题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=2586
题意: 给出一棵 n 个节点的带边权的树, 有 m 个形如 x y 的询问, 要求输出所有 x, y节点之间的最短距离.
思路: dis[i] 存储 i 节点到根节点的最短距离, lca 为 x, y 的最近公共祖先, 那么 x, y 之间的最短距离为: dis[x] + dis[y] - 2 * dis[lca] .
解法1: tarjan离线算法
关于该算法
1 Tarjan(u)//marge和find为并查集合并函数和查找函数 2 { 3 for each(u,v) //访问所有u子节点v 4 { 5 Tarjan(v); //继续往下遍历 6 marge(u,v); //合并v到u上 7 标记v被访问过; 8 } 9 for each(u,e) //访问所有和u有询问关系的e 10 { 11 如果e被访问过; 12 u,e的最近公共祖先为find(e); 13 } 14 }
详解见: http://www.cnblogs.com/ECJTUACM-873284962/p/6613379.html
代码:
1 #include <iostream> 2 #include <stdio.h> 3 #include <string.h> 4 using namespace std; 5 6 const int MAXN = 4e4 + 10; 7 struct node{ 8 int u, v, w, lca, next; 9 }edge1[MAXN << 1], edge2[810];//edge1记录树, edge2记录询问 10 11 int vis[MAXN], pre[MAXN], dis[MAXN];//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 12 int head1[MAXN], head2[MAXN], ance[MAXN], ip1, ip2; 13 14 void init(void){ 15 memset(vis, 0, sizeof(vis)); 16 memset(dis, 0, sizeof(dis)); 17 memset(head1, -1, sizeof(head1)); 18 memset(head2, -1, sizeof(head2)); 19 ip1 = ip2 = 0; 20 } 21 22 void addedge1(int u, int v, int w){//前向星 23 edge1[ip1].v = v; 24 edge1[ip1].w = w; 25 edge1[ip1].next = head1[u]; 26 head1[u] = ip1++; 27 } 28 29 void addedge2(int u, int v){ 30 edge2[ip2].u = u; 31 edge2[ip2].v = v; 32 edge2[ip2].lca = -1; 33 edge2[ip2].next = head2[u]; 34 head2[u] = ip2++; 35 } 36 37 int find(int x){ 38 return pre[x] == x ? x : pre[x] = find(pre[x]); 39 } 40 41 void jion(int x, int y){ 42 x = find(x); 43 y = find(y); 44 if(x != y) pre[y] = x; 45 } 46 47 void tarjan(int u){ 48 vis[u] = 1; 49 ance[u] = pre[u] = u; 50 for(int i = head1[u]; i != -1; i = edge1[i].next){ 51 int v = edge1[i].v; 52 int w = edge1[i].w; 53 if(!vis[v]){ 54 dis[v] = dis[u] + w; 55 tarjan(v); 56 jion(u, v); 57 } 58 } 59 for(int i = head2[u]; i != -1; i = edge2[i].next){ 60 int v = edge2[i].v; 61 if(vis[v]) edge2[i].lca = edge2[i ^ 1].lca = ance[find(v)]; 62 } 63 } 64 65 int main(void){ 66 int t, n, m, x, y, z; 67 scanf("%d", &t); 68 while(t--){ 69 init(); 70 scanf("%d%d", &n, &m); 71 for(int i = 1; i < n; i++){ 72 scanf("%d%d%d", &x, &y, &z); 73 addedge1(x, y, z); 74 addedge1(y, x, z); 75 } 76 for(int i = 0; i < m; i++){ 77 scanf("%d%d", &x, &y); 78 addedge2(x, y); 79 addedge2(y, x); 80 } 81 dis[1] = 0; 82 tarjan(1); 83 for(int i = 0; i < m; i++){ 84 int cc = i << 1; 85 int u = edge2[cc].u; 86 int v = edge2[cc].v; 87 int lca = edge2[cc].lca; 88 printf("%d ", dis[u] + dis[v] - 2 * dis[lca]); 89 } 90 } 91 return 0; 92 }
解法2: lca转RMQ
关于该算法
ver[] 存储树的 dfs 路径
first[u] 为顶点 u 在 ver 数组中第一次出现时的下标
deep[indx] 为顶点 ver[indx] 的深度
对于求 x, y 的 lca, 先令 l = first[x], r = first[y], 即 l, r 分别为 x, y 第一次在 ver 数组中出现时对应的下标
在 deep[] 数组中找到区间 [l, r] 中的最小值, 其下标对应的 ver 值即为 x, y 的 lca. (区间最值可以用 RMQ 处理)
详解见: http://blog.csdn.net/u013076044/article/details/41870751
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 MAXN = 4e4 + 10; 8 struct node{ 9 int v, w, next; 10 }edge[MAXN << 1]; 11 12 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 13 int first[MAXN], ver[MAXN << 1], deep[MAXN << 1]; 14 int vis[MAXN], head[MAXN], dis[MAXN], ip, indx; 15 16 inline void init(void){ 17 memset(vis, 0, sizeof(vis)); 18 memset(head, -1, sizeof(head)); 19 ip = 0; 20 indx = 0; 21 } 22 23 void addedge(int u, int v, int w){ 24 edge[ip].v = v; 25 edge[ip].w = w; 26 edge[ip].next = head[u]; 27 head[u] = ip++; 28 } 29 30 void dfs(int u, int h){ 31 vis[u] = 1; //标记已搜索过的点 32 ver[++indx] = u; //记录dfs路径 33 first[u] = indx; //记录顶点u第一次出现时对应的ver数组的下标 34 deep[indx] = h; //记录ver数组中对应下标的点的深度 35 for(int i = head[u]; i != -1; i = edge[i].next){ 36 int v = edge[i].v; 37 if(!vis[v]){ 38 dis[v] = dis[u] + edge[i].w; 39 dfs(v, h + 1); 40 ver[++indx] = u; 41 deep[indx] = h; 42 } 43 } 44 } 45 46 void ST(int n){ 47 for(int i = 1; i <= n; i++){ 48 dp[i][0] = i; 49 } 50 for(int j = 1; (1 << j) <= n; j++){ 51 for(int i = 1; i + (1 << j) - 1 <= n; i++){ 52 int x = dp[i][j - 1], y = dp[i + (1 << (j - 1))][j - 1]; 53 dp[i][j] = deep[x] < deep[y] ? x : y; 54 } 55 } 56 } 57 58 int RMQ(int l, int r){ 59 int len = log2(r - l + 1); 60 int x = dp[l][len], y = dp[r - (1 << len) + 1][len]; 61 return deep[x] < deep[y] ? x : y; 62 } 63 64 int LCA(int x, int y){ 65 int l = first[x], r = first[y]; 66 if(l > r) swap(l, r); 67 int pos = RMQ(l, r); 68 return ver[pos]; 69 } 70 71 int main(void){ 72 int t, n, m, x, y, z; 73 scanf("%d", &t); 74 while(t--){ 75 init(); 76 scanf("%d%d", &n, &m); 77 for(int i = 1; i < n; i++){ 78 scanf("%d%d%d", &x, &y, &z); 79 addedge(x, y, z); 80 addedge(y, x, z); 81 } 82 dis[1] = 0; 83 dfs(1, 1); 84 ST(2 * n - 1); 85 for(int i = 0; i < m; i++){ 86 scanf("%d%d", &x, &y); 87 int lca = LCA(x, y); 88 printf("%d ", dis[x] + dis[y] - 2 * dis[lca]); 89 } 90 } 91 return 0; 92 }