HDU 2586 LCA

　　　　算是LCA裸题吧，d[x]记录从根节点到x节点的距离然后求出x,y的LCA z

　　　　结果就是d[x]+d[y]-2*d[z]

　　　　我是用Tarjan写的，麻烦的地方就是怎么离线ans

　　　　　刚开是学LCA，cheerup！！

  1 #include <cstdio>
  2 #include <cstring>
  3 #include <vector>
  4 #include <algorithm>
  5 #include <iostream>
  6 using namespace std;
  7 const int maxn = 100005;
  8 typedef long long LL;
  9 struct node
 10 {
 11     int x,w;
 12     node(){}
 13     node(int a,int b)
 14     {
 15         x = a;w = b;
 16     }
 17 }ans[maxn];
 18 vector<int>son[maxn];
 19 vector<node>f[maxn];
 20 vector<node>q[maxn];
 21 int d[maxn],vis[maxn],fa[maxn],anc[maxn];
 22 void dfs(int rt)
 23 {
 24     vis[rt] = 1;
 25     int len = f[rt].size();
 26     for(int i = 0;i<len;++i)
 27     {
 28         int u = f[rt][i].x;
 29         if(vis[u])continue;
 30         son[rt].push_back(u);
 31         d[u] = d[rt]+f[rt][i].w;
 32         dfs(u);
 33     }
 34 }
 35 int findd(int x)
 36 {
 37     int ret = x;
 38     while(ret!=fa[ret])ret=fa[ret];
 39     while(x!=ret)
 40     {
 41         int t = fa[x];
 42         fa[x] = ret;
 43         x = t;
 44     }
 45     return ret;
 46 }
 47 int cnt;
 48 void lca(int rt)
 49 {
 50     for(int i = 0;i<son[rt].size();++i)
 51     {
 52         lca(son[rt][i]);
 53         fa[son[rt][i]] = rt;
 54     }
 55     vis[rt] = 1;
 56     for(int i = 0;i<q[rt].size();++i)
 57     {
 58         int u = q[rt][i].w;
 59         if(vis[u]==0)continue;
 60         ans[cnt].x = q[rt][i].x;
 61         ans[cnt++].w = d[rt]+d[u]-2*d[anc[findd(u)]];
 62     }
 63 }
 64 bool cmp(node a,node b)
 65 {
 66     return a.x<b.x;
 67 }
 68 int main()
 69 {
 70 //    freopen("in.txt","r",stdin);
 71     int T;scanf("%d",&T);
 72     while(T--)
 73     {
 74         int n,m;scanf("%d%d",&n,&m);
 75         memset(f,0,sizeof(f));
 76         memset(son,0,sizeof(son));
 77         memset(vis,0,sizeof(vis));
 78         memset(q,0,sizeof(q));
 79         memset(ans,0,sizeof(ans));
 80         for(int i = 1;i<=n;++i)fa[i] = i,anc[i] = i;
 81         for(int i = 1;i<n;++i)
 82         {
 83             int x,y,w;scanf("%d%d%d",&x,&y,&w);
 84             f[x].push_back(node(y,w));
 85             f[y].push_back(node(x,w));
 86         }
 87         d[0] = 0;dfs(1);memset(vis,0,sizeof(vis));
 88         for(int i = 1;i<=m;++i)
 89         {
 90             int x,y;scanf("%d%d",&x,&y);
 91             q[x].push_back(node(i,y));
 92             q[y].push_back(node(i,x));
 93         }
 94         cnt = 0;
 95         lca(1);
 96         sort(ans,ans+m,cmp);
 97         for(int i = 0;i<cnt;++i)
 98             printf("%d\n",ans[i].w);
 99     }
100     return 0;
101 }

 

posted on 2015-05-07 00:34  round_0  阅读(116)  评论(0)  编辑  收藏  举报

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