BZOJ 4553 [Tjoi2016&Heoi2016]序列 ——CDQ分治 树状数组

考虑答案的构成，发现是一个有限制条件的偏序问题。

然后三个维度的DP，可以排序、CDQ、树状数组各解决一维。

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#include <map> #include <cmath> #include <queue> #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; #define F(i,j,k) for (int i=j;i<=k;++i) #define D(i,j,k) for (int i=j;i>=k;--i) #define ll long long #define mp make_pair #define maxn 200005    struct node{     int a,b,c,id,ans;     void print()     {printf("ID is %d Upper %d Middle %d Lower %d\n",c,b,a);} }q[maxn];    int n,m;    bool cmp1(node x,node y){return x.b<y.b;} bool cmp2(node x,node y){return x.a<y.a;} bool cmp3(node x,node y){return x.id<y.id;}    int bit[maxn];    void add(int x,int f) {     for (;x<maxn;x+=x&(-x)) bit[x]=max(bit[x],f); }    void del(int x) {     for (;x<maxn;x+=x&(-x)) bit[x]=0; }    int query(int x) {     int ret=0;     for (;x;x-=x&(-x)) ret=max(ret,bit[x]);     return ret; }    void CDQ(int l,int r) {     if (l==r) return;     int mid=l+r>>1;     sort(q+l,q+r+1,cmp3);     CDQ(l,mid);     sort(q+l,q+mid+1,cmp1);     sort(q+mid+1,q+r+1,cmp2);     int now=l;     F(i,mid+1,r)     {         while (now<=mid&&q[now].b<=q[i].a) add(q[now].c,q[now].ans),now++;         q[i].ans=max(q[i].ans,query(q[i].b)+1);     }     F(i,l,now-1) del(q[i].c);     CDQ(mid+1,r); }    int main() {     scanf("%d%d",&n,&m);     F(i,1,n)     {         scanf("%d",&q[i].b);         q[i].a=q[i].c=q[i].b;         q[i].id=i; q[i].ans=1;     }     F(i,1,m)     {         int x,y;         scanf("%d%d",&x,&y);         q[x].a=min(q[x].a,y);         q[x].c=max(q[x].c,y);     }     CDQ(1,n);     int ans=0;     F(i,1,n) ans=max(ans,q[i].ans);     printf("%d\n",ans); }