bzoj 4991: [Usaco2017 Feb]Why Did the Cow Cross the Road III

题意：

两列n的排列，相同的数连边，假如一对数的差>k且有交叉，ans++，求答案。

题解：

容易想到，按数从小到大加入，统计现在有多少数与当前数交差，再加i-k。
至于统计乱搞一下，正解是cdq，但本蒟蒻只会树套树，反正空间放开也过的了。官方数据都过了，精神AC。
为什么每次树套树都出SB错误啊啊啊啊。
code：

#include<cstdio>#include<cstdlib>#include<iostream>#include<cstring>#define LL long longusing namespace std;struct node{    int a,b;}a[100010];struct trnode{    int lc,rc,c;}tr[40000000];int root[200010],tot=0;int lson[200010],rson[200010],len=0;int n,k;int lowbit(int x){return x&(-x);}void update(int &x,int l,int r,int k,int c){    if(!x) x=++tot;    if(l==r){tr[x].c+=c;return;}    int mid=(l+r)/2;    if(k<=mid) update(tr[x].lc,l,mid,k,c);    else update(tr[x].rc,mid+1,r,k,c);    tr[x].c=tr[tr[x].lc].c+tr[tr[x].rc].c;}int findans(int x,int l,int r,int fl,int fr){    if(!x) return 0;    if(l==fl&&r==fr) return tr[x].c;    int mid=(l+r)/2;    if(fr<=mid) return findans(tr[x].lc,l,mid,fl,fr);    if(fl>mid) return findans(tr[x].rc,mid+1,r,fl,fr);    return findans(tr[x].lc,l,mid,fl,mid)+findans(tr[x].rc,mid+1,r,mid+1,fr);}void change(int x,int l,int r,int k1,int k2,int c){    update(root[x],1,n,k2,c);    if(l==r) return;    int mid=(l+r)/2;    if(k1<=mid) change(lson[x],l,mid,k1,k2,c);    else change(rson[x],mid+1,r,k1,k2,c);}int get(int x,int l,int r,int fl,int fr,int k1,int k2){    if(l==fl&&r==fr) return findans(root[x],1,n,k1,k2);    int mid=(l+r)/2;    if(fr<=mid) return get(lson[x],l,mid,fl,fr,k1,k2);    if(fl>mid) return get(rson[x],mid+1,r,fl,fr,k1,k2);    return get(lson[x],l,mid,fl,mid,k1,k2)+get(rson[x],mid+1,r,mid+1,fr,k1,k2);}int bt(int l,int r){    int x=++len;    if(l!=r)    {        int mid=(l+r)/2;        lson[x]=bt(l,mid);        rson[x]=bt(mid+1,r);    }    return x;}int main(){    scanf("%d %d",&n,&k);    for(int i=1;i<=n;i++)    {        int x;scanf("%d",&x);        a[x].a=i;    }    for(int i=1;i<=n;i++)    {        int x;scanf("%d",&x);        a[x].b=i;    }    bt(1,n);    LL ans=0;    for(int i=k+1;i<=n;i++)    {        ans+=(LL)get(1,1,n,1,a[i].a,a[i].b,n);        ans+=(LL)get(1,1,n,a[i].a,n,1,a[i].b);        change(1,1,n,a[i-k].a,a[i-k].b,1);    }    printf("%lld",ans);}

cdq（抄的）

#include <cstdio>#include <cstring>#include <cstdlib>#include <algorithm>#define LL long longusing namespace std;const int Maxn = 100010;struct node {    int num, a, b;}list[Maxn], b1[Maxn], b2[Maxn]; int bl1, bl2;bool cmp1(node x, node y) { return x.a < y.a; }bool cmp2(node x, node y) { return x.b > y.b; }int n, K;LL ans;int s[Maxn];int lowbit(int x) { return x & (-x); }void add(int x, int k) { while(x <= n){ s[x] += k; x += lowbit(x); } }int qry(int x) { int ret = 0; while(x > 0){ ret += s[x]; x -= lowbit(x); } return ret; }void cdq(int l, int r) {    if(l == r) return;    int mid = (l+r)>>1;    bl1 = bl2 = 0;    int i, j, k;    for(i = l; i <= mid; i++) b1[++bl1] = list[i];    for(i = mid+1; i <= r; i++) b2[++bl2] = list[i];    sort(b1+1, b1+bl1+1, cmp2);    sort(b2+1, b2+bl2+1, cmp2);    j = 1;    for(i = 1; i <= bl2; i++){        while(j <= bl1 && b1[j].b > b2[i].b){            add(b1[j].num, 1);            j++;        }        ans += qry(b2[i].num-K-1);        if(b2[i].num+K < n) ans += qry(n) - qry(b2[i].num+K);    }    j--;    while(j) add(b1[j].num, -1), j--;    cdq(l, mid);    cdq(mid+1, r);}int main() {    //freopen("b.in", "r", stdin);    //freopen("b.out", "w", stdout);    int i, j, k;    scanf("%d%d", &n, &K);    for(i = 1; i <= n; i++) list[i].num = i;    for(i = 1; i <= n; i++){        int x;        scanf("%d", &x);        list[x].a = i;    }    for(i = 1; i <= n; i++){        int x;        scanf("%d", &x);        list[x].b = i;    }    sort(list+1, list+n+1, cmp1);    cdq(1, n);    printf("%lld\n", ans);    return 0;}

数据生成器：

#include<cstdio>#include<cstdlib>#include<iostream>#include<ctime>using namespace std;int a[100010];int main(){    srand(time(0));    int n=rand()%5+1,k=rand()%(n+1);    printf("%d %d\n",n,k);    for(int i=1;i<=n;i++) a[i]=i;    for(int i=1;i<=100;i++)    {        int x=rand()%n+1,y=rand()%n+1;        swap(a[x],a[y]);    }    for(int i=1;i<=n;i++) printf("%d ",a[i]);printf("\n");    for(int i=1;i<=n;i++) a[i]=i;    for(int i=1;i<=100;i++)    {        int x=rand()%n+1,y=rand()%n+1;        swap(a[x],a[y]);    }    for(int i=1;i<=n;i++) printf("%d ",a[i]);printf("\n");}