fzu 2216 The Longest Straight

Problem 2216 The Longest Straight
Accept: 178 Submit: 536

Time Limit: 1000 mSec Memory Limit : 32768 KB
ProblemDescription
ZB is playing a card game where the goal is to makestraights. Each card in the deck has a number between 1 and M(including 1 andM). A straight is a sequence of cards with consecutive values. Values do notwrap around, so 1 does not come after
M. In addition to regular cards, the deckalso contains jokers. Each joker can be used as any valid number (between 1 andM, including 1 and M).
You will be given N integers card[1] .. card
referringto the cards in your hand. Jokers are represented by zeros, and other cards arerepresented by their values. ZB wants to know the number of cards in thelongest straight that can be formed
using one or more cards from his hand.
Input
The first line contains an integer T, meaning the numberof the cases.
For each test case:
The first line there are two integers N and M in thefirst line (1 <= N, M <= 100000), and the second line contains N integerscard[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
SampleOutput
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）

一道令我伤心了好久的题目，省赛那时我非常弱鸡，这题没做出来，与奖失之交臂。。。

题目大意：给你一些牌（0表示鬼牌，可以代替任何牌）求可以组成的顺子的最大长度。（斗地主的顺子知道吧）

思路：二分+枚举，每次枚举左端点，求这个左端点能组成的最大长度。

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
const int maxn = 100005;
int n, m, zero;
int num[maxn];			//num表示从1到i成为顺子所需鬼牌的个数
int card[maxn];

int main()
{
int T;
scanf("%d", &T);
while (T--)
{
zero = 0;
memset(num, 0, sizeof(num));
memset(card, 0, sizeof(card));
scanf("%d%d", &n, &m);
for (int i = 0; i < n; i++)
{
int x;
scanf("%d", &x);
if (!x)
zero++;
else
card[x] = 1;
}

for (int i = 1; i <= m; i++)
num[i] = num[i - 1] + (1-card[i]);

int ans = 0;
for (int i = 1; i <= m; i++)
{
int k = upper_bound(num + i, num + m + 1, zero + num[i-1]) - (num + i);
ans = max(ans, k);
}
printf("%d\n", ans);
}
return 0;
}

当然，也有更快的做法，时间复制度为O(N)

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
const int maxn = 1e5 + 5;
int card[maxn], num[maxn], pos[maxn];
int main()
{
int T;
scanf("%d", &T);
while (T--)
{
memset(pos, 0, sizeof(pos));
memset(num, 0, sizeof(num));
memset(card, 0, sizeof(card));
int m, n, zero = 0;
scanf("%d%d", &m, &n);
for (int i = 1; i <= m; i++)
{
int x;
scanf("%d", &x);
if (x == 0) zero++;
else
card[x] = 1;
}
int ans = 0,k;
pos[0] = 1;
for (int i = 1; i <= n; i++)
{
num[i] = num[i - 1] + (1 - card[i]);
if (!card[i])
pos[num[i]] = i;
k = num[i] - zero;
if (k <= 0)
{
ans = i;
continue;
}
ans = max(ans, i - pos[k]);
}
printf("%d\n", ans);
}
return 0;
}

/*

22
7 11
0 6 5 3 0 10 11
8 1000
100 100 100 101 100 99 97 103
3 5
1 2 3
3 2
1 2 3
3 5
1 2 0
3 5
2 3 4
3 5
2 0 4

*/