bzoj3262: 陌上花开

来源：互联网发布：外贸大数据应用编辑：程序博客网时间：2024/06/10 18:41

链接：http://www.lydsy.com/JudgeOnline/problem.php?id=3262

题意：中文题。

分析：cdq分治练习题，详细分析戳这里，不过这题没有好的时间戳z，因为这题的三个值都是在1~k范围内的，那就是说有的值不唯一有的不存在，那么我们在分治的时候就要注意边界了，所以我改成了4个参数的分治，详见代码。O(nlogk^2)。

代码：

#include<map>#include<set>#include<cmath>#include<queue>#include<bitset>#include<math.h>#include<cstdio>#include<vector>#include<string>#include<cstring>#include<iostream>#include<algorithm>#pragma comment(linker, "/STACK:102400000,102400000")using namespace std;const int N=100010;const int MAX=1000000100;const int mod=100000000;const int MOD1=1000000007;const int MOD2=1000000009;const double EPS=0.00000001;typedef long long ll;const ll MOD=998244353;const ll INF=10000000010;typedef double db;typedef unsigned long long ull;struct node {    int x,y,z,id;}a[N],b[N];int k,f[2*N],ans[N],p[N];int cmp(node a,node b) {    if (a.x!=b.x) return a.x<b.x;    else if (a.y!=b.y) return a.y<b.y;        else if (a.z!=b.z) return a.z<b.z;            else return a.id<b.id;}void add(int a,int b) {    for (;a<=k;a+=a&(-a)) f[a]+=b;}int getsum(int a) {    int ret=0;    for (;a;a-=a&(-a)) ret+=f[a];    return ret;}void cdq(int l,int r,int L,int R) {    if (l>=r) return ;    int i,mid=(L+R)>>1,w=l,z;    if (L==R) {        for (i=l;i<=r;i++) { ans[a[i].id]+=getsum(a[i].y);add(a[i].y,1); }        for (i=l;i<=r;i++) add(a[i].y,-1);        return ;    }    for (i=l;i<=r;i++)    if (a[i].z<=mid) add(a[i].y,1);    else ans[a[i].id]+=getsum(a[i].y);    for (i=l;i<=r;i++)    if (a[i].z<=mid) add(a[i].y,-1);    for (i=l;i<=r;i++)    if (a[i].z<=mid) { b[w]=a[i];w++; }    z=w-1;    for (i=l;i<=r;i++)    if (a[i].z>mid) { b[w]=a[i];w++; }    for (i=l;i<=r;i++) a[i]=b[i];    cdq(l,z,L,mid);cdq(z+1,r,mid+1,R);}int main(){    int i,n;    scanf("%d%d", &n, &k);    for (i=1;i<=n;i++) {        scanf("%d%d%d", &a[i].x, &a[i].y, &a[i].z);a[i].id=i;    }    sort(a+1,a+n+1,cmp);    a[n+1].x=a[n+1].y=a[n+1].z=0;    memset(f,0,sizeof(f));    memset(ans,0,sizeof(ans));    for (i=n;i;i--)    if (a[i].x==a[i+1].x&&a[i].y==a[i+1].y&&a[i].z==a[i+1].z) {        f[i]=f[i+1]+1;ans[a[i].id]+=f[i];    }    memset(f,0,sizeof(f));    cdq(1,n,1,k);    memset(p,0,sizeof(p));    for (i=1;i<=n;i++) p[ans[i]]++;    for (i=0;i<n;i++) printf("%d\n", p[i]);    return 0;}

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