uva 11920 0 s, 1 s and ? Marks

最新推荐文章于 2021-06-26 10:48:28 发布

原创最新推荐文章于 2021-06-26 10:48:28 发布 · 847 阅读

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本文探讨了一个字符串处理问题，目标是通过将问号替换为0或1，来最小化字符串中最大组的长度。通过实例分析，解释了如何通过贪心算法和分类讨论来解决这一问题。

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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 ( T $\le$ 5000) 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

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

Sample Input

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

Sample Output

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

最大值最小，二分。关键是已知最大长度，怎么判断是否可行。贪心+分类讨论，各种yy就可以了，这题还是比较简单的。

#include <cstdio>
#include <algorithm>
#include <vector>
#include <map>
#include <queue>
#include <iostream>
#include <stack>
#include <set>
#include <cstring>
#include <stdlib.h>
#include <cmath>
using namespace std;
typedef long long LL;
typedef pair<int, int> P;
const int maxn = 500 + 5;

vector<P> v;
int n, from, to;
string s;

bool judge(int x){
    int sum = 1, last_pos = from, total = 0;
    for(int i = from+1;i <= to;i++){
        if(s[i] == '?'){
            total++;
        }
        else{
            if(total == 0){
                if(s[i] == s[i-1]) sum++;
                else sum = 1;
            }
            else if(total == 1){
                if(s[i] == s[last_pos]){
                    sum = 1;
                }
                else{
                    if(sum + 1 <= x){
                        sum = 1;
                    }
                    else{
                        sum = 2;
                    }
                }
            }
            else{
                if(s[i] == s[last_pos]){
                    if(total%2 == 0 && x < 2){
                        return false;
                    }
                }
                else{
                    if(total%2 == 1 && x < 2){
                        return false;
                    }
                }
                sum = 1;
            }
            last_pos = i;
            total = 0;
        }
        if(sum > x) return false;
    }
    return true;
}

int main(){
    int t, kase = 0;
    scanf("%d", &t);
    while(t--){
        kase++;
        cin >> s;
        n = s.length();
        from = n, to = -1;
        for(int i = 0;i < n;i++){
            if(s[i] != '?'){
                from = i;
                break;
            }
        }
        for(int i = n-1;i >= 0;i--){
            if(s[i] != '?'){
                to = i;
                break;
            }
        }

        int l = 1, r = n+1;
        int ans = 1;
        while(l <= r){
            int mid = (l+r) / 2;
            if(judge(mid)){
                r = mid-1;
                ans = mid;
            }
            else{
                l = mid+1;
            }
        }
        printf("Case %d: %d\n", kase, ans);
    }
    return 0;
}