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();
    //}
}