生活随笔
收集整理的這篇文章主要介紹了
hdu 2874(LCA + 节点间距离)
小編覺得挺不錯的,現在分享給大家,幫大家做個參考.
解題思路：Tarjan離線處理
一篇介紹LCA的很好的博客：http://www.cppblog.com/menjitianya/archive/2015/12/10/212447.html
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;const int maxn = 10005;
struct Edge
{int k,next,cost;
}edge[maxn<<1];
struct Ask
{int k,next,id;
}ask[maxn<<1];
int n,m,c,cnt1,cnt2,pre[maxn],head[maxn];
int color[maxn],ans[maxn],dis[maxn],fa[maxn];
bool vis[maxn];int find(int x)
{if(fa[x] == x) return x;else return fa[x] = find(fa[x]);
}void addedge(int u,int v,int cost)
{edge[cnt1].k = v;edge[cnt1].cost = cost;edge[cnt1].next = pre[u];pre[u] = cnt1++;
}void addask(int u,int v,int id)
{ask[cnt2].id = id;ask[cnt2].k = v;ask[cnt2].next = head[u];head[u] = cnt2++;
}void Tarjan(int u,int root,int len)
{vis[u] = true;fa[u] = u;dis[u] = len;for(int i = pre[u]; i != -1; i = edge[i].next){int v = edge[i].k;if(vis[v]) continue;Tarjan(v,root,len + edge[i].cost);fa[v] = u;}color[u] = root;for(int i = head[u]; i != -1; i = ask[i].next){int v = ask[i].k;if(color[v] == root){int ancestor = find(v);ans[ask[i].id] = dis[u] + dis[v] - 2*dis[ancestor];}}
}int main()
{int u,v,cost;while(scanf("%d%d%d",&n,&m,&c)!=EOF){memset(pre,-1,sizeof(pre));memset(head,-1,sizeof(head));cnt1 = cnt2 = 0;for(int i = 1; i <= m; i++){scanf("%d%d%d",&u,&v,&cost);addedge(u,v,cost);addedge(v,u,cost);}for(int i = 1; i <= c; i++){scanf("%d%d",&u,&v);addask(u,v,i);addask(v,u,i);}memset(vis,false,sizeof(vis));memset(color,0,sizeof(color));memset(ans,-1,sizeof(ans));for(int i = 1; i <= n; i++)if(vis[i] == false)Tarjan(i,i,0);for(int i = 1; i <= c; i++){if(ans[i] == -1)printf("Not connected\n");else printf("%d\n",ans[i]);}}return 0;
}
總結
以上是生活随笔為你收集整理的hdu 2874(LCA + 节点间距离)的全部內容，希望文章能夠幫你解決所遇到的問題。
如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。
