【HDOJ】3601 Coach Yehr’s punishment

RMQ+dp+二分。
最好还是离散化一下再处理，通过dp求得每个位置的上一次出现的位置pre数组，从而求得不重复的长度len。
然后RMQ可以预处理区间的最大值，pre是个单调非递减数列。每次查询时，二分可以找到超过l的位置。
注意这个位置k可能超过r。因此最大值为max(k-l, RMQ(k, r) && k<=r)。

 /* 3603 */
 #include <iostream>
 #include <sstream>
 #include <string>
 #include <map>
 #include <queue>
 #include <set>
 #include <stack>
 #include <vector>
 #include <deque>
 #include <algorithm>
 #include <cstdio>
 #include <cmath>
 #include <ctime>
 #include <cstring>
 #include <climits>
 #include <cctype>
 #include <cassert>
 #include <functional>
 #include <iterator>
 #include <iomanip>
 using namespace std;
 //#pragma comment(linker,"/STACK:102400000,1024000")

 #define sti                set<int>
 #define stpii            set<pair<int, int> >
 #define mpii            map<int,int>
 #define vi                vector<int>
 #define pii                pair<int,int>
 #define vpii            vector<pair<int,int> >
 #define rep(i, a, n)     for (int i=a;i<n;++i)
 #define per(i, a, n)     for (int i=n-1;i>=a;--i)
 #define clr                clear
 #define pb                 push_back
 #define mp                 make_pair
 #define fir                first
 #define sec                second
 #define all(x)             (x).begin(),(x).end()
 #define SZ(x)             ((int)(x).size())
 #define lson            l, mid, rt<<1
 #define rson            mid+1, r, rt<<1|1

 const int maxn = 1e6+;
 int pre[maxn], len[maxn], pos[maxn];
 int dp[][maxn];
 int a[maxn];
 int n;

 void init_RMQ(int n) {
     int i, j;

     for (i=; i<=n; ++i)
         dp[][i] = len[i];
     for (j=; (<<j)<=n; ++j)
         for (i=; i+(<<j)-<=n; ++i)
             dp[j][i] = max(dp[j-][i], dp[j-][i+(<<(j-))]);
 }

 int RMQ(int l, int r) {
     if (l > r)
         swap(l, r);

     int k = ;

     while (<<(k+) <= r-l+)
         ++k;

     return max(dp[k][l], dp[k][r-(<<k)+]);
 }

 void solve() {
     memset(pos, , sizeof(pos));
     rep(i, , n+) {
         if (pos[a[i]] == )
             pre[i] = ;
         else
             pre[i] = pos[a[i]];
         pre[i] = max(pre[i], pre[i-]);
         pos[a[i]] = i;
         len[i] = i - pre[i];
     }

     init_RMQ(n);

     int q;
     int l, r;
     int ans;

     scanf("%d", &q);
     while (q--) {
         scanf("%d %d", &l, &r);
         if (l > r)
             swap(l, r);
         int k = lower_bound(pre+l, pre+r+, l) - pre;
         ans = k - l;
         if (k <= r)
             ans = max(ans, RMQ(k, r));
         printf("%d\n", ans);
     }
 }

 int main() {
     ios::sync_with_stdio(false);
     #ifndef ONLINE_JUDGE
         freopen("data.in", "r", stdin);
         freopen("data.out", "w", stdout);
     #endif

     while (scanf("%d", &n)!=EOF) {
         rep(i, , n+)
             scanf("%d", &a[i]);
         solve();
     }

     #ifndef ONLINE_JUDGE
         printf("time = %d.\n", (int)clock());
     #endif

     return ;
 }

数据生成器。

 from random import randint, shuffle
 import shutil
 import string

 def GenDataIn():
     with open("data.in", "w") as fout:
         t = 1
         bound = 10**3
         for tt in xrange(t):
             n = randint(100, 200)
             fout.write("%d\n" % (n))
             dataList = []
             for i in xrange(n):
                 x = randint(1, 100)
                 dataList.append(x)
             fout.write(" ".join(map(str, dataList)) + "\n")
             q = randint(100, 200)
             fout.write("%d\n" % (q))
             for i in xrange(q):
                 l = randint(1, n)
                 r = randint(l, n)
                 fout.write("%d %d\n" % (l, r))

 def MovDataIn():
     desFileName = "F:\eclipse_prj\workspace\hdoj\data.in"
     shutil.copyfile("data.in", desFileName)

 if __name__ == "__main__":
     GenDataIn()
     MovDataIn()