UVA - 11572 Unique Snowflakes（唯一的雪花）（滑动窗口）

题意：输入一个长度为n(n <= 10^6)的序列A，找到一个尽量长的连续子序列AL~AR，使得该序列中没有相同的元素。

分析：

法一：从r=0开始不断增加r，当a[r+1]在子序列a[l~r]中出现过，只需增大l，并继续延伸r，因为a[l~r]为可行解，则l增大后必然还是可行解。用set判断a[r+1]是否出现过，并进行a[l]的删除操作。

#pragma comment(linker, "/STACK:102400000, 102400000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define Min(a, b) ((a < b) ? a : b)
#define Max(a, b) ((a < b) ? b : a)
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const double eps = 1e-8;
const int MAXN = 1000000 + 10;
const int MAXT = 10000 + 10;
using namespace std;
int a[MAXN];
set<int> s;
int main(){
    int T;
    scanf("%d", &T);
    while(T--){
        s.clear();
        int n;
        scanf("%d", &n);
        for(int i = 0; i < n; ++i){
            scanf("%d", &a[i]);
        }
        int l = 0, r = 0, ans = 0;
        while(r < n){
            while(r < n && !s.count(a[r])){
                s.insert(a[r++]);
            }
            ans = Max(ans, r - l);
            s.erase(a[l++]);
        }
        printf("%d\n", ans);
    }
    return 0;
}

法二：mp[cur]记录的是元素cur所对应的下标。只要与a[r+1]相同的上一个元素下标小于l，则可延伸，++r，否则++l。

#pragma comment(linker, "/STACK:102400000, 102400000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define Min(a, b) ((a < b) ? a : b)
#define Max(a, b) ((a < b) ? b : a)
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const double eps = 1e-8;
const int MAXN = 1000000 + 10;
const int MAXT = 10000 + 10;
using namespace std;
int a[MAXN], last[MAXN];
map<int, int> mp;
int main(){
    int T;
    scanf("%d", &T);
    while(T--){
        mp.clear();
        int n;
        scanf("%d", &n);
        for(int i = 0; i < n; ++i){
            scanf("%d", &a[i]);
            if(!mp.count(a[i])){
                last[i] = -1;
            }
            else{
                last[i] = mp[a[i]];//前一个大小为a[i]的元素下标
            }
            mp[a[i]] = i;//不断更新大小为a[i]的当前下标
        }
        int l = 0, r = 0, ans = 0;
        while(r < n){
            while(r < n && last[r] < l) ++r;
            ans = Max(ans, r - l);
            ++l;
        }
        printf("%d\n", ans);
    }
    return 0;
}