多一条边的树的最短路径（倍增lca）
https://codeforces.com/gym/101808/problem/K
题意：给出n个点n条边的无向连通图，m次询问u、v间的最短路径。
解法：n条边减去一条即为树。所以找出成环的边为a , b , l ;
u 、 v两点间有三种路径取最短即可：
1、dis(u , v)
2、dis(u , a)+l+dis(b , v)
3、dis(u , b)+l+dis(a , v)；
倍增求最短路。
//#include<bits/stdc++.h>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <string>
#include <stdio.h>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <string.h>
#include <vector>
typedef long long ll ;
#define mod 1000000007
#define gcd __gcd
#define rep(i , j , n) for(int i = j ; i <= n ; i++)
#define red(i , n , j) for(int i = n ; i >= j ; i--)
#define ME(x , y) memset(x , y , sizeof(x))
//ll lcm(ll a , ll b){return a*b/gcd(a,b);}
//ll quickpow(ll a , ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;b>>=1,a=a*a%mod;}return ans;}
//int euler1(int x){int ans=x;for(int i=2;i*i<=x;i++)if(x%i==0){ans-=ans/i;while(x%i==0)x/=i;}if(x>1)ans-=ans/x;return ans;}
//const int N = 1e7+9; int vis[n],prime[n],phi[N];int euler2(int n){ME(vis,true);int len=1;rep(i,2,n){if(vis[i]){prime[len++]=i,phi[i]=i-1;}for(int j=1;j<len&&prime[j]*i<=n;j++){vis[i*prime[j]]=0;if(i%prime[j]==0){phi[i*prime[j]]=phi[i]*prime[j];break;}else{phi[i*prime[j]]=phi[i]*phi[prime[j]];}}}return len}
#define SC scanf
#define INF 0x3f3f3f3f
#define PI acos(-1)
#define pii pair<int,int>
#define fi first
#define se second
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define int ll
using namespace std;
const int maxn = 1e5+9;
int head[maxn] , tol , fa[maxn][22] , de[maxn] , dis[maxn] , f[maxn];
int a , b , l;
int n , m;
struct node{
int to , next , w;
}g[maxn<<1];
void add(int u , int v , int w){
g[++tol] = {v , head[u] , w};
head[u] = tol;
}
int find(int x){
return x == f[x] ? x : find(f[x]);
}
void unite(int x , int y){
x = find(x) , y = find(y);
f[x] = y;
}
void init(){
ME(head , 0);
ME(fa , 0);
ME(de , 0);
tol = 0;
rep(i , 1 , n){
f[i] = i ;
}
}
void dfs(int u , int pre , int d){
de[u] = de[pre]+1;
dis[u] = d ;
fa[u][0] = pre;
for(int i = 1 ; (1<<i) <= de[u] ; i++){
fa[u][i] = fa[fa[u][i-1]][i-1];
}
for(int i = head[u] ; i ; i = g[i].next){
int v = g[i].to;
if(v == pre) continue;
dfs(v , u , d+g[i].w);
}
}
int lca(int u , int v){
if(de[u] < de[v]) swap(u , v);
red(i , 20 , 0){
if(de[u] - (1<<i) >= de[v]){
u = fa[u][i];
}
}
if(u == v) return v ;
red(i , 20 , 0){
if(fa[u][i] != fa[v][i]){
u = fa[u][i];
v = fa[v][i];
}
}
return fa[u][0];
}
int get_dis(int u , int v){
return dis[u] + dis[v] - 2 * dis[lca(u , v)];
}
void solve(){
cin >> n >> m ;
init();
rep(i , 1 , n){
int u , v , w;
cin >> u >> v >> w ;
if(find(u) != find(v)){
add(u , v , w);
add(v , u , w);
unite(u , v);
}else{
a = u , b = v ;
l = w ;
}
}
dfs(1 , 0 , 0);
rep(i , 1 , m){
int u , v ;
cin >> u >> v ;
int ans = get_dis(u , v);
ans = min(ans , get_dis(u , a)+l+get_dis(b , v));
ans = min(ans , get_dis(u , b)+l+get_dis(a , v));
cout << ans << endl;
}
}
signed main()
{
