BZOJ 4777
Description
Farmer John has recently been experimenting with cultivating different types of grass on his farm, r
ealizing that different types of cows like different types of grass. However, he must be careful to
ensure that different types of grass are planted sufficiently far away from each-other, in order to
prevent them from being inextricably mixed.FJ's farm consists of Nfields (1≤N≤200,000), where M pa
irs of fields are connected by bi-directional pathways (1≤M≤200,000). Using these pathways, it is
possible to walk from any field to any other field. Each pathway has an integer length in the range
1…1,000,000. Any pair of fields will be linked by at most one direct pathway.In each field, FJ init
ially plants one of Ktypes of grass (1≤K≤N). Over time, however, he might decide to switch the gra
ss in some field to a different type. He calls this an "update" operation. He might perform several
updates over the course of time, which are all cumulative in nature.After each update, FJ would like
to know the length of the shortest path between two fields having different grass types. That is, a
mong all pairs of fields having different grass types, he wants to know which two are closest. Ideal
ly, this number is large, so he can prevent grass of one type from mixing with grass of another type
. It is guaranteed that the farm will always have at least two fields with different grass types.In
30 percent of the input cases, each field will be directly connected to at most 10 pathways.
给定一张带权无向图，每个点有一个颜色，每次改变一个点的颜色，要求你在操作后输出这个图中最近异色点对之
间的距离最近异色点对定义为：一对点颜色不同，且距离最小
Input
The first line of input contains four integers, N, M, K, and Q, where Q is the number of updates (1
≤Q≤200,000). The next M lines describe the paths; each one contains three integers A, B, and L, in
dicating a path from field A to field B (both integers in the range 1…N) of length L. The next line
indicates the initial type of grass growing in each field (N integers in the range 1…K). Finally,
the last Q lines each describe an update, specified by two integers A and B, where the grass in fiel
d A is to be updated to type B
Output
For each update, print the length of the shortest path between two fields with different types of grass
after the update is applied.
Sample Input
3 2 3 4
1 2 3
2 3 1
1 1 2
3 3
2 3
1 2
2 2
1 2 3
2 3 1
1 1 2
3 3
2 3
1 2
2 2
Sample Output
1
3
3
1
3
3
1
这题可真TM调的我一逼。。校内测试时用sort水了60正解需要知道两个性质。。
性质1：最后解一定是一条边。因为如果有3个点A,B,C且A,C通过B相连，且A，C，颜色不同，那么B和A，C中一个点颜色一定不同（显然）然而w（A,C）>w(B,C)||w(A,B)
性质2：最后解一定在最小生成树里
假如答案边不在最小生成树上，那么最小生成树上两点之间的路径上一定能找到一条合法的答案边，且权值不会比它大
有了性质后就可以先Kruskal最小生成树然后再对每一个节点建树套树或树套set
再维护一个Ans的平衡树或set
还有这题zz到卡内存到丧心病狂。。自行体会一下吧。
树套树：
#define MAXN 205005
#define inf (0x3f3f3f3f)
