POJ 1986 — Distance Queries

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原题：http://poj.org/problem?id=1986

题意：有n个点，m条边，下面m行给定一棵树（后面那个字母并没有什么卵用）；

Q个询问，求两点间最短距离；

话说，我并没有考虑不连通的情况，然而也过了……另外，注意数组大小；

#include<stdio.h>#include<string.h>#include<queue>#include<iostream>#include<algorithm>using namespace std;const int N = 100005;int head[N], fa[N][20], dep[N];int dis[N];int n, m, e;queue<int>q;struct node{int to, nex, val;}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++;}void bfs(int root){fa[root][0] = root;dis[root] = 0;dep[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;dis[v] = dis[u]+edge[i].val;fa[v][0] = u;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", &n, &m)!=EOF){e = 0;memset(head, -1, sizeof(head));while(m--){int u, v, w;char c;scanf("%d%d%d%c", &u, &v, &w, &c);getchar();add(u, v, w);add(v, u, w);}bfs(1);int Q;scanf("%d", &Q);while(Q--){int u, v;scanf("%d%d", &u, &v);printf("%d\n", dis[u]+dis[v]-2*dis[lca(u, v)]);}}return 0;}

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