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; }

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