LCA

題意：LCA模板題，輸入n和m，表示n個點m條邊，下面m行是邊的信息，兩端點和權(quán)，后面的那個字母無視掉，沒用的。接著k，下面k個詢問lca，輸出即可

有人說要考慮不連通的情況，我沒考慮AC了，另外可能有u，u這樣的詢問，不過這不影響，照樣是寫模板，沒有特判，一樣能過

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還是Tarjan快一些

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LCA轉(zhuǎn)RMQ在線算法

#include <iostream> #include <cstdio> #include <cstring> #include <cmath> using namespace std; #define N 40010 #define M 25int tot; int __pow[M]; int head[N]; struct edge{int u,v,w,next; }e[2*N]; int ver[2*N],R[2*N],first[N],dir[N]; int dp[2*N][25]; bool vis[N];inline void add(int u , int v ,int w ,int &k) {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++; }void dfs(int u ,int dep) {vis[u] = true; first[u] = ++tot; ver[tot] = u; R[tot] = dep;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;dfs(v,dep+1);ver[++tot] = u; R[tot] = dep;} }void ST(int len) {int K = (int)(log((double)(len)) / log(2.0));for(int i=1; i<=len; i++) dp[i][0] = i;for(int j=1; j<=K; j++)for(int i=1; i+__pow[j]-1 <= len; i++){int a = dp[i][j-1] , b = dp[i+__pow[j-1]][j-1];if(R[a] < R[b]) dp[i][j] = a;else dp[i][j] = b;} }int RMQ(int x ,int y) {int K = (int)(log((double)(y-x+1)) / log(2.0));int a = dp[x][K] , b = dp[y-__pow[K]+1][K];if(R[a] < R[b]) return a;else return b; }int LCA(int u ,int v) {int x = first[u] , y = first[v];if(x > y) swap(x,y);int index = RMQ(x,y);return ver[index]; }int main() {for(int i=0; i<M; i++) __pow[i] = (1<<i);int n,m,k,str[10];while(scanf("%d%d",&n,&m)!=EOF){k = 0;memset(head,-1,sizeof(head));memset(vis,false,sizeof(vis));while(m--){int u,v,w;scanf("%d%d%d%s",&u,&v,&w,str);add(u,v,w,k);}tot = dir[1] = 0;dfs(1,1);ST(tot);int q;scanf("%d",&q);while(q--){int u,v,lca;scanf("%d%d",&u,&v);lca = LCA(u,v);printf("%d\n",dir[u] + dir[v] - 2*dir[lca]);}}return 0; }

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Tarjan離線算法

#include <iostream> #include <cstdio> #include <cstring> using namespace std; #define N 40010 #define M 20010int head[N]; struct edge{int u,v,w,next; }e[2*N]; int __head[N]; struct ask{int u,v,lca,next; }ea[M]; int fa[N],ance[N],dir[N]; bool vis[N];inline void add_edge(int u ,int v ,int w ,int &k) {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) {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;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 Tarjan(int u) {vis[u] = true;ance[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);fa[v] = u;}for(int k=__head[u]; k!=-1; k=ea[k].next)if( vis[ea[k].v] )ea[k].lca = ea[k^1].lca = ance[find(ea[k].v)]; }int main() {int n,m,q,k; char str[10];while(scanf("%d%d",&n,&m)!=EOF){memset(head,-1,sizeof(head));memset(__head,-1,sizeof(__head));memset(vis,false,sizeof(vis));k = 0;while(m--){int u,v,w;scanf("%d%d%d%s",&u,&v,&w,str);add_edge(u,v,w,k);}scanf("%d",&q);k = 0;for(int i=0; i<q; i++){int u ,v;scanf("%d%d",&u,&v);add_ask(u,v,k);}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; }

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轉(zhuǎn)載于:https://www.cnblogs.com/scau20110726/archive/2013/05/27/3102068.html

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