題目連接
題意：
求一棵樹上任意兩點的最短距離，n個點，m次詢問
數據範圍：
2 < n < 40000, 2 < m < 200
思路：(Tarjan)
LCA板子題，需要一個dis [ ] 數組， dis[ x ] 記錄一下該點 ( x ) 到父親節點 ( find(x) ) 的距離 ，當找到一個匹配點 ( to ) 時，兩個點的最短距離 = （dis[ x ] + dis[ to ] - dis[ find(x) ] * 2）;
補習：點擊補習LCA
TarjanAC:
#include<iostream>
#include<cstring>
#include<math.h>
#include<algorithm>
using namespace std;
const int MAXN = 4e4 + 10;
int n, m;
int head[MAXN], head_ask[MAXN], cnt, cnt_ask;
int ans[MAXN], fa[MAXN];
bool vis[MAXN];
int dis[MAXN]; //保存該節點到父親的距離
struct Edge {
int to, next, dis;
int num;
}edge[MAXN << 1], ask[MAXN << 1];
void add_edge(int u, int v, int dis){
edge[++cnt].to = v;
edge[cnt].next = head[u];
edge[cnt].dis = dis;
head[u] = cnt;
}
void add_ask(int u, int v, int num) {
ask[++cnt_ask].to = v;
ask[cnt_ask].num = num;
ask[cnt_ask].next = head_ask[u];
head_ask[u] = cnt_ask;
}
int find(int x) {
return fa[x] == x ? x : fa[x] = find(fa[x]);
}
void init() {
cnt = 1;
cnt_ask = 1;
memset(vis, 0, sizeof(vis));
memset(dis, 0, sizeof(dis));
memset(head,0, sizeof(head));
memset(head_ask,0, sizeof(head_ask));
fa[n] = n;
}
void Tarjan(int x) {
vis[x] = true;
for(int i = head[x]; i; i = edge[i].next) {
int to = edge[i].to;
if( !vis[to] ) {
dis[to] = dis[x] + edge[i].dis;
Tarjan(to);
fa[to] = find(x);
}
}
for(int i = head_ask[x]; i; i = ask[i].next) {
int to = ask[i].to;
if (vis[to]) {
ans[ask[i].num] = dis[to] + dis[x] - dis[find(to)] * 2;
}
}
}
int main () {
int T;
cin >> T;
while(T--) {
cin >> n >> m;
int x, y, z;
init();
for(int i = 1; i < n; ++i) {
fa[i] = i;
cin >> x >> y >> z;
add_edge(x, y, z);
add_edge(y, x, z);
}
for(int i = 1; i <= m; ++i) {
cin >> x >> y;
add_ask(x, y, i);
add_ask(y, x, i);
}
Tarjan(1);
for(int i = 1; i <= m; ++i) {
cout << ans[i] << endl;
}
}
return 0;
}
倍增：
#include<bits/stdc++.h>
using namespace std;
const int MAXN = 40000 + 10;
int head[MAXN], cnt;
int fa[MAXN][20], depth[MAXN], dis[MAXN]; //記錄到根節點的距離
int n, m;
struct Edge{
int to, dis, next;
}edge[MAXN << 1];
void add_edge(int u, int v, int dis) {
edge[++cnt].to = v;
edge[cnt].dis = dis;
edge[cnt].next = head[u];
head[u] = cnt;
}
void init () {
cnt = 1;
memset(head, 0, sizeof(head));
memset(fa, 0, sizeof(fa));
memset(depth, 0, sizeof(depth));
memset(dis, 0, sizeof(dis));
}
void DFS(int now, int pre) {
depth[now] = depth[pre] + 1;
fa[now][0] = pre;
for (int i = 1; (1 << i) <= depth[now]; ++i) {
fa[now][i] = fa[fa[now][i-1]][i-1];
}
for(int i = head[now]; i; i = edge[i].next) {
int to = edge[i].to;
if(to != pre){
dis[to] = dis[now] + edge[i].dis;
DFS(edge[i].to, now);
}
}
}
int LCA(int a, int b) {
if(depth[a] < depth[b]) swap(a, b);
int dep = depth[a] - depth[b];
for(int i = 0; (1 << i) <= dep; ++i) {
if ((1 << i) & dep)
a = fa[a][i];
}
if(a == b) return a;
for (int i = log2(depth[a]); i >= 0; --i) {
//如果父親不同就向上跳， 如果父親相同就減小距離再比較，直到不相同在跳。
if (fa[a][i] != fa[b][i]) {
a = fa[a][i];
b = fa[b][i];
}
}
return fa[a][0];
}
int main () {
int T;
cin >> T;
while(T--) {
init();
cin >> n >> m;
int x, y, z;
for(int i = 1; i <= n - 1; ++i) {
cin >> x >> y >> z;
add_edge(x, y, z);
add_edge(y, x, z);
}
DFS(1, 0);
for(int i = 1; i <= m; ++i) {
cin >> x >> y;
int temp = LCA(x, y);
int ans = dis[x] + dis[y] - 2 * dis[temp];
cout << ans << endl;
}
}
return 0;
}