hdu2874_Connections between cities(tarjan/lca/边表)

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#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <vector>#include <map>#define INF 0x3f3f3f3f#define rep0(i, n) for (int i = 0; i < n; i++)#define rep1(i, n) for (int i = 1; i <= n; i++)#define rep_0(i, n) for (int i = n - 1; i >= 0; i--)#define rep_1(i, n) for (int i = n; i > 0; i--)#define MAX(x, y) (((x) > (y)) ? (x) : (y))#define MIN(x, y) (((x) < (y)) ? (x) : (y))#define mem(x, y) memset(x, y, sizeof(x))#define MAXN (10000 + 10)#define MAXC (1000000 + 10)/**题目大意给一个森林，多次询问节点之间的距离思路节点数的1e4 询问数1e6离线tarjan的话怎么存询问很关键用邻接表的边表存很省空间同时查询的复杂度很低，很巧妙当访问一个节点和它有询问关系的节点时很方便一个询问要存两次 但存放时是有序的q[0, 1,    2, 3,    4, 5,    6, 7...]没计算出一个答案可以通过 i >> 1 存入时询问的答案仍然有序HINT每次用宏定义表示数组大小总是被坑 一定记得加括号明明时下表越界用G++交却报称tle 很迷惑啊*/using namespace std;typedef long long LL;struct query{    int to;    int next;} q[MAXC * 2];LL head[MAXN], ans[MAXC], t;struct Edge{    int to, w;    int next;} edges[MAXN * 2];int g[MAXN];LL dis[MAXN];int n, root, pa[MAXN];int vis[MAXN];void addEdge(int u, int v, int w){    edges[t].to = v;    edges[t].w = w;    edges[t].next = g[u];    g[u] = t++;    edges[t].to = u;    edges[t].w = w;    edges[t].next = g[v];    g[v] = t++;}int find_set(int x){    if (pa[x] == x)        return x;    pa[x] = find_set(pa[x]);    return pa[x];}void tarjan(int u, int fa){    vis[u] = root;    for (int i = g[u]; i != -1; i = edges[i].next)    {        Edge e = edges[i];        int v = e.to;        if (v == fa)            continue;        dis[v] = dis[u] + e.w;        tarjan(v, u);        pa[v] = u;    }    for (int i = head[u]; i != -1; i = q[i].next)    {        int v = q[i].to;        if (vis[v] != root)            continue;        int lca = find_set(v);        ans[i >> 1] = dis[u] + dis[v] -  2 * dis[lca];    }}int main(){    #ifndef ONLINE_JUDGE        freopen("in.txt", "r", stdin);    #endif // ONLINE_JUDGE    int m;    LL c;    while (scanf("%d %d %lld", &n, &m, &c) != EOF)    {        mem(head, -1);        for (int i = 1; i <= n; i++)        {            g[i] = -1;            pa[i] = i;            vis[i] = 0;        }        int u, v;        int w;        t = 0;        for (int i = 0; i < m; i++)        {            scanf("%d %d %d", &u, &v, &w);            addEdge(u, v, w);        }        t = 0;        for (int i = 0; i < c; i++)        {            ans[i] = -1;            scanf("%d %d", &u, &v);            q[t].to = v;            q[t].next = head[u];            head[u] = t++;            q[t].to = u;            q[t].next = head[v];            head[v] = t++;        }        for (int i = 1; i <= n; i++)        {            if (vis[i] > 0)                continue;            root = i;            dis[i] = 0;            tarjan(i, -1);        }        for (int i = 0; i < c; i++)        {            if (ans[i] >= 0)                printf("%lld\n", ans[i]);            else                printf("Not connected\n");        }    }    return 0;}
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