HDU5583 Kingdom of Black and White （暴力）

题解：

其实此题还是暴力求解的,首先我们把相邻相同的字符分块,然后对于第i个分块与第i+1个分块,枚举(第i个分块字符个数+1和第i+1个分块字符个数-1)的情况与(第i个分块字符个数-1和第i+1个分块字符个数+1)的情况,需注意的是当第i+1个分块若只有一个字符,那么第i个分块、第i+1个分块、第i+2个分块会合成一个分块,同理,当第i个分块只有一个字符时,第i-1个分块、第i个分块、第i+1个分块会合成一个分块,我们更新最大值就可以了

代码

#include <cstdio>
#include <queue>
#include <cstring>
#include <iostream>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <map>
#include <string>
#include <set>
#include <ctime>
#include <cmath>
#include <cctype>
using namespace std;
#define MAX 100005
#define LL long long
long long s[MAX];
char kuai[MAX];
int cas=1,T;
int main()
{
int t,j,i;
LL ans,maxn,k;
scanf("%d",&t);
while (t--)
{

scanf("%s",kuai);
for (ans=0, j = 1,k=1,i=1;kuai[i]!='\0';i++)
if (kuai[i]!=kuai[i-1])
{
s[j++]=k;
ans+=k*k;
k=1;
}
else
k++;
s[j++]=k,ans+=k*k;s[j++]=0;
maxn=ans;
for (int i = 1;i<j-2;i++)
{
if(s[i+1]==1)
maxn=max(maxn,ans-s[i]*s[i]-s[i+1]*s[i+1]-s[i+2]*s[i+2]+(s[i]+s[i+1]+s[i+2])*(s[i]+s[i+1]+s[i+2]));
else
maxn=max(maxn,ans-s[i]*s[i]-s[i+1]*s[i+1]+(s[i]+1)*(s[i]+1)+(s[i+1]-1)*(s[i+1]-1));
if(s[i]==1)
maxn=max(maxn,ans-s[i]*s[i]-s[i+1]*s[i+1]-s[i-1]*s[i-1]+(s[i]+s[i+1]+s[i-1])*(s[i]+s[i+1]+s[i-1]));
else
maxn=max(maxn,ans-s[i]*s[i]-s[i+1]*s[i+1]+(s[i]-1)*(s[i]-1)+(s[i+1]+1)*(s[i+1]+1));
}
printf("Case #%d: %lld\n",cas++,maxn);
}
return 0;
}

题目

Kingdom of Black and White

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 501 Accepted Submission(s): 175

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