poj3368Frequent values(RMQ)

http://poj.org/problem?id=3368

追完韩剧 想起这题来了 想用线段树搞定来着 结果没想出来。。然后想RMQ 想出来了

算是离散吧 把每个数出现的次数以及开始的位置及结束的位置记录下来 以次数进行RMQ 再特殊处理下区间端点就OK 了

WA一次 盯了半小时 原来少了个=号  好吧好吧。。

#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<stdlib.h>
#include<cmath>
using namespace std;
#define N 200005
struct node
{
    int s,e,a;
}p[N];
int b[N],f[N],fm[N][40];
void init(int n)
{
    int i,j,o = floor(log10(double(n))/log10(double(2)));
    for(j = 1; j <= o ; j++)
        for(i = 1; i <= n+1-(1<<j) ; i++)
        {
            fm[i][j] = max(fm[i][j-1],fm[i+(1<<(j-1))][j-1]);
        }
}
int main()
{
    int i,n,q;
    while(scanf("%d",&n)&&n)
    {
        scanf("%d",&q);
        for(i = 1; i <= n ;i++)
        {
            scanf("%d",&b[i]);
        }
        int g = 0;
        p[++g].a = b[1];
        p[g].s = 1;f[1] = 1;
        int pre = b[1];
        for(i = 2; i <= n ; i++)
        {
            if(b[i]!=pre)
            {
                p[g].e = i-1;
                p[++g].a = b[i];
                p[g].s = i;
                pre = b[i];
            }
            f[i] = g;
        }
        p[g].e = n;
        for(i = 1; i <= g ; i++)
        fm[i][0] = p[i].e-p[i].s+1;
        init(g);
        while(q--)
        {
            int a,b,t;
            scanf("%d%d",&a,&b);
            if(a>b)
            {
                t = a;a=b;b=t;
            }
            int x = f[a],y = f[b];
            if(x==y)
            printf("%d
",b-a+1);
            else if(y-x==1)
            {
                int maxz = max(b-p[y].s+1,p[x].e-a+1);
                printf("%d
",maxz);
            }
            else
            {
                int maxz = max(b-p[y].s+1,p[x].e-a+1);
                x++;y--;
                int o = floor(log10(double(y-x+1))/log10(double(2)));
                int mm = max(fm[x][o],fm[y-(1<<o)+1][o]);
                printf("%d
",max(mm,maxz));
            }
        }
    }
    return 0;
}