Educational Codeforces Round 5 D Longest k-Good Segment 贪心

题意：（话说比赛的时候看错题了，英文水平太渣）。找一个最长的连续的子串，里面包含<=k个不相同的数字。

思路：贪心即可。

http://codeforces.com/contest/616/problem/D

/*********************************************
    Problem : Codeforces
    Author  : NMfloat
    InkTime (c) NM . All Rights Reserved .
********************************************/

#include <map>
#include <set>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>

#define rep(i,a,b)  for(int i = (a) ; i <= (b) ; i ++) //遍历
#define rrep(i,a,b) for(int i = (b) ; i >= (a) ; i --) //反向遍历
#define repS(it,p) for(auto it = p.begin() ; it != p.end() ; it ++) //遍历一个STL容器
#define repE(p,u) for(Edge * p = G[u].first ; p ; p = p -> next) //遍历u所连接的点
#define cls(a,x)   memset(a,x,sizeof(a))
#define eps 1e-8

using namespace std;

const int MOD = 1e9+7;
const int INF = 0x3f3f3f3f;
const int MAXN = 1e5+5;
const int MAXE = 2e5+5;

typedef long long LL;
typedef unsigned long long ULL;

int T,n,m,k;

int fx[] = {0,1,-1,0,0};
int fy[] = {0,0,0,-1,1};

int a[1000005];
int vis[1000005];

void input() {
    rep(i,1,n) scanf("%d",&a[i]);
}

void solve() {
    cls(vis,0);
    int len = 0;//连续的长度;
    int tmplen = 0;
    int l,r; //左值，右值
    int tmpl , tmpr ;
    tmpl = 1;
    int num = 0; //当前存在几个数
    int st = 1;
    rep(i,1,n) {
        tmplen ++;
        if(vis[a[i]]) {
            tmpr = i;
            if(tmplen > len) {
                len = tmplen;
                l = tmpl ; r = tmpr;
            }
        }
        else {
            num ++;
            //printf("num : %d\n", num);
            if(num <= k) {
                tmpr = i;
                if(tmplen > len) {
                    len = tmplen;
                    l = tmpl ; r = tmpr;
                }
            }
            else {
                while(--vis[a[st++]]) {

                }
                tmplen = i - st + 1;
                tmpl = st;//新的起点。
                num --;
            }
        }
        vis[a[i]] ++;
    }
    printf("%d %d\n",l,r);
}

int main(void) {
    //freopen("a.in","r",stdin);
    while(~scanf("%d %d",&n,&k)) {
        input();
        solve();
    }
    return 0;
}