(b, val));
    80             Map[b].PB(MP(a, val));
    81         }
    82         scanf("%d", &q);
    83         for(int i=0; i<q; i++){
    84             scanf("%d%d", &a, &b);
    85             qv[i] = MP(a, b);
    86             que[a].PB(MP(b, i));
    87             que[b].PB(MP(a, i));
    88         }
    89         dis[1] = 0;
    90         tarjan(1);
    91         for(int i=0; i<q; i++){
    92             printf("%d
    ", dis[qv[i].first]+dis[qv[i].second]-2*dis[ans[i]]);
    93         }
    94     }
    95     return 0;
    96 }
    View Code
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  • 原文地址:https://www.cnblogs.com/shu-xiaohao/p/3541336.html
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