hdu2874(lca / tarjan离线 + RMQ在线)

Posted 2020-09-28 ygeloutingyu

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=2874

 

题意: 给出 n 个顶点 m 条边的一个森林, 有 k 个形如 x y 的询问, 输出 x, y 之间的最短路径.

 

思路: 如果将森林换成一棵树的话就是一道 lca 模板题了, 不过本题需要稍作改动.

 

解法1: tarjan

只需要先判断一下 x, y 是否在一颗树里面就 ok 了, 不过这道题的询问有点多, 很容易 mle.

代码:

  1 #include <iostream>
  2 #include <stdio.h>
  3 #include <string.h>
  4 using namespace std;
  5 
  6 const int MAXN = 1e4 + 10;
  7 const int MAX = 1e6 + 10;
  8 struct node1{
  9     int v, w, next;
 10 }edge1[MAXN << 1];
 11 
 12 struct node2{
 13     int u, v, next;
 14 }edge2[MAX << 1];//edge1记录树, edge2记录询问
 15 
 16 int vis[MAXN], pre[MAXN], dis[MAXN], sol[MAX];//vis[i]标记i是否已经搜索过, pre[i]记录i的根节点, dis[i]记录i到根节点的距离
 17 int head1[MAXN], head2[MAXN], ip1, ip2, cnt;
 18 
 19 void init(void){
 20     memset(sol, 0, sizeof(sol));
 21     memset(vis, 0, sizeof(vis));
 22     memset(dis, 0, sizeof(dis));
 23     memset(head1, -1, sizeof(head1));
 24     memset(head2, -1, sizeof(head2));
 25     ip1 = ip2 = cnt = 0;
 26 }
 27 
 28 void addedge1(int u, int v, int w){//前向星
 29     edge1[ip1].v = v;
 30     edge1[ip1].w = w;
 31     edge1[ip1].next = head1[u];
 32     head1[u] = ip1++;
 33 }
 34 
 35 void addedge2(int u, int v){
 36     edge2[ip2].u = u;
 37     edge2[ip2].v = v;
 38     edge2[ip2].next = head2[u];
 39     head2[u] = ip2++;
 40 }
 41 
 42 int find(int x){
 43     return pre[x] == x ? x : pre[x] = find(pre[x]);
 44 }
 45 
 46 void jion(int x, int y){
 47     x = find(x);
 48     y = find(y);
 49     if(x != y) pre[y] = x;
 50 }
 51 
 52 void tarjan(int u){
 53     pre[u] = u;
 54     vis[u] = cnt;//将不同的联通块标记成不同的数字
 55     for(int i = head1[u]; i != -1; i = edge1[i].next){
 56         int v = edge1[i].v;
 57         int w = edge1[i].w;
 58         if(!vis[v]){
 59             dis[v] = dis[u] + w;
 60             tarjan(v);
 61             jion(u, v);
 62         }
 63     }
 64     for(int i = head2[u]; i != -1; i = edge2[i].next){
 65         int v = edge2[i].v;
 66         if(vis[v] == cnt) sol[i >> 1] = find(v);//只有同一个联通块的顶点才存在lca
 67     }
 68 }
 69 
 70 int main(void){
 71     int t, n, m, x, y, z;
 72     while(~scanf("%d%d%d", &n, &m, &t)){
 73         init();
 74         for(int i = 0; i < m; i++){
 75             scanf("%d%d%d", &x, &y, &z);
 76             addedge1(x, y, z);
 77             addedge1(y, x, z);
 78         }
 79         for(int i = 0; i < t; i++){
 80             scanf("%d%d", &x, &y);
 81             addedge2(x, y);
 82             addedge2(y, x);
 83         }
 84         for(int i = 1; i <= n; i++){
 85             if(!vis[i]){
 86                 cnt++;
 87                 dis[i] = 0;//顶点到自己的距离为0
 88                 tarjan(i);
 89             }
 90         }
 91         for(int i = 0; i < t; i++){
 92             int cc = i << 1;
 93             int u = edge2[cc].u;
 94             int v = edge2[cc].v;
 95             int lca = sol[i];//sol开两倍空间会超内存
 96             if(!lca) puts("Not connected");
 97             else printf("%d\n", dis[u] + dis[v] - 2 * dis[lca]);
 98         }
 99     }
100     return 0;
101 }

View Code

 

解法2: lca 转 RMQ

可以先拟个虚根, 另外输出前先判一下 x, y 是否在同一棵树里面即可.

