11235 - Frequent values

《算法竞赛入门经典-训练指南》P198
记录一下区间的左右边界就可以了
#include <iostream>
#include <stack>
#include <queue>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <set>
#include <vector>
#include <cstring>
#include <algorithm>

#define INF 0x7fffffff
#define N 100010
#define M 1000010
#define LL long long
#define mod 95041567

using namespace std;

int arr[N];
int num[N][3];
int dp[N][20];

void RMQ_init(int len){
    for(int j = 1; (1 << j) <= len; ++ j)
        for(int i = 1; i + (1 << j) - 1 < len; ++ i)
            dp[i][j] = max(dp[i][j - 1], dp[i + (1 << (j - 1))][j - 1]);
}

int RMQ(int L, int R){
    if(R < L) return 0;
    else if(R == L) return dp[R][0];
    int k = 0;
    while((1 << (k + 1)) <= R - L + 1) ++ k;
    return max(dp[L][k], dp[R - (1 << k) + 1][k]);
}

int main()
{
    int n, q;
   // freopen("in.txt","r",stdin);
    while(scanf("%d", &n) != EOF){
        if(! n) break;
        scanf("%d", &q);
        for(int i = 0; i < n; ++ i) scanf("%d", &arr[i]);
        for(int i = 0; i <= n + 2; ++ i)
            for(int j = 0; j <= 20; ++ j)
                dp[i][j] = 0;
        arr[n] = INF;
        int p = 0;
        int len = 0;
        for(int i = 1; i <= n; ++ i)
            if(arr[i] != arr[i - 1]){
                for(int j = p; j <= i - 1; ++ j){
                    num[j][0] = i - 1;
                    num[j][1] = p;
                    num[j][2] = len;
                }
                dp[len ++][0] = i - p;
                p = i;
            }
        RMQ_init(len);
        int L, R;
        for(int i = 0; i < q; ++ i){
            scanf("%d %d", &L, &R);
            if(arr[L - 1] == arr[R - 1]){
                printf("%d\n", R - L + 1);
                continue;
            }
            //printf("%s\n", "++");
            p = num[L - 1][0] - (L - 1) + 1;
            p = max(p, R - num[R - 1][1]);
            p = max(p, RMQ(num[L - 1][2] + 1, num[R - 1][2] - 1));
            printf("%d\n", p);
        }
    }
    return 0;
}


posted @ 2013-10-14 09:48  pangbangb  阅读(235)  评论(0编辑  收藏  举报