BZOJ4419 SHOI2013发微博（平衡树）

　　好友状态的变化次数不会超过m，于是考虑暴力，对每个人记录其好友关系的变化，通过前缀和计算贡献。这需要查询一段前缀时间内某人发的微博数量，可以离线建一棵绝对平衡的平衡树。事实上完全可以线性。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
    while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
#define N 200010
#define M 500010
int n,m,ans[N],root[N],cnt=0,flag[N];
vector<int> t[N],h[N],op[N],a[N];
struct data{int ch[2],x,s;
}tree[M];
void build(int i,int &k,int l,int r)
{
    if (l>r) return;
    int mid=l+r>>1;
    k=++cnt;tree[k].x=a[i][mid];
    if (l==r) {tree[k].s=1;return;}
    build(i,tree[k].ch[0],l,mid-1);
    build(i,tree[k].ch[1],mid+1,r);
    tree[k].s=tree[tree[k].ch[0]].s+tree[tree[k].ch[1]].s+1;
}
int query(int k,int x)
{
    if (k==0) return 0;
    if (x<tree[k].x) return query(tree[k].ch[0],x);
    else return tree[tree[k].ch[0]].s+1+query(tree[k].ch[1],x);
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj4419.in","r",stdin);
    freopen("bzoj4419.out","w",stdout);
    const char LL[]="%I64d
";
#else
    const char LL[]="%lld
";
#endif
    srand(20020509);
    n=read(),m=read();
    for (int i=1;i<=m;i++)
    {
        char c=getchar();
        while (c!='!'&&c!='+'&&c!='-') c=getchar();
        if (c=='!') a[read()].push_back(i);
        else
        {
            int x=read(),y=read();
            t[x].push_back(i),h[x].push_back(y),op[x].push_back(c=='+'?-1:1);
            t[y].push_back(i),h[y].push_back(x),op[y].push_back(c=='+'?-1:1);
        }
    }
    for (int i=1;i<=n;i++)
    if (a[i].size())
    {
        sort(a[i].begin(),a[i].end());
        build(i,root[i],0,a[i].size()-1);
    }
    for (int i=1;i<=n;i++)
    {
        for (int j=0;j<t[i].size();j++)
        ans[i]+=query(root[h[i][j]],t[i][j])*op[i][j],flag[h[i][j]]+=op[i][j];
        for (int j=0;j<t[i].size();j++)
        if (flag[h[i][j]]) flag[h[i][j]]=0,ans[i]+=tree[root[h[i][j]]].s;
    }
    for (int i=1;i<=n;i++) printf("%d ",ans[i]);
    return 0;
}