题目链接:HDU2586
【题意】给定一颗有权树,求任意两点之间的最短路。
【分析】是LCA的基本应用,树上任意两点(u,v)的最短路 = dis[u]+dis[v]-2*dis[LCA(u,v)];dis[i]为根到i的距离,我写了两个版本,离线和用线段树的在线(因为这里查询很少,用线段树比ST快,因为线段树初始化为O(n)的,ST是O(nlogn)的)
【AC CODE(离线)】46ms
#include <cstdio> #include <cstring> #include <cctype> #include <cmath> #include <map> //#include <unordered_map> #include <queue> #include <stack> #include <vector> #include <string> #include <algorithm> using namespace std; typedef long long LL; #define rep(i,a,n) for(int i = a; i < n; i++) #define repe(i,a,n) for(int i = a; i <= n; i++) #define per(i,n,a) for(int i = n; i >= a; i--) #define clc(a,b) memset(a,b,sizeof(a)) const int INF = 0x3f3f3f3f, MAXN = 40000+10, MAXM = 40000*2+10; int head[MAXN], tol, to[MAXM], nxt[MAXM], cost[MAXM]; struct Q{ int v, id; }; vector<Q> q[MAXN]; void add_edge(int u, int v, int c) { nxt[tol] = head[u], to[tol] = v, cost[tol] = c; head[u] = tol++; } int f[MAXN]; int find(int x){return f[x] == x?x:f[x] = find(f[x]);} void bing(int a, int b) { int x = find(a), y = find(b); if(x != y) f[y] = x; } bool vis[MAXN]; int ans[200+10], dis[MAXN]; void lca(int u) { vis[u] = true; for(int i = head[u]; ~i; i = nxt[i]) { int v = to[i]; if(vis[v]) continue; dis[v] = dis[u]+cost[i]; lca(v); bing(u,v); } int sz = q[u].size(); rep(i,0,sz) { int v = q[u][i].v; if(vis[v]) { int fa = find(v); ans[q[u][i].id] = dis[u]+dis[v]-2*dis[fa]; } } } int main() { #ifdef SHY freopen("d:\\1.txt", "r", stdin); #endif int t; scanf("%d%*c", &t); while(t--) { int n,m; scanf("%d %d%*c", &n, &m); clc(head,-1); tol = 0; rep(i,1,n) { int u,v,c; scanf("%d %d %d%*c", &u, &v, &c); add_edge(u,v,c); add_edge(v,u,c); } repe(i,1,n) q[i].clear(), f[i] = i; rep(i,0,m) { int u,v; scanf("%d %d%*c", &u, &v); q[u].push_back(Q{v,i}); q[v].push_back(Q{u,i}); } clc(vis,0); dis[1] = 0; lca(1); rep(i,0,m) printf("%d\n", ans[i]); } return 0; }
【AC CODE(在线)】31ms
#include <cstdio> #include <cstring> #include <cctype> #include <cmath> #include <map> //#include <unordered_map> #include <queue> #include <stack> #include <vector> #include <string> #include <algorithm> using namespace std; typedef long long LL; #define rep(i,a,n) for(int i = a; i < n; i++) #define repe(i,a,n) for(int i = a; i <= n; i++) #define per(i,n,a) for(int i = n; i >= a; i--) #define clc(a,b) memset(a,b,sizeof(a)) const int INF = 0x3f3f3f3f, MAXN = 40000+10, MAXM = 40000*2+10; int head[MAXN], tol, to[MAXM], nxt[MAXM], cost[MAXM]; void add_edge(int u, int v, int c) { nxt[tol] = head[u], to[tol] = v, cost[tol] = c; head[u] = tol++; } int dis[MAXN],ft[MAXN],d[MAXN<<1],num[MAXN<<1],cnt; bool vis[MAXN]; void dfs(int u, int deep) { ft[u] = cnt; d[cnt] = deep, num[cnt++] = u; vis[u] = true; for(int i = head[u]; ~i; i = nxt[i]) { int v = to[i]; if(vis[v]) continue; dis[v] = dis[u]+cost[i]; dfs(v,deep+1); d[cnt] = deep, num[cnt++] = u; } } int mi_num[MAXN<<2]; inline int id(int x, int y){return x+y|x!=y;} inline void push_up(int x, int y, int m) { int u = id(x,y), l = id(x,m), r = id(m+1,y); if(d[mi_num[l]] < d[mi_num[r]]) mi_num[u] = mi_num[l]; else mi_num[u] = mi_num[r]; } void bulid(int x, int y) { if(x == y) { mi_num[id(x,y)] = x; return; } int m = (x+y)>>1; bulid(x,m); bulid(m+1,y); push_up(x,y,m); } int query(int x, int y, int ql, int qr) { if(ql <= x && y <= qr) return mi_num[id(x,y)]; int m = (x+y)>>1, ans, mi = INF; if(ql <= m) { int p = query(x,m,ql,qr); if(mi > d[p]) mi = d[p], ans = p; } if(qr > m) { int p = query(m+1,y,ql,qr); if(mi > d[p]) mi = d[p], ans = p; } return ans; } void lca_init() { cnt = 0; dis[1] = 0; clc(vis,0); dfs(1,0); bulid(0,cnt-1); } int lca_query(int x, int y) { int l = ft[x], r = ft[y]; if(l > r) swap(l,r); int fa = num[query(0,cnt-1,l,r)]; return dis[x]+dis[y]-2*dis[fa]; } int main() { #ifdef SHY freopen("d:\\1.txt", "r", stdin); #endif int t; scanf("%d%*c", &t); while(t--) { int n,m; scanf("%d %d%*c", &n, &m); clc(head,-1); tol = 0; rep(i,1,n) { int u,v,c; scanf("%d %d %d%*c", &u, &v, &c); add_edge(u,v,c); add_edge(v,u,c); } lca_init(); rep(i,0,m) { int u,v; scanf("%d %d%*c", &u, &v); printf("%d\n", lca_query(u,v)); } } return 0; }