尺取法 POJ 3320 Jessica's Reading Problem

题目传送门

/*
    尺取法：先求出不同知识点的总个数tot，然后以获得知识点的个数作为界限， 更新最小值
*/
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <set>
#include <map>
using namespace std;

const int MAXN = 1e6 + 10;
const int INF = 0x3f3f3f3f;
int a[MAXN];

int main(void)        //POJ 3320 Jessica's Reading Problem
{
    int n;
    while (scanf ("%d", &n) == 1)
    {
        set<int> S;
        for (int i=1; i<=n; ++i)
        {
            scanf ("%d", &a[i]);    S.insert (a[i]);
        }

        map<int, int> cnt;
        int tot = S.size ();    int ans = n, num = 0;    int i = 1, j = 1;
        while (1)
        {
            while (j <= n && num < tot)    if (cnt[a[j++]]++ == 0)    num++;
            if (num < tot)    break;
            ans = min (ans, j - i);
            if (--cnt[a[i++]] == 0) num--;
        }

        printf ("%d
", ans);
    }

    return 0;
}