CodeChef

题目链接：http://vjudge.net/problem/174774/origin

题意：给定一个长度为n的序列a[],序列的值不大于m，现在有k个询问，每个询问给定(l,r).让你求出max{|x − y| : Li ≤ x, y ≤ Ri and Ax = Ay}。即区间[L,R]中值相同时，位置差的最大值

思路：分块，因为不带修改，所以我们就可以做预处理。求出last,first,Ans。 last[i][j]:值i在第j块最后出现的位置。first[i][j]:值i在第j块最早出现的位置。Ans[i][j]:第i块到第j块的值相同的最大位置差。

#define _CRT_SECURE_NO_DEPRECATE
#include<stdio.h>  
#include<string.h>  
#include<cstring>
#include<algorithm>  
#include<queue>  
#include<math.h>  
#include<time.h>
#include<vector>
#include<iostream>
#include<map>
using namespace std;
typedef long long int LL;
const int MAXN = 100000 + 10;
int belong[MAXN], block, num, L[MAXN], R[MAXN];
int n, m, q;
int a[MAXN], last[MAXN][500], first[MAXN][500], Ans[500][500];
int tim[MAXN], times, Pos[MAXN];
int cal(int st, int ed){
    int ans = 0;
    for (int i = L[st]; i <= R[st]; i++){
        ans = max(ans, last[a[i]][ed] - i);
    }
    return ans;
}
void build(){
    block = (int)sqrt(n + 0.5);
    num = n / block; if (n%block){ num++; }
    for (int i = 1; i <= num; i++){
        L[i] = (i - 1)*block + 1; R[i] = i*block;
    }
    R[num] = n;
    for (int i = 1; i <= n; i++){
        belong[i] = ((i - 1) / block) + 1;
    }
    memset(last, 0, sizeof(last)); memset(first, 0, sizeof(first));
    memset(Ans, 0, sizeof(Ans)); times = 0;
    for (int i = 1; i <= n; i++){
        last[a[i]][belong[i]] = i;
    }
    for (int i = n; i>0; i--){
        first[a[i]][belong[i]] = i;
    }
    for (int i = 1; i <= m; i++){
        for (int j = 1; j <= num; j++){
            if (!last[i][j]){ last[i][j] = last[i][j - 1]; }
        }
        for (int j = num; j; j--){
            if (!first[i][j]){ first[i][j] = first[i][j + 1]; }
        }
    }
    for (int i = num; i; i--){
        for (int j = i; j <= num; j++){
            Ans[i][j] = max(max(Ans[i + 1][j], Ans[i][j - 1]), cal(i, j));
        }
    }
}
int query(int st, int ed){
    int ans = 0; times++;
    if (belong[st] == belong[ed]){
        for (int i = st; i <= ed; i++){
            if (tim[a[i]] != times){ Pos[a[i]] = i; tim[a[i]] = times; }
            else{ ans = max(ans, i - Pos[a[i]]); }
        }
        return ans;
    }
    for (int i = st; i <= R[belong[st]]; i++){
        if (tim[a[i]] != times){ Pos[a[i]] = i; tim[a[i]] = times; }
        else{ ans = max(ans, i - Pos[a[i]]); }
        ans = max(ans, last[a[i]][belong[ed] - 1] - i);
    }
    ans = max(ans, Ans[belong[st] + 1][belong[ed] - 1]);
    for (int i = L[belong[ed]]; i <= ed; i++){
        if (tim[a[i]] != times){ Pos[a[i]] = i; tim[a[i]] = times; }
        else{ ans = max(ans, i - Pos[a[i]]); }
        ans = max(ans, i - first[a[i]][belong[st] + 1]);
    }
    return ans;
}
int main(){
    //#ifdef kirito
    //    freopen("in.txt", "r", stdin);
    //    freopen("out.txt", "w", stdout);
    //#endif
    //    int start = clock();
    while (~scanf("%d%d%d", &n, &m, &q)){
        for (int i = 1; i <= n; i++){ scanf("%d", &a[i]); }
        build(); int st, ed;
        for (int i = 1; i <= q; i++){
            scanf("%d%d", &st, &ed);
            printf("%d
", query(st, ed));
        }
    }
    //#ifdef LOCAL_TIME
    //    cout << "[Finished in " << clock() - start << " ms]" << endl;
    //#endif
    return 0;
}