POJ 1986 Distance Queries(LCA Tarjan法)

Distance Queries

【题目链接】Distance Queries

【题目类型】LCA Tarjan法

&题意：

输入n和m，表示n个点m条边，下面m行是边的信息，两端点和权，后面的那个字母无视掉，没用的。接着k，下面k个询问lca，输出即可

&题解：

首先看的这个 http://www.cnblogs.com/JVxie/p/4854719.html 大致懂了方法,之后又找了这个代码 http://blog.csdn.net/lianai911/article/details/42300301 就懂了一些.

【时间复杂度】(O(n+q))

&代码：

#include <cstdio>
#include <bitset>
#include <iostream>
#include <set>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <map>
#include <queue>
#include <vector>
using namespace std;
#define INF 0x3f3f3f3f
#define ll long long
#define fo(i,a,b) for(int i=(a);i<=(b);i++)
#define fd(i,a,b) for(int i=(a);i>=(b);i--)
const int maxn = 80000 + 7;
int fa[maxn], Head[maxn], QHead[maxn], Dist[maxn];
struct Edge {
	int to, next, w;
} edges[maxn], Qedges[maxn];
int find(int x) { return fa[x] == x ? x : fa[x] = find(fa[x]); }
bool vis[maxn];
void LCA(int u) {
	fa[u] = u;
	vis[u] = true;
	for(int k = Head[u]; k != -1; k = edges[k].next) {
		if(!vis[edges[k].to]) {
			Dist[edges[k].to] = Dist[u] + edges[k].w;
			LCA(edges[k].to);
			fa[edges[k].to] = u;
		}
	}
	for(int k = QHead[u]; k != -1; k = Qedges[k].next) {
		if(vis[Qedges[k].to]) {
			Qedges[k].w = Dist[u] + Dist[Qedges[k].to] - 2 * Dist[find(Qedges[k].to)];
		}
	}
}
int id, iq;
void ori() {
	memset(fa, 0, sizeof(fa));
	memset(vis, 0, sizeof(vis));
	memset(Head, -1, sizeof(Head));
	memset(QHead, -1, sizeof(QHead));
	memset(edges, 0, sizeof(edges));
	memset(Qedges, 0, sizeof(Qedges));
	id = 0;
	iq = 0;
}
void Add(int x, int y, int z, int Head[], int& id, Edge edges[]) {
	edges[id].to = y;
	edges[id].w = z;
	edges[id].next = Head[x];
	Head[x] = id++;
}
int main() {
	freopen("E:1.in", "r", stdin);
	int n, m, u, v, w;
	char c;
	while(~scanf("%d%d", &n, &m)) {
		ori();
		fo(i, 1, m) {
			scanf("%d%d%d %c", &u, &v, &w, &c);
			Add(u, v, w, Head, id, edges);
			Add(v, u, w, Head, id, edges);
		}
		scanf("%d", &m);
		fo(i, 1, m) {
			scanf("%d%d", &u, &v);
			Add(u, v, 0, QHead, iq, Qedges);
			Add(v, u, 0, QHead, iq, Qedges);
		}
		LCA(1);
		for(int i = 0; i < iq; i += 2)
			printf("%d
", max(Qedges[i].w, Qedges[i + 1].w));
	}
	return 0;
}