【bzoj4241】 历史研究

http://www.lydsy.com/JudgeOnline/problem.php?id=4241 (题目链接)

  看到题目就联想到了【bzoj2809】 Apio2012—dispatching。想了想权值分块+莫队,发现不好维护块内最值,又看了看80s的时间,于是怒水一发线段树+莫队,结果先WA后TLE,不断TLE,无论怎么改常数都不行,难道nlogn*sqrt(n)就是过不了吗!!不爽,蒯个题解,再见!

题意

  求区间加权众数。

solution

  貌似是分块,离散化之后,用mx[i][j]表示第i块到第j块的答案,cnt[i][j]表示前i块数字j的个数。

代码

// bzoj4241
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<cmath>
#define LL long long
#define inf 2147483640
#define Pi acos(-1.0)
#define free(a) freopen(a".in","r",stdin),freopen(a".out","w",stdout);
using namespace std;
inline LL getint() {
    int f,x=0;char ch=getchar();
    while (ch<='0' || ch>'9') {if (ch=='-') f=-1;else f=1;ch=getchar();}
    while (ch>='0' && ch<='9') {x=x*10+ch-'0';ch=getchar();}
    return x*f;
}

const int maxn=100010;
struct data {int w,id;}a[maxn];
int b[maxn],pos[maxn],L[400],R[400],cnt[400][maxn],num[maxn],st[maxn];
LL mx[400][400],ans;
int n,m,block,tot,top;

bool cmp(data a,data b) {
    return a.w<b.w;
}
LL cal(int l,int r) {
    LL ans=0;
    top=0;
    for (int i=l;i<=r;i++) {
        if (!num[b[i]]) st[++top]=b[i];
        num[b[i]]++;
        ans=max(ans,(LL)num[b[i]]*a[b[i]].w);
    }
    while (top) num[st[top--]]=0;
    return ans;
}    
void build() {
    block=sqrt(n);tot=(n-1)/block+1;
    for (int i=1;i<=n;i++) {
        pos[i]=(i-1)/block+1;
        if (!L[pos[i]]) L[pos[i]]=i;
        R[pos[i]]=i;
    }
    for (int i=1;i<=n;i++) cnt[pos[i]][b[i]]++;
    for (int i=2;i<=tot;i++)
        for (int j=1;j<=n;j++) cnt[i][j]+=cnt[i-1][j];
}
int main() {
    scanf("%d%d",&n,&m);
    for (int i=1;i<=n;i++) scanf("%d",&a[i].w),a[i].id=i;
    sort(a+1,a+1+n,cmp);
    b[a[1].id]=1;
    for (int i=2;i<=n;i++) {
        if (a[i].w==a[i-1].w) b[a[i].id]=b[a[i-1].id];
        else b[a[i].id]=i;
    }
    build();
    for (int i=1;i<=tot;i++) {
        LL ans=0;
        memset(num,0,sizeof(num));
        for (int j=L[i];j<=n;j++) {
            num[b[j]]++;
            ans=max(ans,(LL)num[b[j]]*a[b[j]].w);
            if (j==R[pos[j]]) mx[i][pos[j]]=ans;
        }
    }
    memset(num,0,sizeof(num));
    for (int i=1;i<=m;i++) {
        int l,r;
        scanf("%d%d",&l,&r);
        ans=0;
        if (pos[l]==pos[r]) printf("%lld\n",cal(l,r));
        else {
            ans=mx[pos[l]+1][pos[r]-1];
            top=0;
            for (int i=l;i<=R[pos[l]];i++)
                if (!num[b[i]]) st[++top]=b[i],num[b[i]]=cnt[pos[r]-1][b[i]]-cnt[pos[l]][b[i]];
            for (int i=L[pos[r]];i<=r;i++)
                if (!num[b[i]]) st[++top]=b[i],num[b[i]]=cnt[pos[r]-1][b[i]]-cnt[pos[l]][b[i]];
            for (int i=l;i<=R[pos[l]];i++) num[b[i]]++;
            for (int i=L[pos[r]];i<=r;i++) num[b[i]]++;
            while (top) {
                ans=max(ans,(LL)a[st[top]].w*num[st[top]]);
                num[st[top--]]=0;
            }
            printf("%lld\n",ans);
        }
    }
    return 0;
}

  

This passage is made by MashiroSky.
posted @ 2016-09-27 22:39  MashiroSky  阅读(272)  评论(0编辑  收藏  举报