HDU 2874 Connections between cities(LCA离线算法)
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该题用离线算法的时候要注意会MLE, 内存卡的很紧, 所以要想办法优化内存, 利用存储边的数组就行了。
LCA是利用了并查集在树上进行的操作, 由于该题可能不形成一棵树,所以要对所有子树进行LCA。 然后不在一个集合中的两个点不能联通。
下面简单说一下我对LCA的理解: LCA就是dfs+并查集优化。 用dfs深搜, 当其回溯到结点u时, u的子树已经全部搜寻完了, 并且用并查集将其子树合并到了一个集合之中。 这时, 其子树的最近公共祖先就是当前结点u。 当然,其实我们也可以省略掉数组ancestor, 直接将最近公共祖先这个信息维护成并查集的根。
细节参见代码:
#include<cstdio>#include<cstring>#include<algorithm>using namespace std;typedef long long ll;const int maxn = 10000+5;const int maxq = 1000000+5;int n,u,m,v,dist[maxn],k,answer[maxq],f[maxn],h[maxn],tt,q,head[maxn],tot;int _find(int x) { return f[x] == x ? x : f[x] = _find(f[x]); }void bing(int u, int v) { int t1 = _find(u); int t2 = _find(v); if(t1 != t2) f[t1] = t2;}bool vis[maxn];struct Edge { int to, next, dist;}edge[maxn*2];void addedge(int u, int v, ll dist) { edge[tot].to = v; edge[tot].dist = dist; edge[tot].next = head[u]; head[u] = tot++;}struct Query { int q, next, index;}query[maxq*2];void add_query(int u, int v, int index) { query[tt].q = v; query[tt].next = h[u]; query[tt].index = index; h[u] = tt++; query[tt].q = u; query[tt].next = h[v]; query[tt].index = index; h[v] = tt++;}void init() { tot = tt = 0; for(int i=1;i<=n;i++) { h[i] = head[i] = -1; f[i] = i; vis[i] = false; }}void LCA(int u) { vis[u] = true; for(int i = head[u]; i != -1; i = edge[i].next) { int v = edge[i].to; if(vis[v]) continue; dist[v] = dist[u] + edge[i].dist; LCA(v); bing(v, u); } for(int i = h[u]; i != -1; i = query[i].next) { int v = query[i].q; if(vis[v]) { answer[query[i].index] = _find(v); } }}int main() { while(~scanf("%d%d%d",&n,&m,&q)) { init(); for(int i=0;i<m;i++) { scanf("%d%d%d",&u,&v,&k); addedge(u, v, k); addedge(v, u, k); } for(int i=0;i<q;i++) { scanf("%d%d",&u,&v); add_query(u,v,i); } for(int i=1;i<=n;i++) { if(!vis[i]) { dist[i] = 0; LCA(i); } } for(int i=0;i<2*q;i+=2) { u = _find(query[i+1].q); v = _find(query[i].q); if(u != v) printf("Not connected\n"); else printf("%d\n",dist[query[i+1].q]-dist[answer[i/2]]+dist[query[i].q]-dist[answer[i/2]]); } } return 0;}
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