FZU-2216 The Longest Straight(尺取法)

 Problem 2216 The Longest Straight

Accept: 523    Submit: 1663
Time Limit: 1000 mSec    Memory Limit : 32768 KB

 Problem Description

ZB is playing a card game where the goal is to make straights. Each card in the deck has a number between 1 and M(including 1 and M). A straight is a sequence of cards with consecutive values. Values do not wrap around, so 1 does not come after M. In addition to regular cards, the deck also contains jokers. Each joker can be used as any valid number (between 1 and M, including 1 and M).

You will be given N integers card[1] .. card[n] referring to the cards in your hand. Jokers are represented by zeros, and other cards are represented by their values. ZB wants to know the number of cards in the longest straight that can be formed using one or more cards from his hand.

 Input

The first line contains an integer T, meaning the number of the cases.

For each test case:

The first line there are two integers N and M in the first line (1 <= N, M <= 100000), and the second line contains N integers card[i] (0 <= card[i] <= M).

 Output

For each test case, output a single integer in a line -- the longest straight ZB can get.

 Sample Input

2 7 11 0 6 5 3 0 10 11 8 1000 100 100 100 101 100 99 97 103

 Sample Output

5 3

 Source

第六届福建省大学生程序设计竞赛-重现赛（感谢承办方华侨大学）

Submit  Back  Status  Discuss

原题地址：

题意:

你有n张牌在1~~m之间取值，0可以表示成任意一张牌（也就是王牌），

现在问你你手里的牌最长有几张牌是连续的。

思路：

记录卡牌的值，和零的值，求出1至M之间缺少的个数是否小于等于零的个数，求出最大值，

队友竟然用了尺取法！

代码

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
using namespace std;
int main()
{
        std::ios::sync_with_stdio(false);
        int n,m,t,i,j,l,r,a,num[100001],b[100001],ans;
        cin>>t;
        while(t--)
        {
                ans=0;
                memset(num,0,sizeof(num));
                memset(b,0,sizeof(b));
                cin>>n>>m;
                for(i=0;i<n;i++)
                {
                        cin>>a;
                        num[a]++;
                }
                ans=0;
                for(i=1;i<=m;i++)
                        if(num[i]==0)b[i]=b[i-1]+1;else b[i]=b[i-1];
                int l=0,r=0;
                while(r<=m)
                {
                        if(b[r]-b[l]<=num[0])
                        {
                                ans=max(ans,r-l);
                                r++;
                        }
                        else l++;
                }
                cout<<ans<<endl;
        }

}