poj 1986 Distance Queries(LCA离线Tarjan算法)

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题目同poj 2586相似，查询两点之间最短距离。
#include <iostream>#include <string.h>#include <stdio.h>#define NN 40002 // number of house#define MM 40002   // number of queryusing namespace std;typedef struct node{    int v;    int d;    struct node *nxt;}NODE;NODE *Link1[NN];NODE edg1[NN * 2]; // 树中的边NODE *Link2[NN];NODE edg2[NN * 2]; // 询问的点对int idx1, idx2, N, M,Q;int res[MM][3]; // 记录结果，res[i][0]: u   res[i][1]: v  res[i][2]: lca(u, v)int fat[NN];int vis[NN];int dis[NN];void Add(int u, int v, int d, NODE edg[], NODE *Link[], int &idx){//头插法建立邻接表    edg[idx].v = v;    edg[idx].d = d;    edg[idx].nxt = Link[u];    Link[u] = edg + idx++;    edg[idx].v = u;    edg[idx].d = d;    edg[idx].nxt = Link[v];    Link[v] = edg + idx++;}int find(int x){ // 并查集路径压缩    if(x != fat[x]){        return fat[x] = find(fat[x]);    }    return x;}void Tarjan(int u){    vis[u] = 1;    fat[u] = u;    for (NODE *p = Link2[u]; p; p = p->nxt){        if(vis[p->v]){            res[p->d][2] = find(p->v); // 存的是最近公共祖先结点        }    }    for (NODE *p = Link1[u]; p; p = p->nxt){        if(!vis[p->v]){            dis[p->v] = dis[u] + p->d;            Tarjan(p->v);            fat[p->v] = u;        }    }}int main() {    int T, i, u, v, d;    char c;      scanf("%d%d", &N, &M);        idx1 = 0;        memset(Link1, 0, sizeof(Link1));        for (i = 1; i <=M; i++){            scanf("%d%d%d%c", &u, &v, &d,&c);            getchar();            Add(u, v, d, edg1, Link1, idx1);        }         scanf("%d",&Q);        idx2 = 0;        memset(Link2, 0, sizeof(Link2));        for (i = 1; i <= Q; i++){            scanf("%d%d", &u, &v);            Add(u, v, i, edg2, Link2, idx2);            res[i][0] = u;            res[i][1] = v;        }        memset(vis, 0, sizeof(vis));        dis[1] = 0;        Tarjan(1);        for (i = 1; i <= Q; i++){            printf("%d\n", dis[res[i][0]] + dis[res[i][1]] - 2 * dis[res[i][2]]);        }    return 0;}
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