HDU - 2874 Connections between cities(LCA)

题目大意：给出N个点，M条线，Q个询问，询问的是两点之间的最短距离

解题思路：恶心的数据量，一不小心就超空间了
这题给图不是张连通图，是森林，所以计算两点之间的最短距离时还要考虑一下是否在同一棵树中

剩下的就是裸LCA了

#include <cstdio>
#include <cstring>

#define N 10010
#define M 20010
#define C 2000010

struct Edge{
    int to, next, dis;
}E[M];

struct Edge2{
    int to, next, w;
}E2[C];

int n, m, c, tot, tot2;
int head[N], head2[N], dist[N], f[N], vis[N];

void AddEdge(int u, int v, int dis) {
    E[tot].to = v; E[tot].next = head[u]; E[tot].dis = dis; head[u] = tot++;
    u = u ^ v; v = u ^ v; u = u ^ v;
    E[tot].to = v; E[tot].next = head[u]; E[tot].dis = dis; head[u] = tot++;
}

void AddEdge2(int u, int v) {
    E2[tot2].to = v; E2[tot2].next = head2[u]; E2[tot2].w = -1; head2[u] = tot2++;
    u = u ^ v; v = u ^ v; u = u ^ v;
    E2[tot2].to = v; E2[tot2].next = head2[u]; E2[tot2].w = -1; head2[u] = tot2++;
}

void init() {
    memset(head, -1, sizeof(head));
    memset(head2, -1, sizeof(head2));
    tot = tot2 = 0;

    int u, v, d;
    for (int i = 0; i < m; i++) {
        scanf("%d%d%d", &u, &v, &d);
        AddEdge(u, v, d);
    }

    for (int i = 0; i < c; i++) {
        scanf("%d%d", &u, &v);
        AddEdge2(u, v);
    }

    memset(vis, 0, sizeof(vis));
}

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

void tarjan(int u, int time) {
    vis[u] = time;
    f[u] = u;

    int v;
    for (int i = head[u]; i != -1; i = E[i].next) {
        v = E[i].to;
        if (vis[v])
            continue;
        dist[v] = dist[u] + E[i].dis;
        tarjan(v, time);
        f[v] = u;
    }

    for (int i = head2[u]; i != -1; i = E2[i].next) {
        v = E2[i].to;
        if (vis[v] == time) 
            E2[i].w = E2[i^1].w = dist[u] + dist[v] - 2 * dist[find(v)];
    }

}

void solve() {

    int cnt = 1;
    for (int i = 1; i <= n; i++) {
        if (!vis[i]) {
            dist[i] = 0;
            tarjan(i, cnt);
        }
        cnt++;
    }
    for (int i = 0; i < tot2; i += 2) {
        if (E2[i].w == -1)
            printf("Not connected\n");
        else  
            printf("%d\n", E2[i].w);
    }
}

int main() {
    while (scanf("%d%d%d", &n, &m, &c) != EOF) {
        init();
        solve();
    }
    return 0;
}