#include <iostream>
#include <queue>
#include <cstring>
#include <stdio.h>
#include <vector>
#include <algorithm>
using namespace std;
int first[MAXN],e=1,n,m,k,q,val[MAXN],cost[MAXN],f[MAXN],last[MAXN];
template<typename _t>
inline _t read(){
_t x=0;
int f=1;
char ch=getchar();
for(;ch>'9'||ch<'0';ch=getchar())if(ch=='-')f=-f;
for(;ch>='0'&&ch<='9';ch=getchar())x=(x<<3)+(x<<1)+ch-48;
return (_t)x*f;
}
struct edge{
int u,v,next,w;
bool operator < (const edge &a)const{
return w<a.w;
}
}a[MAXN<<1],b[MAXN<<1];
void push(int u,int v,int w){
a[e].u=u;
a[e].v=v;
a[e].w=w;
a[e].next=first[u];
first[u]=e++;
}
int fa[MAXN];
int find(int x){
if(fa[x]!=x)
fa[x]=find(fa[x]);
return fa[x];
}
void Bfs(){
bool vis[MAXN];
memset(vis,0,sizeof vis);
queue<int>q;
q.push(1);
vis[1]=1;
while(!q.empty()){
int k = q.front();
q.pop();
for(int i=first[k];i;i=a[i].next){
if(!vis[a[i].v]){
f[a[i].v]=k;
cost[a[i].v]=a[i].w;
vis[a[i].v]=1;
q.push(a[i].v);
}
}
}
}
void Kr(){
sort(b+1,b+1+m);
int cnt = 0;
for(int i=1;i<=n;i++)fa[i]=i;
for(int i=1;i<=m;i++){
int u = find(b[i].u),v=find(b[i].v);
if(u==v)continue;
fa[u]=v;
push(b[i].u,b[i].v,b[i].w);
push(b[i].v,b[i].u,b[i].w);
if(cnt==n-1)break;
}
Bfs();
}
struct Treap_node{
int v,r;
Treap_node *ch[2];
Treap_node(){}
Treap_node(int x){
v = x;r=rand();
ch[0]=ch[1]=NULL;
}
void *operator new (size_t);
void operator delete (void* p);
int cmp(int x)const{
if(x==v)return -1;
return x<v?0:1;
}
}*C,*mempool,*Ans;
vector<Treap_node*>Stack;
void* Treap_node :: operator new (size_t){
Treap_node *p;
if(!Stack.empty()){
p = Stack.back();
Stack.pop_back();
}
else{
if(C==mempool){
C=new Treap_node[1<<15];
mempool = C+(1<<15);
}
p=C++;
}
return p;
}
void Treap_node :: operator delete (void* p){
Stack.push_back((Treap_node*)p);
}
void rotate (Treap_node *&o,int d){
Treap_node *p=o->ch[d^1];
o->ch[d^1]=p->ch[d];
p->ch[d]=o;
o=p;
}
void Treap_insert(Treap_node *&o,int x){
if(!o)o=new Treap_node(x);
else{
int d = x<o->v ? 0:1;
Treap_insert(o->ch[d],x);
if(o->ch[d]->r > o->r)rotate(o,d^1);
}
}
void Treap_erase(Treap_node *&o,int x){
int d = o->cmp(x);
if(d==-1){
Treap_node *tmp = o;
if(o->ch[0]!=NULL&&o->ch[1]!=NULL){
int k = o->ch[0]->r > o->ch[1]->r ? 1:0;
rotate(o,k);
Treap_erase(o->ch[k],x);
}
else{
if(o->ch[1]==NULL)o=o->ch[0];
else o = o->ch[1];
delete tmp;
tmp = NULL;
}
}
else Treap_erase(o->ch[d],x);
}
int Treap_Query(Treap_node *x){
int tot = inf;
while(x){
tot=min(tot,x->v);
x=x->ch[0];
}
return tot;
}
struct Tree{
Tree *ls,*rs;
Treap_node *v;
int min_x;
Tree() { ls=rs=NULL,v=NULL,min_x=inf; }
void Push_up(){
#define MIN(x) ((x)?((x)->min_x):inf)
min_x=min(MIN(ls),MIN(rs));
}
void* operator new (size_t);
void operator delete(void* p);
}*root[MAXN],*S,*T;
