POJ 1986 Distance Queries

http://poj.org/problem?id=1986

题意：一棵树里找到两个点的距离。（不用考虑不联通的情况）

题解：LCA模板题。

  1 #include <iostream>
  2 #include <algorithm>
  3 #include <cstring>
  4 #include <string>
  5 #include <cstdio>
  6 #include <cmath>
  7 #include <queue>
  8 #include <map>
  9 #include <set>
 10 
 11 #define eps 1e-5
 12 #define MAXN 55555
 13 #define MAXM 111111
 14 #define INF 1000000000
 15 using namespace std;
 16 int n, m, q;
 17 struct edge
 18 {
 19     int v, next, w;
 20 }es[MAXM];
 21 int head[MAXN], e;
 22 int ide,tmpdfn;
 23 int f[2 * MAXN], id[MAXN], used[MAXN], pos[MAXN], dis[MAXN];
 24 int mi[2 * MAXN][18];
 25 
 26 void init()
 27 {
 28     memset(head, -1, sizeof(head));
 29     e = 0;
 30     ide = tmpdfn = 0;
 31     memset(used, 0, sizeof(used));
 32     dis[1] = 0;
 33 }
 34 
 35 void add(int u, int v, int w)
 36 {
 37     es[e].v = v;
 38     es[e].w = w;
 39     es[e].next = head[u];
 40     head[u] = e++;
 41 }
 42 void dfs(int u)
 43 {
 44     used[u] = 1;
 45     int tmp = ++tmpdfn;
 46     f[++ide] = tmp;
 47     id[tmp] = u;
 48     pos[u] = ide;
 49     for(int i = head[u]; i != -1; i = es[i].next)
 50     {
 51         int v = es[i].v;
 52         if(!used[v])
 53         {
 54             dis[v] = dis[u] + es[i].w;
 55             dfs(v);
 56             f[++ide] = tmp;
 57         }
 58     }
 59     
 60 }
 61 void rmqinit(int n, int *w)
 62 {
 63     for(int i = 1; i <= n; i++) mi[i][0] = w[i];
 64     int m = (int)(log(n * 1.0) / log(2.0));
 65     for(int i = 1; i <= m; i++)
 66         for(int j = 1; j <= n; j++)
 67         {
 68             mi[j][i] = mi[j][i - 1];
 69             if(j + (1 << (i - 1)) <= n) mi[j][i] = min(mi[j][i], mi[j + (1 << (i - 1))][i - 1]);
 70         }
 71 }
 72 
 73 int rmqmin(int l,int r)
 74 {
 75     int m = (int)(log((r - l + 1) * 1.0) / log(2.0));
 76     return min(mi[l][m] , mi[r - (1 << m) + 1][m]);
 77 }
 78 
 79 int LCA(int l, int r)
 80 {
 81     if(pos[l] > pos[r]) swap(l, r);
 82     int ans = rmqmin(pos[l], pos[r]);
 83     return id[ans];
 84 }
 85 int main()
 86 {
 87    // freopen("/Users/apple/Desktop/题/POJ 1986_1/POJ 1986_1/in","r",stdin);
 88     scanf("%d%d", &n, &m);
 89     int u, v, w, l, r;
 90     init();
 91     while(m--)
 92     {
 93         scanf("%d%d%d%*s", &u, &v, &w);
 94         add(u, v, w);
 95         add(v, u, w);
 96     }
 97     dfs(1);
 98     rmqinit(ide, f);
 99     scanf("%d", &q);
100     while(q--)
101     {
102         scanf("%d%d", &l, &r);
103         printf("%d\n", dis[l] + dis[r] - 2 * dis[LCA(l, r)]);
104     }
105     return 0;
106 }

 

posted @ 2014-10-06 10:09  Der_Z  阅读(197)  评论(0)  编辑  收藏  举报