标签:numa important poj ORC seq 范围 color init .com
这道题相当于将这两题结合:
http://poj.org/problem?id=2763
http://codeforces.com/gym/101808/problem/K
题意:有N各点N条边的带权无向图(相当于一棵树多了一条边),两种操作:修改一条边的权值;求两点间的最短路径。
分析:将任意一条边取出来,其余n-1条边可以结合LCA解最短路。询问时,比较通过取出的边和仅通过树上的边的路径的大小,最小值就是两点的最短路径。
树状数组差分维护点到根节点的距离,根据dfs序来记录需要维护的范围。
#include<iostream> #include<cstdio> #include<algorithm> #include<cstring> #include<cmath> using namespace std; #define maxn 100005 typedef long long LL; struct Edge{ int to,next,id; }edge[maxn<<1]; int n,a[maxn],head[maxn],dep[maxn<<1],cnt,pos[maxn],dfs_seq[maxn<<1],dfn,f[maxn<<1][20]; int L[maxn],R[maxn],dfs_clock,G[maxn]; LL W[maxn],C[maxn]; inline void add(int u,int v,int id){ edge[cnt].to=v; edge[cnt].next=head[u]; edge[cnt].id=id; head[u]=cnt++; } inline int lowbit(int x){return (x)&(-x);} void init(){ memset(head,-1,sizeof(head)); memset(pos,-1,sizeof(pos)); memset(C,0,sizeof(C)); cnt=dfn=0; dfs_clock=0; } void dfs(int u,int deep) { dfs_seq[dfn]=u,dep[dfn]=deep,pos[u]=dfn++; L[u]=++dfs_clock; for(int i=head[u];~i;i=edge[i].next){ int v=edge[i].to; if(pos[v]==-1){ G[edge[i].id]=v; //important dfs(v,deep+1); dfs_seq[dfn]=u,dep[dfn++]=deep; } } R[u]=dfs_clock; } void init_RMQ(int n) { for(int i=1;i<=n;++i) f[i][0]=i; for(int j=1;(1<<j)<=n;++j) for(int i=1;i+(1<<j)-1<=n;++i){ if(dep[f[i][j-1]]<dep[f[i+(1<<(j-1))][j-1]]) f[i][j]=f[i][j-1]; else f[i][j]=f[i+(1<<(j-1))][j-1]; } } inline int RMQ(int L,int R) { int k=0; while(1<<(k+1)<=R-L+1) ++k; if(dep[f[L][k]]<dep[f[R-(1<<k)+1][k]]) return f[L][k]; return f[R-(1<<k)+1][k]; } inline int lca(int u,int v) { if(pos[u]>pos[v]) return dfs_seq[RMQ(pos[v],pos[u])]; return dfs_seq[RMQ(pos[u],pos[v])]; } inline void update(int i,LL x) { for(;i<=n;i+=lowbit(i)) C[i]+=x; } inline LL sum(int i) { LL s=0; for(;i>0;i-=lowbit(i)) s+=C[i]; return s; } inline LL dist(int u,int v) { return sum(L[u])+sum(L[v])-2*sum(L[lca(u,v)]); } int main() { #ifndef ONLINE_JUDGE freopen("in.txt","r",stdin); freopen("out.txt","w",stdout); #endif int i,u,v,k,q,s,T; LL w; scanf("%d",&T); while(T--){ scanf("%d%d",&n,&q); init(); for(i=1;i<=n-1;++i){ scanf("%d%d%lld",&u,&v,&w); add(u,v,i); add(v,u,i); W[i]=w; } dfs(1,0); init_RMQ(dfn-1); int X,Y; LL Z; scanf("%d%d%lld",&X,&Y,&Z); //第n条边 W[n] = Z; for(i=1;i<n;++i){ update(L[G[i]],W[i]); update(R[G[i]]+1,-W[i]); } while(q--){ scanf("%d",&k); if(k==0){ scanf("%d%lld",&u,&w); if(u==n) W[n] = w; else{ update(L[G[u]],w-W[u]); update(R[G[u]]+1,-w+W[u]); W[u]=w; } } else{ scanf("%d%d",&u,&v); LL ans=dist(u,v); ans=min(ans,dist(u,X)+dist(v,X)); ans=min(ans,dist(u,Y)+dist(v,Y)); ans=min(ans,dist(u,X)+dist(v,Y)+Z); ans=min(ans,dist(u,Y)+dist(v,X)+Z); printf("%lld\n",ans); } } } return 0; }
HDU - 6393 Traffic Network in Numazu (LCA+RMQ+树状数组)
标签:numa important poj ORC seq 范围 color init .com
原文地址:https://www.cnblogs.com/xiuwenli/p/9470162.html