vector<Tree*>bin;
void* Tree :: operator new(size_t size){
Tree *o;
if(!bin.empty()){
o=bin.back();
bin.pop_back();
}
else{
if(S==T){
S=new Tree[1<<15];
T=S+(1<<15);
}
o = S++;
}
o->ls=o->rs=NULL;
o->v=NULL;
o->min_x=inf;
return o;
}
void Tree :: operator delete (void* p){
bin.push_back((Tree*)p);
}
void insert(Tree *&o,int l,int r,int x,int va){
if(!o)o = new Tree;
if(l==r){
Treap_insert(o->v,va);
o->min_x=Treap_Query(o->v);
return;
}
int mid=l+r>>1;
if(x<=mid)insert(o->ls,l,mid,x,va);
else insert(o->rs,mid+1,r,x,va);
o->Push_up();
}
void erase(Tree *&o,int l,int r,int x,int va){
if(l==r){
Treap_erase(o->v,va);
o->min_x=Treap_Query(o->v);
return;
}
int mid = l+r>> 1;
if(x<=mid)erase(o->ls,l,mid,x,va);
else erase(o->rs,mid+1,r,x,va);
o->Push_up();
if(o->min_x==inf){
delete o;
o=NULL;
}
}
int query(Tree *o,int l,int r,int x,int y){
if(!o)return inf;
if(x<=l&&r<=y)return o->min_x;
int mid = l+r>>1,ans=inf;
if(x<=mid)ans=min(ans,query(o->ls,l,mid,x,y));
if(mid<y)ans=min(ans,query(o->rs,mid+1,r,x,y));
return ans;
}
int Query(int x){
int ans = inf;
if(val[x]!=1)ans=min(ans,query(root[x],1,k,1,val[x]-1));
if(val[x]!=k)ans=min(ans,query(root[x],1,k,val[x]+1,k));
return ans;
}
int main(){
bin.clear();
n=read<int>();
m=read<int>();
k=read<int>();
q=read<int>();
for(int i=1;i<=m;i++)scanf("%d%d%d",&b[i].u,&b[i].v,&b[i].w);
for(int i=1;i<=n;i++)scanf("%d",&val[i]);Kr();
for(int i=2;i<=n;i++)insert(root[f[i]],1,k,val[i],cost[i]);
for(int i=1;i<=n;i++)Treap_insert(Ans,last[i]=Query(i));
while(q--){
int pos=read<int>(),w=read<int>();
if(f[pos])erase(root[f[pos]],1,k,val[pos],cost[pos]);
val[pos]=w;
if(f[pos])insert(root[f[pos]],1,k,val[pos],cost[pos]);
Treap_erase(Ans,last[pos]);
Treap_insert(Ans,last[pos]=Query(pos));
if(f[pos]){
Treap_erase(Ans,last[f[pos]]);
Treap_insert(Ans,last[f[pos]]=Query(f[pos]));
}
printf("%d\n",Treap_Query(Ans));
}
}树+Set
#define MAXN 205005
#define inf (0x3f3f3f3f)
#include <iostream>
#include <set>
#include <queue>
#include <cstring>
#include <stdio.h>
#include <vector>
#include <algorithm>
using namespace std;
int first[MAXN],e=1,n,m,k,q,val[MAXN],cost[MAXN],f[MAXN],last[MAXN];
multiset<int>Ans;
multiset<int>::iterator it;
template<typename _t>
inline _t read(){
_t x=0;
int f=1;
char ch=getchar();
for(;ch>'9'||ch<'0';ch=getchar())if(ch=='-')f=-f;
for(;ch>='0'&&ch<='9';ch=getchar())x=(x<<3)+(x<<1)+ch-48;
return (_t)x*f;
}
struct edge{
int u,v,next,w;
bool operator < (const edge &a)const{
return w<a.w;
}
}a[MAXN<<1],b[MAXN<<1];
void push(int u,int v,int w){
a[e].u=u;
a[e].v=v;
a[e].w=w;
a[e].next=first[u];
first[u]=e++;
}
int fa[MAXN];
int find(int x){
if(fa[x]!=x)
fa[x]=find(fa[x]);
return fa[x];
}
void Bfs(){
bool vis[MAXN];
memset(vis,0,sizeof vis);
queue<int>q;
