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题意：给定N个节点一棵树，现在要求询问任意两点之间的简单路径的距离，其实也就是最短路径距离。
析：用LCA问题的Tarjan算法，利用并查集的优越性，产生把所有的点都储存下来，然后把所有的询问也储存下来，然后从树根开始搜索这棵树，
在搜索子树的时候，把并查集的父结点不断更新，在搜索时计算答案，d[i] 表示从根结点到 i 结点的距离，然后分别计算到两个结点的距离，
再减去2倍的根结点到他们公共最近的结点距离，就OK了。不过这个题有个坑人的地方，不用输出那个空行，否则就是PE。
因为这个题询问比较少，所以我们可以用spfa来做。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 4e4 + 5;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
return r >= 0 && r < n && c >= 0 && c < m;
}
struct Edge{
int to, w, next;
};
struct node{
int to, id, next;
};
Edge a[maxn<<1];
node b[405];
int p[maxn], head[maxn<<1], cnt, cnt1, head1[maxn];
int rec[205][3], d[maxn];
bool vis[maxn];
int Find(int x){ return x == p[x] ? x : p[x] = Find(p[x]); }
void add(int u, int v, int w){
a[cnt].to = v;
a[cnt].w = w;
a[cnt].next = head[u];
head[u] = cnt++;
}
void add1(int u, int v, int id){
b[cnt1].to = v;
b[cnt1].id = id;
b[cnt1].next = head1[u];
head1[u] = cnt1++;
}
void Tarjan(int u){
vis[u] = true;
for(int i = head1[u]; ~i; i = b[i].next){
int v = b[i].to;
if(vis[v]) rec[b[i].id][2] = Find(v);
}
for(int i = head[u]; ~i; i = a[i].next){
int v = a[i].to;
if(!vis[v]){
d[v] = d[u] + a[i].w;
Tarjan(v);
p[v] = u;
}
}
}
int main(){
int T; cin >> T;
while(T--){
scanf("%d %d", &n, &m);
for(int i = 0; i <= n; ++i) p[i] = i;
int u, v, w;
cnt = cnt1 = 0;
memset(head, -1, sizeof head);
memset(head1, -1, sizeof head1);
for(int i = 1; i < n; ++i){
scanf("%d %d %d", &u, &v, &w);
add(u, v, w);
add(v, u, w);
}
for(int i = 0; i < m; ++i){
scanf("%d %d", &u, &v);
add1(u, v, i);
add1(v, u, i);
rec[i][0] = u;
rec[i][1] = v;
}
memset(vis, false, sizeof vis);
d[1] = 0;
Tarjan(1);
for(int i = 0; i < m; ++i)
printf("%d\n", d[rec[i][0]] + d[rec[i][1]] - 2*d[rec[i][2]]);
//printf("\n");
}
return 0;
}
第二种写法：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 4e4 + 5;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
return r >= 0 && r < n && c >= 0 && c < m;
}
vector<int> G[maxn], w[maxn], q[maxn], id[maxn];
int d[maxn], p[maxn], ans[205];
bool vis[maxn];
int Find(int x){ return x == p[x] ? x : p[x] = Find(p[x]); }
void Tarjan(int u){
vis[u] = true;
for(int i = 0; i < G[u].size(); ++i){
int v = G[u][i];
if(!vis[v]){
d[v] = d[u] + w[u][i];
Tarjan(v);
p[v] = u;
}
}
for(int i = 0; i < q[u].size(); ++i){
int v = q[u][i];
if(vis[v]) ans[id[u][i]] = d[u] + d[v] - 2*d[Find(v)];
}
}
int main(){
int T;; cin >> T;
while(T--){
scanf("%d %d", &n, &m);
for(int i = 1; i <= n; ++i) G[i].clear(), w[i].clear(), q[i].clear(), id[i].clear(), p[i] = i;
int u, v, val;
for(int i = 1; i < n; ++i){
scanf("%d %d %d", &u, &v, &val);
G[u].push_back(v);
G[v].push_back(u);
w[u].push_back(val);
w[v].push_back(val);
}
for(int i = 0; i < m; ++i){
scanf("%d %d", &u, &v);
q[u].push_back(v);
q[v].push_back(u);
id[u].push_back(i);
id[v].push_back(i);
}
memset(vis, false, sizeof vis);
d[1] = 0;
Tarjan(1);
for(int i = 0; i < m; ++i)
printf("%d\n", ans[i]);
}
return 0;
}
spfa:
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 4e4 + 5;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
return r >= 0 && r < n && c >= 0 && c < m;
}
struct node{
int to, w, next;
};
node a[maxn<<1];
int cnt;
int head[maxn];
bool vis[maxn];
int d[maxn];
int q[maxn];
void add(int u, int v, int w){
a[cnt].to = v;
a[cnt].w = w;
a[cnt].next = head[u];
head[u] = cnt++;
}
int spfa(int s, int t){
memset(vis, false, sizeof vis);
fill(d, d+n+1, INF);
int front = 0, rear = 0;
q[rear++] = s;
d[s] = 0;
while(front < rear){
int u = q[front++];
vis[u] = false;
for(int i = head[u]; ~i; i = a[i].next){
int v = a[i].to;
if(d[v] > d[u] + a[i].w){
d[v] = d[u] + a[i].w;
if(!vis[v]) q[rear++] = v, vis[v] = true;
}
}
}
return d[t];
}
int main(){
int T; cin >> T;
while(T--){
scanf("%d %d", &n, &m);
int u, v, w;
cnt = 0;
memset(head, -1, sizeof head);
for(int i = 1; i < n; ++i){
scanf("%d %d %d", &u, &v, &w);
add(u, v, w);
add(v, u, w);
}
while(m--){
scanf("%d %d", &u, &v);
printf("%d\n", spfa(u, v));
}
}
return 0;
}
HDU 2586 How far away ？ (LCA，Tarjan， spfa)
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原文地址：http://www.cnblogs.com/dwtfukgv/p/5828131.html
