POJ 3368 Frequent values 线段树 || RMQ

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题目：http://poj.org/problem?id=3368

题意：给定一个非递减数组，然后有一些查询，询问区间内出现次数最多的数的次数

思路：没想到用RMQ怎么做，用线段树的话，类似于之前做过的一题，要区间合并，好恶心。最后两种方法都写了

RMQ

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;const int N = 100100;const int INF = 0x3f3f3f3f;int dp[20][N];int arr[N], num[N];void ST(int n){    for(int i = 1; i <= n; i++)        dp[0][i] = num[i];    for(int i = 1; (1<<i) <= n; i++)        for(int j = 1; j <= n - (1<<i) + 1; j++)            dp[i][j] = max(dp[i-1][j], dp[i-1][j+(1<<(i-1))]);}int RMQ(int l, int r){    if(l > r) return 0;    int k = (int)log2(r - l + 1);    return max(dp[k][l], dp[k][r-(1<<k)+1]);}int main(){    int n, m, a, b;    while(scanf("%d", &n), n)    {        scanf("%d", &m);        memset(num, 0, sizeof num);//num数组统计到当前位置时某个数字出现的次数        arr[0] = INF;        for(int i = 1; i <= n; i++)        {            scanf("%d", &arr[i]);            if(arr[i] == arr[i-1])                num[i] = num[i-1] + 1;            else num[i]++;        }        ST(n);        while(m--)        {            scanf("%d%d", &a, &b);            int k = a;            while(k <= b && arr[a] == arr[k]) k++;//左起第一个数的次数不能RMQ查询，因为查询区间之外可能有这个数            printf("%d\n", max(k - a, RMQ(k, b)));        }    }    return 0;}

线段树

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;const int N = 100100;struct node{    int l, r, mx, lmx, rmx, lv, rv;//mx为区间中出现的最大次数，lmx为左端点出现次数，rmx为右端点出现次数，lv为左端点值，rv为右端点值}s[N*4];int arr[N];void push_up(int k){    s[k].lv = s[k<<1].lv, s[k].rv = s[k<<1|1].rv;    s[k].lmx = s[k<<1].lmx;    if(s[k<<1].lmx == (s[k<<1].r - s[k<<1].l + 1) && s[k<<1].rv == s[k<<1|1].lv)        s[k].lmx += s[k<<1|1].lmx;    s[k].rmx = s[k<<1|1].rmx;    if(s[k<<1|1].rmx == (s[k<<1|1].r - s[k<<1|1].l + 1) && s[k<<1|1].lv == s[k<<1].rv)        s[k].rmx += s[k<<1].rmx;    if(s[k<<1].rv == s[k<<1|1].lv)        s[k].mx = max(s[k<<1].mx, max(s[k<<1|1].mx, s[k<<1].rmx + s[k<<1|1].lmx));    else s[k].mx = max(s[k<<1].mx, s[k<<1|1].mx);}void build(int l, int r, int k){    s[k].l = l, s[k].r = r;    if(l == r)    {        s[k].lv = s[k].rv = arr[l];        s[k].mx = s[k].lmx = s[k].rmx = 1;        return;    }    int mid = (l + r) >> 1;    build(l, mid, k << 1);    build(mid + 1, r, k << 1|1);    push_up(k);}int query(int l, int r, int k){    if(l <= s[k].l && s[k].r <= r)        return s[k].mx;    int mid = (s[k].l + s[k].r) >> 1;    int res1 = -1, res2 = -1, res3 = -1;    if(l <= mid) res1 = query(l, r, k << 1);    if(r > mid) res2 = query(l, r, k << 1|1);    if(s[k<<1].rv == s[k<<1|1].lv && l <= mid && r > mid)        res3 = min(s[k<<1].r - l + 1, s[k<<1].rmx) + min(r - s[k<<1|1].l + 1, s[k<<1|1].lmx);    return max(res1, max(res2, res3));}int main(){    int n, m, a, b;    while(scanf("%d", &n), n)    {        scanf("%d", &m);        for(int i = 1; i <= n; i++)            scanf("%d", &arr[i]);        build(1, n, 1);        while(m--)        {            scanf("%d%d", &a, &b);            printf("%d\n", query(a, b, 1));        }    }    return 0;}

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