2015福建省赛 fzoj The Longest Straight 2216 （二分&转换）好题

 Problem 2216 The Longest Straight

Accept: 5    Submit: 12
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

//题意：

现在玩一个游戏（棋牌类的），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：

咋一看很迷茫，不知从何下手，但我们可以转换一下，就是将其转换成01串，就相对简单了，但怎么转换呢，定义一个数组a[]，表示对应数，若手中有这张牌，将对应位置标记为1，否则为0，zero记录输入的0的个数。然后定义一个数组b[]记录从0-i之间的0的个数。然后用二分查找，找出从i到m之间的满足0的个数等于zero的值的位置r，那么其连续的长度即为r-i，然后在比较找出最大的r-i就行了。

具体看代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int a[100010];
int b[100010];
int main()
{
	int t,n,m,x;
	int i,j,k;
	scanf("%d",&t);
	while(t--)
	{
		memset(a,0,sizeof(a));
		int zero=0;
		scanf("%d%d",&n,&m);
		for(i=0;i<n;i++)
		{
			scanf("%d",&x);
			if(x)
				a[x]=1;
			else
				zero++;
		}
		b[0]=0;
		for(i=1;i<=m;i++)
		{
			if(a[i])
				b[i]=b[i-1];
			else
				b[i]=b[i-1]+1;
		}
		int mm=0;
		for(i=0;i<=m;i++)
		{
			int l=i,r=m;
			int mid;
			while(l<=r)
			{
				mid=(l+r)/2;
				if(b[mid]-b[i]>zero)
					r=mid-1;
				else
					l=mid+1;
			}
			mm=max(mm,r-i);
		}
		printf("%d\n",mm);
	}
	return 0;
}