Battle over Cities
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 467 Accepted Submission(s): 125
Problem Description
It is vitally important to have all the cities connected by highways in a war, but some of them are destroyed now because of the war. Furthermore,if a city is conquered, all the highways from/toward that city will be closed by the enemy, and we must repair some destroyed highways to keep other cities connected, with the minimum cost if possible.
Given the map of cities which have all the destroyed and remaining highways marked, you are supposed to tell the cost to connect other cities if each city is conquered by the enemy.
Given the map of cities which have all the destroyed and remaining highways marked, you are supposed to tell the cost to connect other cities if each city is conquered by the enemy.
Input
The input contains multiple test cases. The first line is the total number of cases T (T ≤ 10). Each case starts with a line containing 2 numbers N (0 < N ≤ 20000), and M (0 ≤ M ≤ 100000), which are the total number of cities, and the number of highways, respectively. Then M lines follow, each describes a highway by 4 integers: City1 City2 Cost Status where City1 and City2 are the numbers of the cities the highway connects (the cities are numbered from 1 to N), Cost (0 < Cost ≤ 20000) is the effort taken to repair that highway if necessary, and Status is either 0, meaning that highway is destroyed, or 1, meaning that highway is in use.
Note: It is guaranteed that the whole country was connected before the war and there is no duplicated high ways between any two cities.
Note: It is guaranteed that the whole country was connected before the war and there is no duplicated high ways between any two cities.
Output
For each test case, output N lines of integers. The integer in the i-th line indicates the cost to keep the cities connected if the i-th city is conquered by the enemy. In case the cities cannot be connected after the i-th city is conquered by the enemy, output "inf" instead in the corresponding place.
Sample Input
3
4 5
1 2 1 1
1 3 1 1
2 3 1 0
2 4 1 1
3 4 2 0
4 5
1 2 1 1
1 3 1 1
2 3 1 0
2 4 1 1
3 4 1 0
3 2
1 2 1 1
1 3 1 1
Sample Output
1
2
0
0
1
1
0
0
inf
0
0
Author
GUAN, Yao
Source
Recommend
题解：给定一张N个点,M条边的无向连通图,每条边上有边权w，求删去每一个后的最小生成树。
思路：如果暴力删除每个点后重新建图，跑 Kruskal的话 时间复杂度O(NM),显然不行。
我们考虑先求出没有删点的最小生成树，然后如果我们删除一个点，那么和这个点相连的边全都得去掉，只剩下 该点的邻接点（子节点）和它的父亲节点（可能没有），那么我们只需要在我的子节点连向我的父亲节点的边和我的子树之间的连边中跑一遍MST就行了。
怎么预处理呢?
(1)连接子树之间的边。对于每条连接u,v没有在初始最小生成树里面的边，先求出u,v的lca，则这条边就是连接u的dep[u]-dep[lca]-1个father和v的dep[v]-dep[lca]-1个father的子树的边。用倍增求即可。
(2)连向父亲子树外面的边。对于每条连接u,v没有在初始最小生成树里面的边，先求出u,v的lca，则对于u的dep[u]-dep[lca]-2个father到u这条路径上的所有点，这条边都是连到它们父亲的子树外面的。注意，连向父亲子树外面的边只要取最小的一条即可，于是用倍增+树链剖分进行维护连向父亲子树外面的边的最短长度。对于v同理。
但为什么对每个点把所有可供选择的边预处理出来，再对这些边进行一次最小生成树不会超时呢？我们可以这样来想。因为一次Kruskal并查集的find操作是log(n)的，最坏情况下每条边都会用到一次find，因此重点是求出共有多少条边被用到。连接子树之间的边最多m-n条，连向子树外的边最多n条，所以总共进行m次find。对于m-n条不在最小生成树中的边，都进行预处理,一次预处理倍增是log(n)的，树剖是log(n)*log(n)的。因此，总时间复杂度为O(m log n+ m log n log n)。
参考代码：