代码:

  1 #include <iostream>
  2 #include <stdio.h>
  3 #include <string.h>
  4 #include <math.h>
  5 using namespace std;
  6 
  7 const int inf = 1e9;
  8 const int MAXN = 4e4 + 10;
  9 struct node{
 10     int v, w, next;
 11 }edge[MAXN << 1];
 12 
 13 int dp[MAXN << 1][30]; //dp[i][j]存储deep数组中下标i开始长度为2^j的子串中最小值的下标
 14 int first[MAXN], ver[MAXN << 1], deep[MAXN << 1];
 15 int vis[MAXN], head[MAXN], dis[MAXN], tag[MAXN], ip, indx, cnt;
 16 
 17 inline void init(void){
 18     memset(tag, 0, sizeof(tag));
 19     memset(vis, 0, sizeof(vis));
 20     memset(head, -1, sizeof(head));
 21     ip = 0;
 22     indx = 0;
 23     cnt = 0;
 24 }
 25 
 26 void addedge(int u, int v, int w){
 27     edge[ip].v = v;
 28     edge[ip].w = w;
 29     edge[ip].next = head[u];
 30     head[u] = ip++;
 31 }
 32 
 33 void dfs(int u, int h){
 34     vis[u] = 1; //标记已搜索过的点
 35     ver[++indx] = u; //记录dfs路径
 36     first[u] = indx; //记录顶点u第一次出现时对应的ver数组的下标
 37     deep[indx] = h; //记录ver数组中对应下标的点的深度
 38     for(int i = head[u]; i != -1; i = edge[i].next){
 39         int v = edge[i].v;
 40         if(!vis[v]){
 41             dis[v] = dis[u] + edge[i].w;
 42             dfs(v, h + 1);
 43             ver[++indx] = u;
 44             deep[indx] = h;
 45         }
 46     }
 47 }
 48 
 49 void dfs1(int u){
 50     tag[u] = cnt;
 51     for(int i = head[u]; i != -1; i = edge[i].next){
 52         if(!tag[edge[i].v]) dfs1(edge[i].v);
 53     }
 54 }
 55 
 56 void ST(int n){
 57     for(int i = 1; i <= n; i++){
 58         dp[i][0] = i;
 59     }
 60     for(int j = 1; (1 << j) <= n; j++){
 61         for(int i = 1; i + (1 << j) - 1 <= n; i++){
 62             int x = dp[i][j - 1], y = dp[i + (1 << (j - 1))][j - 1];
 63             dp[i][j] = deep[x] < deep[y] ? x : y;
 64         }
 65     }
 66 }
 67 
 68 int RMQ(int l, int r){
 69     int len = log2(r - l + 1);
 70     int x = dp[l][len], y = dp[r - (1 << len) + 1][len];
 71     return deep[x] < deep[y] ? x : y;
 72 }
 73 
 74 int LCA(int x, int y){
 75     int l = first[x], r = first[y];
 76     if(l > r) swap(l, r);
 77     int pos = RMQ(l, r);
 78     return ver[pos];
 79 }
 80 
 81 int main(void){
 82     int n, m, k, x, y, z;
 83     while(~scanf("%d%d%d", &n, &m, &k)){
 84         init();
 85         for(int i = 0; i < m; i++){
 86             scanf("%d%d%d", &x, &y, &z);
 87             addedge(x, y, z);
 88             addedge(y, x, z);
 89         }
 90         for(int i = 1; i <= n; i++){
 91             if(!tag[i]){
 92                 cnt++;
 93                 dfs1(i);
 94                 addedge(i, 0, inf); //拟个虚根0
 95                 addedge(0, i, inf);
 96             }
 97         }
 98         dis[0] = 0;
 99         dfs(0, 1);
100         ST(2 * n - 1);
101         for(int i = 0; i < k; i++){
102             scanf("%d%d", &x, &y);
103             int lca = LCA(x, y);
104             if(tag[x] != tag[y]) puts("Not connected");
105             else printf("%d\n", dis[x] + dis[y] - 2 * dis[lca]);
106         }
107     }
108     return 0;
109 }

View Code