POJ3320 Jessica's Reading Problem

  set用来统计所有不重复的知识点的数,map用来维护区间[s,t]上每个知识点出现的次数,此题很好的体现了map的灵活应用

  

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
#include <vector>
#include <map>
#include <set>
#include <string>
#include <cmath>
using namespace std;
const int INF = 0x3f3f3f3f;
typedef long long ll;
const int MAX_P = ;
int P;
int a[MAX_P];
int main()
{
scanf("%d", &P);
for (int i = ; i < P; ++i)
{
scanf("%d", &a[i]);
}
set <int> all;
for (int i = ; i < P; ++i) {
all.insert(a[i]);
}
int n = all.size();
int s = , t = , num = ;
map<int, int> count;
int res = P;
for (;;)
{
while (t < P && num < n) {
if (count[a[t]]== ) { //出现了新的知识点
num++;
}
count[a[t]]++;
t++;
}
if (num < n) break;
res = min(res, t-s);//更新最小区间长度
count[a[s]]--;
if (count[a[s]] == ) //某个知识点出现次数为0
{
num--;
}
s++;
}
printf("%d\n", res );
return ;
}

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