ios::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
int t ;
cin >> t ;
while(t--){
solve();
}
}
https://codeforces.com/contest/1304/problem/E
题意：给出一n节点的颗树，m此询问：增加一条边a到b，能否满足u经过w条边（每条边可以走任意次）最终到达v。
解法：与上一题一样，只是稍稍该变问的方式，只要三条最短路径的一条dis<=w&&dis%2==w%2，则满足条件。
//#include<bits/stdc++.h>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <string>
#include <stdio.h>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <string.h>
#include <vector>
typedef long long ll ;
#define mod 1000000007
#define gcd __gcd
#define rep(i , j , n) for(int i = j ; i <= n ; i++)
#define red(i , n , j) for(int i = n ; i >= j ; i--)
#define ME(x , y) memset(x , y , sizeof(x))
//ll lcm(ll a , ll b){return a*b/gcd(a,b);}
//ll quickpow(ll a , ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;b>>=1,a=a*a%mod;}return ans;}
//int euler1(int x){int ans=x;for(int i=2;i*i<=x;i++)if(x%i==0){ans-=ans/i;while(x%i==0)x/=i;}if(x>1)ans-=ans/x;return ans;}
//const int N = 1e7+9; int vis[n],prime[n],phi[N];int euler2(int n){ME(vis,true);int len=1;rep(i,2,n){if(vis[i]){prime[len++]=i,phi[i]=i-1;}for(int j=1;j<len&&prime[j]*i<=n;j++){vis[i*prime[j]]=0;if(i%prime[j]==0){phi[i*prime[j]]=phi[i]*prime[j];break;}else{phi[i*prime[j]]=phi[i]*phi[prime[j]];}}}return len}
#define SC scanf
#define INF 0x3f3f3f3f
#define PI acos(-1)
#define pii pair<int,int>
#define fi first
#define se second
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define int ll
using namespace std;
const int maxn = 1e5+9;
int head[maxn] , tol , fa[maxn][22] , de[maxn] , f[maxn];
int n , m;
struct node{
int to , next;
}g[maxn<<1];
void add(int u , int v){
g[++tol] = {v , head[u]};
head[u] = tol;
}
void init(){
ME(head , 0);
ME(fa , 0);
ME(de , 0);
tol = 0;
rep(i , 1 , n){
f[i] = i ;
}
}
void dfs(int u , int pre){
de[u] = de[pre]+1;
fa[u][0] = pre;
for(int i = 1 ; (1<<i) <= de[u] ; i++){
fa[u][i] = fa[fa[u][i-1]][i-1];
}
for(int i = head[u] ; i ; i = g[i].next){
int v = g[i].to;
if(v == pre) continue;
dfs(v , u);
}
}
int lca(int u , int v){
if(de[u] < de[v]) swap(u , v);
red(i , 20 , 0){
if(de[u] - (1<<i) >= de[v]){
u = fa[u][i];
}
}
if(u == v) return v ;
red(i , 20 , 0){
if(fa[u][i] != fa[v][i]){
u = fa[u][i];
v = fa[v][i];
}
}
return fa[u][0];
}
int get_dis(int u , int v){
return de[u] + de[v] - 2 * de[lca(u , v)];
}
void solve(){
cin >> n ;
init();
rep(i , 1 , n-1){
int u , v;
cin >> u >> v ;
add(u , v);
add(v , u);
}
int m ;cin >> m;
dfs(1 , 0);
rep(i , 1 , m){
int u , v , a , b , w;
cin >> a >> b >> u >> v >> w;
int dis1 = get_dis(u , v);
int dis2 = get_dis(u , a)+1+get_dis(b , v);
int dis3 = get_dis(u , b)+1+get_dis(a , v);
if((dis1 <= w && dis1%2==w%2) ||(dis2 <= w && dis2%2==w%2) || (dis3 <= w && dis3%2==w%2)){
cout << "YES" << endl;
}else{
cout << "NO" << endl;
}
}
}
signed main()
{
ios::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
//int t ;
//cin >> t ;
//while(t--){
solve();
//}
}
