HDU 2874 — Connections between cities

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原题：http://acm.hdu.edu.cn/showproblem.php?pid=2874

题意：有n个城市，m条路，接下来m行给出城市间距离；

有c个询问，若两点间不连通，输出 “Not connected”，否则输出两点间的最短距离；

思路：判断两点是否联通用并查集，最短路用LCA；

#include<stdio.h>#include<string.h>#include<queue>#include<iostream>#include<algorithm>using namespace std;const int N = 10005;int p[N];bool vis[N];int head[N], fa[N][20], dep[N], dis[N];int n, m, c, e;queue<int>q;struct node{int to, val, nex;}edge[N<<1];void add(int u, int v, int w){edge[e].to = v;edge[e].val = w;edge[e].nex = head[u];head[u] = e++;}int find(int x){if(x == p[x])return x;p[x] = find(p[x]);return p[x];}void Union(int x, int y){int px = find(x);int py = find(y);if(px!=py)p[px] = py;}void bfs(int root){fa[root][0] = root;dep[root] = 0;dis[root] = 0;q.push(root);while(!q.empty()){int u = q.front();q.pop();for(int i = 1;i<20;i++)fa[u][i] = fa[fa[u][i-1]][i-1];for(int i = head[u];i!=-1;i = edge[i].nex){int v = edge[i].to;if(v == fa[u][0])continue;dep[v] = dep[u]+1;fa[v][0] = u;dis[v] = dis[u]+edge[i].val;q.push(v);}}}int lca(int x, int y){if(dep[x]<dep[y])swap(x, y);for(int i = 0;i<20;i++){if((dep[x]-dep[y])&(1<<i))x = fa[x][i];}if(x == y)return x;for(int i = 19;i>=0;i--){if(fa[x][i]!=fa[y][i]){x = fa[x][i];y = fa[y][i];}}return fa[x][0];}int main(){while(scanf("%d%d%d", &n, &m, &c)!=EOF){e = 0;memset(head, -1, sizeof(head));memset(dis, 0, sizeof(dis));memset(vis, false, sizeof(vis));for(int i = 1;i<=n;i++)p[i] = i;while(m--){int a, b, w;scanf("%d%d%d", &a, &b, &w);add(a, b, w);add(b, a, w);Union(a, b);}for(int i = 1;i<=n;i++)p[i] = find(i);for(int i = 1;i<n;i++)vis[p[i]] = true;for(int i = 1;i<=n;i++)//将离线的树联通； {if(vis[i])add(1, i, 0);}bfs(1);while(c--){int a, b;scanf("%d%d", &a, &b);if(p[a]!=p[b])printf("Not connected\n");elseprintf("%d\n", dis[a]+dis[b]-2*dis[lca(a, b)]);}}return 0;}

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