[LUOGU] P3128 [USACO15DEC]最大流Max Flow

题意:一棵树,多次给指定链上的节点加1,问最大节点权值

n个点,n-1条边很容易惯性想成一条链,幸好有样例..

简单的树剖即可!(划去)

正常思路是树上差分,毕竟它就询问一次..

#include<iostream>
#include<cstring>
#include<cstdio>

using namespace std;

inline int rd(){
    int ret=0,f=1;char c;
    while(c=getchar(),!isdigit(c))f=c=='-'?-1:1;
    while(isdigit(c))ret=ret*10+c-'0',c=getchar();
    return ret*f;
}

const int MAXN=500005<<2;

struct Edge{
    int next,to;
    Edge(int x=0,int 
    y=0){next=x;to=y;}
}e[MAXN];
int ecnt,head[MAXN];
inline void add(int x,int y){
    e[++ecnt]=Edge(head[x],y);
    head[x]=ecnt;
}

int n,m;

int fa[MAXN],dep[MAXN],siz[MAXN],hs[MAXN];
void dfs1(int x,int pre){
    fa[x]=pre;dep[x]=dep[pre]+1;siz[x]=1;
    int mx=0,tmp=0;
    for(int i=head[x];i;i=e[i].next){
        int v=e[i].to;
        if(v==pre) continue;
        dfs1(v,x);
        siz[x]+=siz[v];
        if(siz[v]>mx){mx=siz[v];tmp=v;}
    }
    hs[x]=tmp;
}

int top[MAXN],id[MAXN],tot;
void dfs2(int x,int tp){
    top[x]=tp;id[x]=++tot;
    if(hs[x]) dfs2(hs[x],tp);
    for(int i=head[x];i;i=e[i].next){
        int v=e[i].to;
        if(v==fa[x]||v==hs[x]) continue;
        dfs2(v,v);
    }
}

int lca(int x,int y){
    int ret;
    while(top[x]!=top[y]){
        dep[top[x]]>dep[top[y]]?x=fa[top[x]]:y=fa[top[y]];
    }
    return dep[x]<dep[y]?x:y;
}
        
int ans,sum[MAXN];
void dfs(int cur){
    for(int i=head[cur];i;i=e[i].next){
        int v=e[i].to;
        if(v==fa[cur]) continue;
        dfs(v);
        sum[cur]+=sum[v];
    }
    ans=max(sum[cur],ans);
}
                

int main(){
    n=rd();m=rd();
    int x,y,t;
    for(int i=1;i<=n-1;i++){
        x=rd();y=rd();
        add(x,y);add(y,x);
    }
    dfs1(1,0);
    dfs2(1,1);
    for(int i=1;i<=m;i++){
        x=rd();y=rd();t=lca(x,y);
        sum[x]+=1;sum[y]+=1;
        sum[t]-=1;sum[fa[t]]-=1;
    }
    dfs(1);
    cout<<ans<<endl;
    return 0;
}
树上差分
#include<iostream>
#include<cstring>
#include<cstdio>

using namespace std;

inline int rd(){
    int ret=0,f=1;char c;
    while(c=getchar(),!isdigit(c))f=c=='-'?-1:1;
    while(isdigit(c))ret=ret*10+c-'0',c=getchar();
    return ret*f;
}

const int MAXN=500005<<2;

struct Edge{
    int next,to;
    Edge(int x=0,int 
    y=0){next=x;to=y;}
}e[MAXN];
int ecnt,head[MAXN];
inline void add(int x,int y){
    e[++ecnt]=Edge(head[x],y);
    head[x]=ecnt;
}

int n,m;

struct Seg{
    #define ls (cur<<1)
    #define rs (cur<<1|1)
    #define mid (l+r>>1)
    int mx[MAXN],add[MAXN];
    Seg(){memset(mx,0,sizeof(mx));memset(add,0,sizeof(add));}
    void pushup(int cur){
        mx[cur]=max(mx[ls],mx[rs]);
    }
    void pushdown(int cur,int l,int r){
        int v=add[cur];
        add[ls]+=v;add[rs]+=v;
        mx[ls]+=v;mx[rs]+=v;
        add[cur]=0;
    }
    void build(int cur,int l,int r){
        if(l==r) {mx[cur]=0;return;}
        build(ls,l,mid);build(rs,mid+1,r);
        pushup(cur);
    }
    void update(int L,int R,int cur,int l,int r,int w){
        if(L<=l&&r<=R){mx[cur]+=w;add[cur]+=w;return;}
        pushdown(cur,l,r);
        if(L<=mid) update(L,R,ls,l,mid,w);
        if(mid <R) update(L,R,rs,mid+1,r,w);
        pushup(cur);
    }
}T;

int fa[MAXN],dep[MAXN],siz[MAXN],hs[MAXN];
void dfs1(int x,int pre){
    fa[x]=pre;dep[x]=dep[pre]+1;siz[x]=1;
    int mx=0,tmp=0;
    for(int i=head[x];i;i=e[i].next){
        int v=e[i].to;
        if(v==pre) continue;
        dfs1(v,x);
        siz[x]+=siz[v];
        if(siz[v]>mx){mx=siz[v];tmp=v;}
    }
    hs[x]=tmp;
}

int top[MAXN],id[MAXN],tot;
void dfs2(int x,int tp){
    top[x]=tp;id[x]=++tot;
    if(hs[x]) dfs2(hs[x],tp);
    for(int i=head[x];i;i=e[i].next){
        int v=e[i].to;
        if(v==fa[x]||v==hs[x]) continue;
        dfs2(v,v);
    }
}

void updateLink(int x,int y,int w){
    while(top[x]!=top[y]){
        if(dep[top[x]]<dep[top[y]]) swap(x,y);
        T.update(id[top[x]],id[x],1,1,n,w);
        x=fa[top[x]];
    }
    if(dep[x]>dep[y]) swap(x,y);
    T.update(id[x],id[y],1,1,n,w);
}
        

int main(){
    n=rd();m=rd();
    int x,y;
    for(int i=1;i<=n-1;i++){
        x=rd();y=rd();
        add(x,y);add(y,x);
    }
    dfs1(1,0);
    dfs2(1,1);
    T.build(1,1,n); 
    for(int i=1;i<=m;i++){
        x=rd();y=rd();
        updateLink(x,y,1);
    }
    cout<<T.mx[1];
    return 0;
}
树剖

本文来自博客园,作者:GhostCai,转载请注明原文链接:https://www.cnblogs.com/ghostcai/p/9380643.html

原文地址:https://www.cnblogs.com/ghostcai/p/9380643.html