UVA - 11536 - Smallest Sub-Array（滑动窗口）

题意：从给定字符串中，选最短连续子序列，包含1~k中的所有数。

           尺取法（滑动窗口）解决：在第一次找到1~k的序列之后，向右滑动保证每个数至少存在一次，不断取最小值即可。

#include<cstdio>
#include<cstring>
#include<cctype>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<deque>
#include<queue>
#include<stack>
#include<list>
typedef long long ll;
typedef unsigned long long llu;
const int MAXN = 100 + 10;
const int MAXT = 1000000 + 10;
const int INF = 0x7f7f7f7f;
const double pi = acos(-1.0);
const double EPS = 1e-6;
using namespace std;

int n, m, k, a[MAXT], T, vis[MAXN];

void init(){
    a[0] = 1, a[1] = 2, a[2] = 3;
    for(int i = 3; i < n; ++i)  a[i] = (a[i - 1] + a[i - 2] + a[i - 3]) % m + 1;
}

int solve(){
    int head = 0, num = 0, ans = INF;
    memset(vis, 0, sizeof vis);
    for(int tail = 0; tail < n; ++tail)
        if(a[tail] >= 1 && a[tail] <= k){
            if(!vis[a[tail]])  ++num;
            ++vis[a[tail]];
            if(num == k){
                while((a[head] >= 1 && a[head] <= k && vis[a[head]] > 1) || a[head] < 1 || a[head] > k){
                    if(a[head] >= 1 && a[head] <= k && vis[a[head]] > 1)  --vis[a[head]];
                    ++head;
                }
                ans = min(tail - head + 1, ans);
            }
        }
    return ans == INF ? -1 : ans;
}

int main(){
    int ca = 0;
    scanf("%d", &T);
    while(T--){
        scanf("%d%d%d", &n, &m, &k);
        init();
        int ans = solve();
        if(ans != -1)  printf("Case %d: %d\n", ++ca, ans);
        else  printf("Case %d: sequence nai\n", ++ca);
    }
    return 0;
}