Kingdom of Black and White

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1890    Accepted Submission(s): 588

Problem Description

In the Kingdom of Black and White (KBW), there are two kinds of frogs: black frog and white frog.

Now N frogs are standing in a line, some of them are black, the others are white. The total strength of those frogs are calculated by dividing the line into minimum parts, each part should still be continuous, and can only contain one kind of frog. Then the strength is the sum of the squared length for each part.

However, an old, evil witch comes, and tells the frogs that she will change the color of at most one frog and thus the strength of those frogs might change.

The frogs wonder the maximum possible strength after the witch finishes her job.

Input

First line contains an integer T, which indicates the number of test cases.

Every test case only contains a string with length N, including only 0 (representing
a black frog) and 1 (representing a white frog).

⋅ 1≤T≤50.

⋅ for 60% data, 1≤N≤1000.

⋅ for 100% data, 1≤N≤105.

⋅ the string only contains 0 and 1.

Output

For every test case, you should output "Case #x: y",where x indicates the case number and counts from 1 and y is the answer.

Sample Input

2

000011

0101

Sample Output

Case #1: 26

Case #2: 10

Source

2015ACM/ICPC亚洲区上海站-重现赛（感谢华东理工）

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#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <vector>
#include <iomanip>
#include <math.h>
#include <map>
using namespace std;
#define FIN     freopen("input.txt","r",stdin);
#define FOUT    freopen("output.txt","w",stdout);
#define INF     0x3f3f3f3f
#define INFLL   0x3f3f3f3f3f3f3f
#define lson    l,m,rt<<1
#define rson    m+1,r,rt<<1|1
typedef long long LL;
typedef pair<int,int> PII;

const int MAXN = 1e5 + 5;
int T;
char a[MAXN];
LL dp[MAXN];
LL b[MAXN];

int main()
{
    //FIN
    scanf("%d", &T);
    int cas = 1;
    while(T--) {
        scanf("%s", a);
        int lena = strlen(a);
        dp[0] = 1;
        LL sum = 0;
        int t = 0;
        LL mx = 0;
        memset(b, 0, sizeof(b));
        for(int i = 1; i < lena; i++) {
            if(a[i] == a[i - 1]) dp[i] = dp[i - 1] + 1;
            else {
                mx = max(dp[i - 1], mx);
                b[t++] = dp[i - 1];
                sum += LL(dp[i - 1] * dp[i - 1]);
                dp[i] = 1;
            }
        }

        b[t] = dp[lena - 1];
        mx = max(b[t], mx);
        sum += LL(b[t] * b[t]);
        LL ans = sum;

        if(b[0] == lena) {
            ans = sum;
            printf("Case #%d: %lld\n", cas++, ans);
            continue;
        }

        LL tt;

        //cout << "mx=" << mx << endl;

        for(int i = 0; i <= t; i++) {
            if(b[i] == mx) {
                //cout << "i=" << i << endl;
                if(i == 0) tt = b[1];
                else if(i == t) tt = b[t - 1];
                else tt = min(b[i - 1], b[i + 1]);
                LL now = sum + (mx + 1) * (mx + 1) - (mx * mx) + (tt - 1) * (tt -1) - tt * tt;
                ans = max(ans, now);
            }
        }

        for(int i = 0; i <= t; i++) {
            if(b[i] == 1) {
                LL now = sum - b[i - 1] * b[i - 1] - b[i + 1] * b[i + 1] - 1 + (b[i + 1] + 1 + b[i - 1]) * (b[i + 1] + 1 + b[i - 1]);
                ans = max(ans, now);
            }
        }

        printf("Case #%d: %lld\n", cas++, ans);
    }
    return 0;
}