[UVA][二分greedy] 11920 - 0 s, 1 s and ? Marks＠Morris' Blog｜PChome Online 人新台

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2013-08-11 22:27:36| 人1,448| 回0 | 上一篇 | 下一篇

[UVA][二分greedy] 11920 - 0 s, 1 s and ? Marks

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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 ( 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

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

使用的，事上也可以出的。

最特的是：

*0?1*     -> *0?1*      still don't know

因此考一的情且前後接的不同，二分答案後，
可能 '?' 於前面的那，否只好放後面。

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
using namespace std;
int greedy(int mx, char s[], int v[], int n) {
    int i, ret = 0;
    int prev = 0;
    for(i = 0; i < n; i++) {
        if(prev > mx)   return 0;
        if(s[i] == '?') {
            if(i && i != n-1 && s[i-1] != s[i+1]) {
                if(v[i] == 1) {
                    if(prev < mx) {
                        pre++;
                        prev = 0;
                    } else {
                        prev = 1;
                    }
                } else {
                    prev = 0;
                }
                if(v[i]&1)
                    ret = max(ret, 2);
                else
                    ret = max(ret, 1);
            } else if(i && i != n-1 && s[i-1] == s[i+1]) {
                if(v[i]&1)
                    ret = max(ret, 1);
                else
                    ret = max(ret, 2);
                prev = 0;
            } else {
            }
        } else {
            if(i && s[i-1] != '?')
                prev = 0;
            prev += v[i];
        }
        ret = max(ret, prev);
    }
    return ret <= mx;
}
int main() {
    int testcase, cases = 0;
    char s[1005];
    int v[1005];
    scanf("%d", &testcase);
    while(testcase--) {
        scanf("%s", s);
        int n = strlen(s);
        int l = 1, r = n, m;
        int ret = 1;
        int i, j, k;
        m = 0;
        for(i = 0; i < n; i++) {
            s[m] = s[i], v[m] = 1;
            while(i < n && s[i+1] == s[i])
                i++, v[m]++;
            m++;
        }
        n = m;
        /*for(i = 0; i < n; i++)
            printf("%c %d\n", s[i], v[i]);*/
        while(l <= r) {
            m = (l+r)/2;
            if(greedy(m, s, v, n))
                r = m-1, ret = m;
            else
                l = m+1;
        }
        printf("Case %d: %d\n", ++cases, ret);
    }
    return 0;
}
/*
* if the digits before && after "?" is different, we can change the sequence to
* *0?1*     -> *0?1*      still don't know
* *0??1*    -> *0101*     ans>=1
* *0???1*   -> *01001*    ans>=2
* *0????1* -> *010101*   ans>=1
* ....
* else if they are the same, "0" is consider here, the same to "1"
* *0?0*     -> *010*      ans>=1
* *0??0*    -> *0110*     ans>=2
* *0???0*   -> *01010*    ans>=1
* *0????0* -> *010110*   ans>=2
*/