BZOJ 2333 棘手的操作(离线+线段树+带权并查集)

这题搞了我一天啊。。。拍不出错原来是因为极限数据就RE了啊，竟然返回WA啊。我的线段树要开8倍才能过啊。。。

 

首先可以发现除了那个加边操作，其他的操作有点像线段树啊。如果我们把每次询问的联通块都放在一个区间的话，那么就可以用线段树维护了啊。

于是我们只需要用带权并查集把联通块串成一条链的形式。就可以用区间表示出来了啊。。

 

# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iostream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <set>
# include <cmath>
# include <algorithm>
using namespace std;
# define lowbit(x) ((x)&(-x))
# define pi acos(-1.0)
# define eps 1e-3
# define MOD 1000000000
# define INF 1000000000
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define bug puts("H");
# define lch p<<1,l,mid
# define rch p<<1|1,mid+1,r
# define mp make_pair
# define pb push_back
typedef pair<int,int> PII;
typedef vector<int> VI;
# pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
int Scan() {
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
void Out(int a) {
    if(a<0) {putchar('-'); a=-a;}
    if(a>=10) Out(a/10);
    putchar(a%10+'0');
}
const int N=300005;
//Code begin...

struct Node{char s[5]; int x, y;}node[N];
struct Q{int l, r;}q[N];
int seg[N<<3], tag[N<<3], a[N], fa[N], to[N], tail[N], suc[N], F[N];

int find(int x){
    int tmp;
    if (fa[x]!=x) tmp=find(fa[x]), fa[x]=tmp;
    return fa[x];
}
void union_set(int u, int v){fa[u]=v; suc[tail[v]]=u; tail[v]=tail[u];}
void union_set1(int x, int y){
    int l=min(q[x].l,q[y].l), r=max(q[x].r,q[y].r);
    fa[x]=y, q[y].l=l, q[y].r=r;
}
void push_up(int p){seg[p]=max(seg[p<<1],seg[p<<1|1]);}
void push_down(int p){
    if (!tag[p]) return ;
    seg[p]+=tag[p];
    tag[p<<1]+=tag[p]; tag[p<<1|1]+=tag[p]; tag[p]=0;
}
void init(int p, int l, int r){
    if (l<r) {
        int mid=(l+r)>>1;
        init(lch); init(rch); push_up(p);
    }
    else seg[p]=a[F[l]];
}
int query(int p, int l, int r, int L, int R){
    push_down(p);
    if (L>r||R<l) return -INF;
    if (L<=l&&R>=r) return seg[p];
    int mid=(l+r)>>1;
    return max(query(lch,L,R),query(rch,L,R));
}
void update(int p, int l, int r, int L, int R, int val){
    push_down(p);
    if (L>r||R<l) return ;
    if (L<=l&&R>=r) tag[p]=val, push_down(p);
    else {
        int mid=(l+r)>>1;
        update(lch,L,R,val); update(rch,L,R,val); push_up(p);
    }
}
int main ()
{
    int n, m, u, v;
    scanf("%d",&n);
    FOR(i,1,n) fa[i]=tail[i]=i;
    FOR(i,1,n) scanf("%d",a+i);
    scanf("%d",&m);
    FOR(i,1,m) {
        scanf("%s",node[i].s);
        if (!strcmp(node[i].s,"U")||!strcmp(node[i].s,"A1")||!strcmp(node[i].s,"A2")) scanf("%d%d",&node[i].x,&node[i].y);
        else if(!strcmp(node[i].s,"F3")) continue;
        else scanf("%d",&node[i].x);
    }
    FOR(i,1,m) if(node[i].s[0]=='U') {
        u=find(node[i].x), v=find(node[i].y);
        if (u!=v) union_set(u,v);
    }
    int now=0;
    FOR(i,1,n) if (fa[i]==i) {
        to[i]=++now; F[now]=i;
        int tmp=i;
        while (suc[tmp]) tmp=suc[tmp], to[tmp]=++now, F[now]=tmp;
    }
    //debug
    //FOR(i,1,n) printf(" %d",to[i]); putchar('\n');
    //debug
    init(1,1,n); FOR(i,1,n) fa[i]=q[i].l=q[i].r=i;
    FOR(i,1,m) {
        if (!strcmp(node[i].s,"U")) {
            u=find(to[node[i].x]); v=find(to[node[i].y]);
            if (u!=v) union_set1(u,v);
        }
        else if (!strcmp(node[i].s,"A1")) update(1,1,n,to[node[i].x],to[node[i].x],node[i].y);
        else if (!strcmp(node[i].s,"A2")) u=find(to[node[i].x]), update(1,1,n,q[u].l,q[u].r,node[i].y);
        else if (!strcmp(node[i].s,"A3")) update(1,1,n,1,n,node[i].x);
        else if (!strcmp(node[i].s,"F1")) printf("%d\n",query(1,1,n,to[node[i].x],to[node[i].x]));
        else if (!strcmp(node[i].s,"F2")) u=find(to[node[i].x]), printf("%d\n",query(1,1,n,q[u].l,q[u].r));
        else printf("%d\n",query(1,1,n,1,n));
    }
    return 0;
}