poj 3320 尺取法

       题意：给一个包含n个数的数组，数组中元素可重复，求一个最短的连续子区间，该区间满足整个数组中所有元素至少在其中至少出现一次。

       分析：这也是一道求满足条件的最短区间的问题，并且是可以通过不断推进区间两端点来解决，即区间[s,t]若满足条件，那么对另一个满足条件的区间[s+1,t']，必有t'>=t。对该题而言，若从s页读到t页能覆盖所有知识点，那么当从s+1页开始读到t页的过程中，s页的知识点数量要减一，若减一后为0，那么就把t向后推进直到出现s页的知识点。

   代码如下：

#include <cstdio>#include <stack>#include <set>#include <iostream>#include <string>#include <vector>#include <queue>#include <list>#include <functional>#include <cstring>#include <algorithm>#include <cctype>#include <string>#include <map>#include <iomanip>#include <cmath>#define LL long long#define ULL unsigned long long#define SZ(x) (int)x.size()#define Lowbit(x) ((x) & (-x))#define MP(a, b) make_pair(a, b)#define MS(arr, num) memset(arr, num, sizeof(arr))#define PB push_back#define F first#define S second#define ROP freopen("input.txt", "r", stdin);#define MID(a, b) (a + ((b - a) >> 1))#define LC rt << 1, l, mid#define RC rt << 1|1, mid + 1, r#define LRT rt << 1#define RRT rt << 1|1#define BitCount(x) __builtin_popcount(x)#define BitCountll(x) __builtin_popcountll(x)#define LeftPos(x) 32 - __builtin_clz(x) - 1#define LeftPosll(x) 64 - __builtin_clzll(x) - 1const double PI = acos(-1.0);const int INF = 0x3f3f3f3f;using namespace std;const double eps = 1e-5;const int MAXN = 300 + 10;const int MOD = 1000007;const double M=1e-8;const int N=1e6+10;typedef pair<int, int> pii;typedef pair<int, string> pis;const int d[4][2]={{0,1},{0,-1},{-1,0},{1,0}};int n,m,a[N];set<int> s;int slove(int len){    int i,j;    int s=0,t=0,tot=0,ans=INF;    map<int,int> vis;    while(1) {        while (t<n && tot<len) {            if (vis[a[t++]]++==0) {                tot++;            }        }        if (tot<len) break;        ans=min(ans,t-s);        if (--vis[a[s++]]==0) tot--;    }    return ans;}int main(){    int i,j;    while(~scanf("%d",&n))    {        s.clear();        for (i=0;i<n;i++) {            scanf("%d",a+i);            s.insert(a[i]);        }        int len=s.size();        cout<<slove(len)<<endl;    }}/*71 8 8 8 3 8 1*/