UVA - 11920 0 s, 1 s and ? Marks

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Description

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 contains0, 1 and ? only. The length of the string will be in the range [1,1000].

Output

For each case, output the case number first followed by the size of the minimized largest group.

Sample Input

4011?010??????00011100000000000000

Sample Output

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

題意：給定一個帶問號的01串，把每個問號替換成0或1，讓最長的“連續的相同數字串”盡量短

思路：最大的最小採用二分，至於判斷的時候，每個位置要麼是0要麼是1，根據情況判斷

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 1005;int dp[maxn][2];char str[maxn];int n;int check(int len) {dp[0][0] = dp[0][1] = 0;for (int i = 1; i <= n; i++) {dp[i][0] = dp[i][1] = -1;if (str[i] != '0') {if (dp[i-1][0] >= 0)dp[i][1] = 1;if (dp[i-1][1] >= 0 && dp[i-1][1]+1 <= len && dp[i][1] == -1)dp[i][1] = dp[i-1][1] + 1;}if (str[i] != '1') {if (dp[i-1][1] >= 0)dp[i][0] = 1;if (dp[i-1][0] >= 0 && dp[i-1][0]+1 <= len && dp[i][0] == -1)dp[i][0] = dp[i-1][0] + 1;}if (dp[i][1] == -1 && dp[i][0] == -1)return 0;}return 1;}int main() {int t, cas = 1;scanf("%d", &t);while (t--) {scanf("%s", str+1);n = strlen(str+1);int l = 1, r = n;while (l <= r) {int m = l + r >> 1;if (check(m))r = m-1;else l = m + 1;}printf("Case %d: %d\n", cas++, l);}return 0;}