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TJOI2017 不勤劳的图书管理员
Date: Sat Nov 24 20:12:30 CST 2018 In Category: NoCategory

Description:

给定一个长度为 $n$ 序列,每个位置上有一个编号 $id_i$ 和一个权 $v_i$ ,对于一个逆序对(按编号和位置),贡献是权的和。
有 $m$ 次交换操作,计算每次操作后的总贡献。
$1\leqslant n\leqslant 50000$

Solution:

树状数组套主席树。

Code:


#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<cctype>
#include<cstring>
using namespace std;
int n,m;
#define MAXN 50010
int p[MAXN];
int v[MAXN];
#define MOD 1000000007
struct node
{
int lc,rc;
int sum,tot;
node(){sum = 0;}
}t[MAXN * 250];
int ptr = 0;
int newnode(){return ++ptr;}
#define mid ((l + r) >> 1)
void insert(int &rt,int p,int k,int c,int l,int r)
{
if(rt == 0)rt = newnode();
t[rt].sum = (t[rt].sum + k) % MOD;t[rt].tot += c;
if(l == r)return;
if(p <= mid)insert(t[rt].lc,p,k,c,l,mid);
else insert(t[rt].rc,p,k,c,mid + 1,r);
return;
}
int query_sum(int rt,int L,int R,int l,int r)
{
if(rt == 0)return 0;
if(L <= l && r <= R)return t[rt].sum;
int res = 0;
if(L <= mid)res = (res + query_sum(t[rt].lc,L,R,l,mid)) % MOD;
if(R > mid)res = (res + query_sum(t[rt].rc,L,R,mid + 1,r)) % MOD;
return res;
}
int query_tot(int rt,int L,int R,int l,int r)
{
if(rt == 0)return 0;
if(L <= l && r <= R)return t[rt].tot;
int res = 0;
if(L <= mid)res += query_tot(t[rt].lc,L,R,l,mid);
if(R > mid)res += query_tot(t[rt].rc,L,R,mid + 1,r);
return res;
}
int root[MAXN];
int ans = 0;
int lowbit(int x){return x & (-x);}
void add(int k,int p,int v)
{
for(int i = k;i <= n;i += lowbit(i))insert(root[i],p,v,1,1,n);
return;
}
void del(int k,int p,int v)
{
for(int i = k;i <= n;i += lowbit(i))insert(root[i],p,-v,-1,1,n);
return;
}
int query1(int l1,int r1,int l2,int r2)
{
if(l1 > r1 || l2 > r2)return 0;
int res = 0;
for(int i = r1;i >= 1;i -= lowbit(i))res = (res + query_sum(root[i],l2,r2,1,n)) % MOD;
for(int i = l1 - 1;i >= 1;i -= lowbit(i))res = (res - query_sum(root[i],l2,r2,1,n) + MOD) % MOD;
return res;
}
int query2(int l1,int r1,int l2,int r2)
{
if(l1 > r1 || l2 > r2)return 0;
int res = 0;
for(int i = r1;i >= 1;i -= lowbit(i))res += query_tot(root[i],l2,r2,1,n);
for(int i = l1 - 1;i >= 1;i -= lowbit(i))res -= query_tot(root[i],l2,r2,1,n);
return res;
}
int main()
{
scanf("%d%d",&n,&m);
for(int i = 1;i <= n;++i)scanf("%d%lld",&p[i],&v[i]);
for(int i = 1;i <= n;++i)
{
ans = (ans + query1(1,i - 1,p[i] + 1,n)) % MOD;
ans = (ans + 1ll * query2(1,i - 1,p[i] + 1,n) * v[i] % MOD) % MOD;
ans = (ans + query1(i + 1,n,1,p[i] - 1)) % MOD;
ans = (ans + 1ll * query2(i + 1,n,1,p[i] - 1) * v[i] % MOD) % MOD;
add(i,p[i],v[i]);
}
int x,y;
for(int i = 1;i <= m;++i)
{
scanf("%d%d",&x,&y);
del(x,p[x],v[x]);
ans = (ans - query1(1,x - 1,p[x] + 1,n) + MOD) % MOD;
ans = (ans - 1ll * query2(1,x - 1,p[x] + 1,n) * v[x] % MOD + MOD) % MOD;
ans = (ans - query1(x + 1,n,1,p[x] - 1) + MOD) % MOD;
ans = (ans - 1ll * query2(x + 1,n,1,p[x] - 1) * v[x] % MOD + MOD) % MOD;
del(y,p[y],v[y]);
ans = (ans - query1(1,y - 1,p[y] + 1,n) + MOD) % MOD;
ans = (ans - 1ll * query2(1,y - 1,p[y] + 1,n) * v[y] % MOD + MOD) % MOD;
ans = (ans - query1(y + 1,n,1,p[y] - 1) + MOD) % MOD;
ans = (ans - 1ll * query2(y + 1,n,1,p[y] - 1) * v[y] % MOD + MOD) % MOD;
swap(p[x],p[y]);swap(v[x],v[y]);
add(x,p[x],v[x]);
ans = (ans + query1(1,x - 1,p[x] + 1,n)) % MOD;
ans = (ans + 1ll * query2(1,x - 1,p[x] + 1,n) * v[x] % MOD) % MOD;
ans = (ans + query1(x + 1,n,1,p[x] - 1)) % MOD;
ans = (ans + 1ll * query2(x + 1,n,1,p[x] - 1) * v[x] % MOD) % MOD;
add(y,p[y],v[y]);
ans = (ans + query1(1,y - 1,p[y] + 1,n)) % MOD;
ans = (ans + 1ll * query2(1,y - 1,p[y] + 1,n) * v[y] % MOD) % MOD;
ans = (ans + query1(y + 1,n,1,p[y] - 1)) % MOD;
ans = (ans + 1ll * query2(y + 1,n,1,p[y] - 1) * v[y] % MOD) % MOD;
printf("%lld\n",ans);
}
return 0;
}
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