KGold 51Nod - 1112

https://www.51nod.com/onlineJudge/questionCode.html#!problemId=1112

题目就是给n条直线y=kx+b求前min(n,1e4)个交点 发现x值越大 交点数量越多 有单调性

这样就可以二分找到一个x值 使所有直线从y轴到当前这个位置共产生min(n,1e4)个交点 交点即逆序对 树状数组可求

对一个序列 假设其逆序对数量为num 我们可以O(num)的把这些逆序对求出来 从大到小扫一遍 双端链表维护一下（相见代码 比较麻烦 ） 因为在二分时就是要求正好min(n,1e4)个逆序对的序列 所以求出所有逆序对的复杂度不超过1e4 然后把这些交点排个序即可

总体复杂度是n*logn*logn 网上的方法最好的也是n*n 队友还写了个n*n*logn都过了 就很气

 

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll N=1e5;
const int maxn=1e4+10;

struct node1
{
    int id;
    ll m;
    ll s;
};

struct node2
{
    int id;
    ll val;
};

struct node3
{
    int u;
    int v;
};

node1 ary[maxn];
node2 seg[maxn];
node3 ans[maxn];
ll val[maxn],tmp[maxn],sum[maxn];
int pre[maxn],nxt[maxn];
int n,len,tot;

bool cmpI(node1 n1,node1 n2)
{
    return n1.m<n2.m;
}

bool cmpII(node2 n1,node2 n2)
{
    if(n1.val==n2.val) return n1.id<n2.id;
    else return n1.val>n2.val;
}

bool cmpIII(node3 n1,node3 n2)
{
    if((ary[n1.u].m-ary[n1.v].m)*(ary[n2.u].s-ary[n2.v].s)==(ary[n2.u].m-ary[n2.v].m)*(ary[n1.u].s-ary[n1.v].s))
    {
        if(ary[n1.u].id==ary[n2.u].id) return ary[n1.v].id<ary[n2.v].id;
        else return ary[n1.u].id<ary[n2.u].id;
    }
    else return (ary[n1.u].m-ary[n1.v].m)*(ary[n2.u].s-ary[n2.v].s)>(ary[n2.u].m-ary[n2.v].m)*(ary[n1.u].s-ary[n1.v].s);
}

int lowbit(int x)
{
    return x&(-x);
}

ll query(int tar)
{
    ll res;
    int i;
    res=0;
    for(i=tar;i>=1;i-=lowbit(i)) res+=sum[i];
    return res;
}

void update(int tar)
{
    int i;
    for(i=tar;i<=len;i+=lowbit(i)) sum[i]++;
}

ll getni(ll t)
{
    ll res;
    int i,p;
    for(i=1;i<=n;i++) val[i]=ary[i].m+t*ary[i].s;
    for(i=1;i<=n;i++) tmp[i]=val[i];
    sort(tmp+1,tmp+n+1);
    len=unique(tmp+1,tmp+n+1)-tmp-1;
    memset(sum,0,sizeof(sum));
    res=0;
    for(i=1;i<=n;i++)
    {
        p=lower_bound(tmp+1,tmp+len+1,val[i])-tmp;
        res+=(query(len)-query(p-1));
        update(p);
    }
    return res;
}

void solve(ll t)
{
    int i,j;
    //t=10;
    for(i=1;i<=n;i++)
    {
        seg[i].id=i;
        seg[i].val=ary[i].m+t*ary[i].s;
        //printf("*%lld %lld*\n",ary[i].m,seg[i].val);
    }
    //for(i=1;i<=n;i++) printf("*%d %lld*\n",seg[i].id,seg[i].val);
    sort(seg+1,seg+n+1,cmpII);
    //for(i=1;i<=n;i++) printf("*%d %lld*\n",seg[i].id,seg[i].val);
    for(i=1;i<=n;i++) pre[i]=i-1,nxt[i]=i+1;
    tot=0;
    for(i=1;i<=n;i++)
    {
        for(j=nxt[seg[i].id];j<=n;j=nxt[j])
        {
            tot++;
            ans[tot].u=seg[i].id,ans[tot].v=j;
        }
        nxt[pre[seg[i].id]]=nxt[seg[i].id];
        pre[nxt[seg[i].id]]=pre[seg[i].id];
    }
    sort(ans+1,ans+tot+1,cmpIII);
    tot=min(tot,10000);
    for(i=1;i<=tot;i++) printf("%d %d\n",ary[ans[i].u].id,ary[ans[i].v].id);
    //printf("***%d***\n",tot);
}

int main()
{
    ll res,l,r,m;
    int i;
    scanf("%d",&n);
    for(i=1;i<=n;i++)
    {
        ary[i].id=i;
        scanf("%lld%lld",&ary[i].m,&ary[i].s);
    }
    sort(ary+1,ary+n+1,cmpI);
    res=getni(N);
    if(res==0) printf("No Solution\n");
    else if(res<=10000ll) solve(N);
    else
    {
        l=1,r=N;
        while(l<=r)
        {
            m=(l+r)/2;
            if(getni(m)<10000ll) l=m+1;
            else r=m-1,res=m;
        }
        solve(res);
    }
    return 0;
}