q.push(1);
vis[1]=1;
while(!q.empty()){
int k = q.front();
q.pop();
for(int i=first[k];i;i=a[i].next){
if(!vis[a[i].v]){
f[a[i].v]=k;
cost[a[i].v]=a[i].w;
vis[a[i].v]=1;
q.push(a[i].v);
}
}
}
}
void Kr(){
sort(b+1,b+1+m);
int cnt = 0;
for(int i=1;i<=n;i++)fa[i]=i;
for(int i=1;i<=m;i++){
int u = find(b[i].u),v=find(b[i].v);
if(u==v)continue;
fa[u]=v;
push(b[i].u,b[i].v,b[i].w);
push(b[i].v,b[i].u,b[i].w);
if(cnt==n-1)break;
}
Bfs();
}
struct Tree{
Tree *ls,*rs;
multiset<int>*v;
int min_x;
Tree() { ls=rs=NULL,v=NULL,min_x=inf; }
void Push_up(){
#define MIN(x) ((x)?((x)->min_x):inf)
min_x=min(MIN(ls),MIN(rs));
}
void* operator new (size_t);
void operator delete(void* p);
}*root[MAXN],*S,*T;
vector<Tree*>bin;
void* Tree :: operator new(size_t size){
Tree *o;
if(!bin.empty()){
o=bin.back();
bin.pop_back();
}
else{
if(S==T){
S=new Tree[1<<15];
T=S+(1<<15);
}
o = S++;
}
o->ls=o->rs=NULL;
o->v=NULL;
o->min_x=inf;
return o;
}
void Tree :: operator delete (void* p){
bin.push_back((Tree*)p);
}
void insert(Tree *&o,int l,int r,int x,int va){
if(!o)o = new Tree;
if(l==r){
if(!o->v)o->v=new multiset<int>;
o->v->insert(va);
o->min_x=*(o->v->begin());
return;
}
int mid=l+r>>1;
if(x<=mid)insert(o->ls,l,mid,x,va);
else insert(o->rs,mid+1,r,x,va);
o->Push_up();
}
void erase(Tree *&o,int l,int r,int x,int va){
if(l==r){
o->v->erase(o->v->find(va));
if(o->v->empty()){
delete o->v;
o->v=NULL;
delete o;
o=NULL;
}
else o->min_x=*(o->v->begin());
return;
}
int mid = l+r>> 1;
if(x<=mid)erase(o->ls,l,mid,x,va);
else erase(o->rs,mid+1,r,x,va);
o->Push_up();
if(o->min_x==inf){
delete o;
o=NULL;
}
}
int query(Tree *o,int l,int r,int x,int y){
if(!o)return inf;
if(x<=l&&r<=y)return o->min_x;
int mid = l+r>>1,ans=inf;
if(x<=mid)ans=min(ans,query(o->ls,l,mid,x,y));
if(mid<y)ans=min(ans,query(o->rs,mid+1,r,x,y));
return ans;
}
int Query(int x){
int ans = inf;
if(val[x]!=1)ans=min(ans,query(root[x],1,k,1,val[x]-1));
if(val[x]!=k)ans=min(ans,query(root[x],1,k,val[x]+1,k));
return ans;
}
int main(){
//freopen("grass.in","r",stdin);
//freopen("grass.out","w",stdout);
bin.clear();
n=read<int>();
m=read<int>();
k=read<int>();
q=read<int>();
for(int i=1;i<=m;i++)scanf("%d%d%d",&b[i].u,&b[i].v,&b[i].w);
for(int i=1;i<=n;i++)scanf("%d",&val[i]);Kr();
for(int i=2;i<=n;i++)insert(root[f[i]],1,k,val[i],cost[i]);
for(int i=1;i<=n;i++)Ans.insert(last[i]=Query(i));
while(q--){
int pos=read<int>(),w=read<int>();
if(f[pos])erase(root[f[pos]],1,k,val[pos],cost[pos]);
val[pos]=w;
if(f[pos])insert(root[f[pos]],1,k,val[pos],cost[pos]);
Ans.erase(Ans.find(last[pos]));
Ans.insert(last[pos]=Query(pos));
if(f[pos]){
Ans.erase(Ans.find(last[f[pos]]));
Ans.insert(last[f[pos]]=Query(f[pos]));
}
printf("%d\n",*(Ans.begin()));
}
}
