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POJ 1986 Distance Queries LCA树上两点的距离
题目来源:POJ 1986 Distance Queries
题意:给你一颗树 q次询问 每次询问你两点之间的距离
思路:对于2点 u v dis(u,v) = dis(root,u) + dis(root,v) - 2*dis(roor,LCA(u,v)) 求最近公共祖先和dis数组
#include <cstdio> #include <cstring> #include <vector> using namespace std; const int maxn = 40010; int first[maxn], head[maxn], cnt, sum; struct edge { int u, v, w, next; }e[maxn*2], qe[maxn], Q[maxn]; int ans[maxn]; int f[maxn], vis[maxn]; int d[maxn]; void AddEdge(int u, int v, int w) { e[cnt].u = u; e[cnt].v = v; e[cnt].w = w; e[cnt].next = first[u]; first[u] = cnt++; e[cnt].u = v; e[cnt].v = u; e[cnt].w = w; e[cnt].next = first[v]; first[v] = cnt++; } void AddEdge2(int u, int v, int w) { qe[sum].u = u; qe[sum].v = v; qe[sum].w = w; qe[sum].next = head[u]; head[u] = sum++; qe[sum].u = v; qe[sum].v = u; qe[sum].w = w; qe[sum].next = head[v]; head[v] = sum++; } int find(int x) { if(f[x] != x) return f[x] = find(f[x]); return f[x]; } void LCA(int u, int k) { f[u] = u; d[u] = k; vis[u] = true; for(int i = first[u]; i != -1; i = e[i].next) { int v = e[i].v; if(vis[v]) continue; LCA(v, k + e[i].w); f[v] = u; } for(int i = head[u]; i != -1; i = qe[i].next) { int v = qe[i].v; if(vis[v]) { ans[qe[i].w] = find(v); } } } int main() { int n, m; memset(first, -1, sizeof(first)); memset(head, -1, sizeof(head)); cnt = 0; sum = 0; scanf("%d %d", &n, &m); for(int i = 0; i < m; i++) { int u, v, w; char s[10]; scanf("%d %d %d %s", &u, &v, &w, s); AddEdge(u, v, w); } int q; scanf("%d", &q); for(int i = 0; i < q; i++) { int u, v; scanf("%d %d", &u, &v); Q[i].u = u, Q[i].v = v; AddEdge2(u, v, i); AddEdge2(v, u, i); } memset(vis, 0, sizeof(vis)); d[1] = 0; LCA(1, 0); for(int i = 0; i < q; i++) { int u = Q[i].u, v = Q[i].v; printf("%d\n", d[u] + d[v] - 2*d[ans[i]]); } return 0; }
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