0 s, 1 s and ? Marks

Given a string consisting of 0, 1 and ? only, change all the ? to 0/1, so that the size of the largest group is minimized. A group is a substring that contains either all zeros or all ones.

Consider the following example:

0 1 1 ? 0 1 0 ? ? ?

We can replace the question marks (?) to get

0 1 1 0 0 1 0 1 0 0

The groups are (0) (1 1) (0 0) (1) (0) (1) (0 0) and the corresponding sizes are 1, 2, 2, 1, 1, 1, 2. That means the above replacement would give us a maximum group size of 2. In fact, of all the 2⁴ possible replacements, we won‘t get any maximum group size that is smaller than 2.

Input 

The first line of input is an integer T ( T5000) that indicates the number of test cases. Each case is a line consisting of a string that contains 0, 1 and ? only. The length of the string will be in the range [1,1000].

Output 

Sample Input 

4
011?010???
???
000111
00000000000000

Sample Output 

Case 1: 2
Case 2: 1
Case 3: 3
Case 4: 14

题意：给定一个串有0,1,?。?可以变成0或1，要求构造出一个串，使得连续的0,1的最大值最小。

思路：贪心，把连续的0,1,?保存在一起，然后去判断，如果是0,1就直接算最大值，如果是?，就看前面和后面是0还是1，如果前后相同且?个数为奇数或前后不同?个数为偶数，那么肯定可以交替01是最小的，反之，就看前面一串多1还是后面一串多1或者中间串多1比较好，注意如果中间只有一个?，那么就不用考虑中间多1的情况。

代码：

#include <stdio.h>
#include <string.h>
#define max(a,b) (a)>(b)?(a):(b)
const int N = 1005;

int t, pn;
char str[N];
struct Point{
	int v, num;
} p[N];

void init() {
	scanf("%s", str);
	memset(p, 0, sizeof(p));
	int len = strlen(str); pn = 0;
	char c = ‘#‘;
	for (int i = 0; i < len; i++) {
		if (str[i] == c) {
			p[pn - 1].num++;
		}
		else {
			c = str[i];
			p[pn].v = str[i] - ‘0‘;
			p[pn++].num++;
		}
	}
}

int solve() {
	int ans = 1;
	for (int i = 0; i < pn; i++) {
		if (p[i].v == 0 || p[i].v == 1) {
			ans = max(ans, p[i].num);
		}
		else {
			if (ans >= 2 && p[i].num != 1) continue;
			if (i == 0 || i == pn - 1) continue;
			if (p[i - 1].v == p[i + 1].v && p[i].num % 2) continue;
			if (p[i - 1].v != p[i + 1].v && p[i].num % 2 == 0) continue;
			if (p[i - 1].num <= p[i + 1].num) {
				p[i - 1].num++;
				ans = max(ans, p[i - 1].num);
			}
			else {
				p[i + 1].num++;
				ans = max(ans, p[i + 1].num);	
			}
		}
	}
	return ans;
}

int main() {
	int cas = 0;
	scanf("%d", &t);
	while (t--) {
		init();
		printf("Case %d: %d\n", ++cas, solve());
	}
	return 0;
}

UVA 11920 - 0 s, 1 s and ? Marks（贪心）