LCA离线算法Tarjan的模板
hdu 2586:题意：输入n个点的n-1条边的树，m组询问任意点 a b之间的最短距离
思路：LCA中的Tarjan算法,RMQ还不会。。
#include <stdio.h> #include <iostream> #include <string.h> using namespace std; const int N = 40010; const int M = 410; int head[N]; //树边邻接表的表头 int __head[N]; //保存询问的邻接表的表头 struct edge{ //保存边 int u,v,w,next; }e[2*N]; struct ask{ //保存询问 int u,v,lca,next; }ea[M]; int dir[N]; //保存点到树根的距离 int fa[N]; //并查集，保存集合的代表元素 int ance[N]; //保存集合的组合，注意对象是集合而不是元素 bool vis[N]; //遍历时的标记数组 inline void add_edge(int u,int v,int w,int &k) //保存边 { e[k].u = u; e[k].v = v; e[k].w = w; e[k].next = head[u]; head[u] = k++; u = u^v; v = u^v; u = u^v; e[k].u = u; e[k].v = v; e[k].w = w; e[k].next = head[u]; head[u] = k++; } inline void add_ask(int u ,int v ,int &k) //保存询问 { ea[k].u = u; ea[k].v = v; ea[k].lca = -1; ea[k].next = __head[u]; __head[u] = k++; u = u^v; v = u^v; u = u^v; ///看上去深奥。。其实就是swap(u,v); ea[k].u = u; ea[k].v = v; ea[k].lca = -1; ea[k].next = __head[u]; __head[u] = k++; } int Find(int x) { return x == fa[x] ? x : fa[x] = Find(fa[x]); } void Union(int u ,int v) { fa[v] = fa[u]; //可写为 fa[Find(v)] = fa[u]; } void Tarjan(int u) { vis[u] = true; ance[u] = fa[u] = u; //可写为 ance[Find(u)] = fa[u] = u; for(int k=head[u]; k!=-1; k=e[k].next) if( !vis[e[k].v] ) { int v = e[k].v , w = e[k].w; dir[v] = dir[u] + w; Tarjan(v); Union(u,v); ance[Find(u)] = u; } for(int k=__head[u]; k!=-1; k=ea[k].next) if( vis[ea[k].v] ) { int v = ea[k].v; ea[k].lca = ea[k^1].lca = ance[Find(v)]; } } int main() { int tcase; scanf("%d",&tcase); while(tcase--){ int n,q; scanf("%d%d",&n,&q); memset(head,-1,sizeof(head)); memset(__head,-1,sizeof(__head)); int tot = 0; for(int i=1; i<n; i++) //建树 { int u,v,w; scanf("%d%d%d",&u,&v,&w); add_edge(u,v,w,tot); } tot = 0; for(int i=0; i<q; i++) //拆开保存询问 { int u,v; scanf("%d%d",&u,&v); add_ask(u,v,tot); } memset(vis,0,sizeof(vis)); dir[1] = 0; Tarjan(1); for(int i=0; i<q; i++) { int s = i * 2 , u = ea[s].u , v = ea[s].v , lca = ea[s].lca; printf("%d\n",dir[u]+dir[v]-2*dir[lca]); } } return 0; }
hdu 2874：和上题一样都是求两点之间的最短距离，但是有多棵树，所以存在不连通的情况（用father判断一下就OK），，然后华丽丽的 超内存，拿别人的代码也是MLE
#include <stdio.h> #include <iostream> #include <string.h> using namespace std; const int N = 10001; const int M = 1000001; int head[N]; int __head[N]; struct edge{ int u,v,w,next; }e[2*N]; struct ask{ int u,v,lca,next; }ea[2*M]; int dir[N]; int fa[N]; int ance[N]; bool vis[N]; inline void add_edge(int u,int v,int w,int &k) //保存边 { e[k].u = u; e[k].v = v; e[k].w = w; e[k].next = head[u]; head[u] = k++; u = u^v; v = u^v; u = u^v; e[k].u = u; e[k].v = v; e[k].w = w; e[k].next = head[u]; head[u] = k++; } inline void add_ask(int u ,int v ,int &k) //保存询问 { ea[k].u = u; ea[k].v = v; ea[k].lca = -1; ea[k].next = __head[u]; __head[u] = k++; u = u^v; v = u^v; u = u^v; ///看上去深奥。。其实就是swap(u,v); ea[k].u = u; ea[k].v = v; ea[k].lca = -1; ea[k].next = __head[u]; __head[u] = k++; } int Find(int x) { return x == fa[x] ? x : fa[x] = Find(fa[x]); } void Union(int u ,int v) { int x = Find(u); int y = Find(v); fa[x] = y; } void Tarjan(int u) { vis[u] = true; ance[u] = fa[u] = u; //可写为 ance[Find(u)] = fa[u] = u; for(int k=head[u]; k!=-1; k=e[k].next) if( !vis[e[k].v] ) { int v = e[k].v , w = e[k].w; dir[v] = dir[u] + w; Tarjan(v); Union(u,v); ance[Find(u)] = u; //可写为ance[u] = u; //甚至不要这个语句都行 } for(int k=__head[u]; k!=-1; k=ea[k].next) if( vis[ea[k].v] ) { int v = ea[k].v; ea[k].lca = ea[k^1].lca = ance[Find(v)]; } } int main() { int k,n,q; while(scanf("%d%d%d",&n,&k,&q)!=EOF){ memset(head,-1,sizeof(head)); memset(__head,-1,sizeof(__head)); int tot = 0; for(int i=0; i<k; i++) //建树 { int u,v,w; scanf("%d%d%d",&u,&v,&w); add_edge(u,v,w,tot); } tot = 0; for(int i=0; i<q; i++) //拆开保存询问 { int u,v; scanf("%d%d",&u,&v); add_ask(u,v,tot); } memset(vis,0,sizeof(vis)); for(int i=1;i<=n;i++){ ///访问每个节点 if(!vis[i]){ dir[i]=0; Tarjan(i); } } for(int i=0; i<q; i++) { int s = i * 2 , u = ea[s].u , v = ea[s].v , lca = ea[s].lca; ///s代表偶数边，奇偶只是方向不同罢了，所以取一个就行 if(fa[u]!=fa[v]) printf("Not connected\n"); ///父亲结点不同当然就不是同一棵子树了 else printf("%d\n",dir[u]+dir[v]-2*dir[lca]); } } return 0; }