#include<bits/stdc++.h> using namespace std; const int N=2e4+10; const int M=1e5+10; const int inf=2139062143; int T,n,m,ns,cnt,tim,head[N],to[N<<1],nxt[N<<1]; int dep[N],fa[N][20],siz[N],son[N],dfn[N],pos[N],top[N]; int minn[N],minv[N<<2],tag[N<<2]; bool use[M]; int s,mst,smst,pa[N],w[N],sont[N],num[N]; struct edge{ int u,v,d; bool operator < (const edge &x)const{return d<x.d;} } e[M]; vector<edge> link[N],edges; void Init() { smst=tim=cnt=0; memset(use,0,sizeof(use)); memset(head,0,sizeof(head)); memset(dep,0,sizeof(dep)); memset(fa,0,sizeof(fa)); memset(son,0,sizeof(son)); memset(sont,0,sizeof(sont)); memset(w,0,sizeof(w)); memset(minv,127,sizeof(minv)); memset(tag,127,sizeof(tag)); for(int i=1;i<=n;i++) pa[i]=i,link[i].clear(); } void AddEdge(int u,int v) { to[++cnt]=v; nxt[cnt]=head[u]; head[u]=cnt; } int Find(int u){return u==pa[u]?u:pa[u]=Find(pa[u]);} void dfs1(int u) { for(int i=1;(1<<i)<=dep[u];++i) fa[u][i]=fa[fa[u][i-1]][i-1]; siz[u]=1; for(int i=head[u];i;i=nxt[i]) { int v=to[i]; if(v==fa[u][0]) continue; fa[v][0]=u; dep[v]=dep[u]+1; num[v]=++sont[u]; dfs1(v); siz[u]+=siz[v]; if(!son[u]||siz[son[u]]<siz[v]) son[u]=v; } } void dfs2(int u,int tp) { dfn[u]=++tim; pos[tim]=u; top[u]=tp; if(son[u]) dfs2(son[u],tp); for(int i=head[u];i;i=nxt[i]) { int v=to[i]; if(v!=fa[u][0]&&v!=son[u]) dfs2(v,v); } } void pushdown(int rt) { minv[rt<<1]=min(minv[rt<<1],tag[rt]); minv[rt<<1|1]=min(minv[rt<<1|1],tag[rt]); tag[rt<<1]=min(tag[rt<<1],tag[rt]); tag[rt<<1|1]=min(tag[rt<<1|1],tag[rt]); tag[rt]=inf; } void upd(int rt,int l,int r,int L,int R,int x) { if(L<=l&&R>=r) { minv[rt]=min(minv[rt],x); tag[rt]=min(tag[rt],x); return; } if(tag[rt]!=inf) pushdown(rt); int mid=(l+r)/2; if(L<=mid) upd(rt<<1,l,mid,L,R,x); if(R>mid) upd(rt<<1|1,mid+1,r,L,R,x); minv[rt]=min(minv[rt<<1],minv[rt<<1|1]); } void update(int u,int v,int x) { while(top[u]!=top[v]) { upd(1,1,n,dfn[top[u]],dfn[u],x); u=fa[top[u]][0]; } upd(1,1,n,dfn[v],dfn[u],x); } void getmin(int rt,int l,int r) { if(l==r) { minn[pos[l]]=minv[rt]; return; } if(tag[rt]!=inf) pushdown(rt); int mid=(l+r)/2; getmin(rt<<1,l,mid); getmin(rt<<1|1,mid+1,r); } void work(edge &e) { int u=e.u,v=e.v,d; d=dep[u]-dep[v]; for(int i=0;(1<<i)<=d;i++) if(d&(1<<i)) u=fa[u][i]; if(u==v) { u=e.u; d=dep[u]-dep[v]-2; if(d<0) return; for(int i=0;(1<<i)<=d;i++) if(d&(1<<i)) u=fa[u][i]; update(e.u,u,e.d); return; } int tmpu=u,tmpv=v; for(int i=15;i>=0;i--) { if(fa[tmpu][i]!=fa[tmpv][i]) { tmpu=fa[tmpu][i]; tmpv=fa[tmpv][i]; } } link[fa[tmpu][0]].push_back((edge){tmpu,tmpv,e.d}); d=dep[e.u]-dep[tmpu]-1; if(d>=0) { u=e.u; for(int i=0;(1<<i)<=d;i++) if(d&(1<<i)) u=fa[u][i]; update(e.u,u,e.d); } d=dep[e.v]-dep[tmpv]-1; if(d>=0) { v=e.v; for(int i=0;(1<<i)<=d;i++) if(d&(1<<i)) v=fa[v][i]; update(e.v,v,e.d); } } int main() { scanf("%d",&T); while(T--) { scanf("%d%d",&n,&m); Init(); int d,f; for(int i=1;i<=m;i++) { scanf("%d%d%d%d",&e[i].u,&e[i].v,&d,&f); e[i].d=d*(1-f); } sort(e+1,e+m+1); ns=0; for(int i=1;i<=m&&ns<n-1;i++) { int u=Find(e[i].u),v=Find(e[i].v); if(u!=v) { use[i]=true; pa[v]=u; ns++; AddEdge(e[i].u,e[i].v); AddEdge(e[i].v,e[i].u); w[e[i].u]+=e[i].d; w[e[i].v]+=e[i].d; smst+=e[i].d; } } dfs1(1); dfs2(1,1); for(int i=1;i<=m;i++) { if(!use[i]) { if(dep[e[i].u]<dep[e[i].v]) swap(e[i].u,e[i].v); work(e[i]); } } getmin(1,1,n); for(int i=1;i<=n;i++) { edges.clear(); s=sont[i]; if(fa[i][0])//把所有连向i上面的边加进去 { ++s; for(int j=head[i];j;j=nxt[j]) { int v=to[j]; if(v!=fa[i][0]&&minn[v]!=inf) edges.push_back((edge){num[v],s,minn[v]}); } } for(int j=0;j<link[i].size();j++)//把i的邻接点之间的连边加进去 edges.push_back((edge){num[link[i][j].u],num[link[i][j].v],link[i][j].d}); sort(edges.begin(),edges.end()); mst=0; for(int j=1;j<=s;j++) pa[j]=j; ns=0; for(int j=0;j<edges.size()&&ns<s-1;j++) { int u=Find(edges[j].u),v=Find(edges[j].v); if(u!=v) { pa[v]=u; ns++; mst+=edges[j].d; } } if(ns<s-1) puts("inf"); else printf("%d ",smst-w[i]+mst); } } return 0; }