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  • UVALive 5881 Unique Encryption Keys （DP）

    Unique Encryption Keys

    题目链接：

    http://acm.hust.edu.cn/vjudge/problem/26633

    Description

    http://7xjob4.com1.z0.glb.clouddn.com/bced90d15c27e75270cc759651dcfa63

    Input

    The input contains several cipher descriptions. Each description starts with one line containing two integer numbers M and Q separated by a space. M (1 ≤ M ≤ 1000000) is the number of encrypted messages, Q is the number of queries (0 ≤ Q ≤ 1000000). Each of the following M lines contains one number Ki (0 ≤ Ki ≤ 2 30) specifying the identifier of a key used to encrypt the i-th message. The next Q lines then contain one query each. Each query is specified by two integer numbers Bj and Ej , 1 ≤ Bj ≤ Ej ≤ M, giving the interval of messages we want to check. There is one empty line after each description. The input is terminated by a line containing two zeros in place of the numbers M and Q.

    Output

    For each query, print one line of output. The line should contain the string “OK” if all keys used to encrypt messages between Bj and Ej (inclusive) are mutually different (that means, they have different identifiers). If some of the keys have been used repeatedly, print one identifier of any such key. Print one empty line after each cipher description.

    Sample Input

    10 5 3 2 3 4 9 7 3 8 4 1 1 3 2 6 4 10 3 7 2 6 5 2 1 2 3 1 2 2 4 1 5 0 0

    Sample Output

    3 OK 4 3 OK OK 1

    Source

    2016-HUST-线下组队赛-1
    ##题意： 给定一组序列并进行M次询问，每次询问一个区间是否有重复的数字，若有输出任意重复.
    ##题解： dp[i]：区间以i为左端点时，最远能够达到并无重复数字的右端点. 对于i要找到最近的出现重复的位置，要么是i+1以后的重复，要么是num[i]本身出现重复. 转移方程： dp[i] = min(dp[i+1], last_occ[num[i]]); 另外，感觉UVA上对着道题的特判有点问题.
    ##代码： ``` cpp #include #include #include #include #include #include #include #include #include #include #include #define maxn 1010000 #define inf 0x3f3f3f3f #define mod 1000000007 #define LL long long #define mid(a,b) ((a+b)>>1) #define IN freopen("unique.in", "r", stdin); #define OUT freopen("unique.out", "w", stdout); using namespace std;

    int n,m;
    int num[maxn];
    int dp[maxn];
    map<int,int> last;

    int main()
    {
    //IN;
    //OUT;

    while(scanf("%d %d",&n,&m) != EOF && (m||n))
{
    for(int i=1; i<=n; i++) {
        scanf("%d", &num[i]);
    }

    last.clear();
    dp[n+1] = n + 1;
    for(int i=n; i>=1; i--) {
        dp[i] = dp[i+1];
        if(last.count(num[i])) dp[i] = min(dp[i], last[num[i]]);
        last[num[i]] = i;
    }

    while(m--) {
        int l,r; scanf("%d %d", &l,&r);
        if(dp[l] > r) {
            printf("OK
");
        } else {
            printf("%d
", num[dp[l]]);
        }
    }
    printf("
");
}

return 0;

    }
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  • 原文地址：https://www.cnblogs.com/Sunshine-tcf/p/5791290.html
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