POJ - 1986 Distance Queries (Tarjian求LCA）

​​POJ - 1986 Distance Queries​​

題目

分析

代碼

#include <cstdio>
#include <iostream>
#include <cmath>
#include <vector>
#include <algorithm>
#include <cstring>
using namespace std;
#define d(x) cout<<(x)<<endl;
const int N = 1e5 + 10;

int n, m, q, x, y, z, r;
char str[10];

struct node{
    int v, l;
    node(int v, int l):v(v),l(l){}
};

vector<node> tr[N];         // 存圖
vector<node> ask[N];        // 存詢問， l 代表順序

int dis[N], fa[N], ans[N], vis[N], rt[N];
int flag[N];

int find(int x){
    return fa[x] = fa[x] == x ? x : find(fa[x]);
}

void tarjian(int rt){
    for (vector<node>::iterator it = tr[rt].begin(); it != tr[rt].end(); it++){
        node x = *it;
        if(flag[x.v])            //不通路父親節點
            continue;
        flag[x.v] = 1;
        dis[x.v] = dis[rt] + x.l;
        tarjian(x.v);
        fa[x.v] = rt;
    }
    vis[rt] = 1;
    for(vector<node>::iterator it = ask[rt].begin(); it != ask[rt].end(); it++){
        node x = *it;
        if(vis[x.v]){
            ans[x.l] = dis[x.v] + dis[rt] - 2 * dis[find(x.v)];
        }
    }
}

int main()
{
    scanf("%d%d", &n, &m);
    for (int i = 0; i < m; i++){
        scanf("%d%d%d%s", &x, &y, &z, str);
        tr[x].push_back(node(y, z));
        tr[y].push_back(node(x, z));
        // rt[y]++;
    }
    scanf("%d", &q);
    for (int i = 0; i < q; i++){
        scanf("%d%d", &x, &y);
        ask[x].push_back(node(y, i));
        ask[y].push_back(node(x, i));
    }
    for(int i = 1; i <= n; i++){
        fa[i] = i;
        if(!rt[i])
            r = i;
    }
    flag[1] = 1;
    tarjian(1);             //轉有向樹
    for(int i = 0; i < q; i++){
        printf("%d\n", ans[i]);
    }
    return 0;
}