FZU 2216——The Longest Straight——————【二分、枚举】

Problem 2216 The Longest Straight

Accept: 17    Submit: 39
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）

 

 

题目大意：Zb在玩卡牌游戏，给你n张牌，一个m，每个牌面是在1~m之间的一个数，包括1和m。但是可能会有王牌，王牌可以变成1~m中的任意一个，这里王牌用0表示。问你最长的连续上升的牌的长度。

解题思路：首先明确，这几个0是应该连续填放在空里，这样能连成最长的。我们首先记录哪些牌是有的，哪些没有。然后我们枚举左端点，然后通过二分去查找需要0的个数小于总的0的个数的最长的右端点。 mlogm的复杂度。

#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<iostream>
using namespace std;
const int maxn = 1e6+200;
int card[maxn],presum[maxn];
int BinSearch(int l,int r, int k,int key){
    int mi = (l+r)/2;
    while(l < r){
        mi = (l+r)/2;
        if(presum[mi]-k > key){
            r = mi;
        }else{
            l = mi+1;
        }
    }
    if(presum[l]-k > key)
        l--;
    return l;
}
int main(){
    int T,n,m;
    scanf("%d",&T);
    while(T--){
        int a, Zero = 0;
        scanf("%d%d",&n,&m);
        memset(card,0,sizeof(card));
        for(int i = 1; i <= n; i++){
            scanf("%d",&a);
            card[a] = 1;
            if(!a) Zero++;
        }
        for(int i = 1; i <= m; i++){
            presum[i] = presum[i-1] + (!card[i]);
        }
        int ans = 1;
        for(int i = 1; i <= m; i++){
            int idx = BinSearch(i,m,presum[i-1],Zero);
          //  printf("%d ",idx);
            ans = max(ans, idx-i+1);
        }
        printf("%d
",ans);
    }
    return 0;
}

/*

55
11 15
0 0 1 2 4 5 8 9 11 13 14
*/