The Longest Straight

最长顺子牌算法解析

最新推荐文章于 2024-07-15 17:53:06 发布

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本文介绍了一种解决特定纸牌游戏中寻找最长连续序列（顺子）问题的算法。通过输入玩家手中的牌（包括普通牌和可以替代任意值的百搭牌），该算法能够找出能形成的最长顺子的长度。文章详细解释了算法的工作原理，并提供了完整的实现代码。

Description

ZB is playing a card game where the goal is tom make straights. Each card in the deck has a number between 1 and M, inclusive. 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, inclusive).

You will be giben 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 <= 10⁵), and the second line contains N integers card[i](0 <= card[i] <= M).

Output

For each test case, output a single integer 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

#include <stdio.h>  
#include <string.h>  
int main(){  
    int T;  
    int a[100001], N, M, flag[100001], joker;  
    scanf("%d", &T);  
    while(T--){  
        joker = 0;  
        memset(a, 0, sizeof(a));  
        scanf("%d %d", &N, &M);  
        for(int i = 0; i < N; i++){  
            int temp;  
            scanf("%d", &temp);  
            if(temp)  
                a[temp] = 1;  
            else
                ++joker;  
        }  
        flag[0] = 0;  
        for(int i = 1; i <= M; i++){  
            if(a[i]){  
                flag[i] = flag[i-1];  
            }  
            else{  
                flag[i] = flag[i-1] + 1;  
            }  
        }  
        int ans = 0;  
        for(int i = 0; i <= M; i++){  
            int left = i, right = M, mid;  
            while(left <= right){  
                mid = (left + right) / 2;  
                if(flag[mid] - flag[i] > joker){  
                    right = mid - 1;  
                }  
                else{  
                    left = mid + 1;  
                }  
            }  
            if(right - i > ans){  
                ans = right - i;  
            }  
        }  
        printf("%d\n", ans);  
    }  
    return 0;  
}